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CHAPTER
5
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Genetics/Ch5
Mendelian Genetics—
patterns of Inheritance
How Do Genes Get passed on from parent
to Child?
KEY CONCEPTS
After completing this chapter you will
be able to
• describe the early experiments
of Gregor Mendel and relate his
conclusions to modern genetic
theory
• solve monohybrid and dihybrid
cross problems using Punnett
squares, and use probabilities
to determine the expected
phenotypes and genotypes in
the offspring
• explainandprovideexamples
of dominance, incomplete
dominance, and multiple alleles
• drawandfollowapedigreechart
in order to track a specific allele
through many generations
• describegeneticdisordersthat
are caused or influenced by
specific genes
• considerthesocialandethical
implications associated with
genetic testing
You have your mother’s eyes, your father’s curly hair, and your grandmother’s
disposition. How are characteristics passed on from one generation to the
next? Why do characteristics disappear from one generation and then reappear in the next? Can you predict which characteristics will be inherited?
Would you like to be able to predict the characteristics you will pass on to
your children, or find out which ones have been passed on to you from your
parents but have not surfaced yet?
Researchers in modern genetics laboratories now study how specific characteristics are passed on from parent to offspring. However, humans have
studied the characteristics of plants and animals ever since the beginning of
organized agriculture. Early farmers realized that selective breeding resulted
in a better food supply. These early farmers did not know about DNA or genes.
They practised genetic selection by choosing to breed plants and animals
based on characteristics that were important to them.
Today we have a better understanding of how genes are passed from parent
to child. Geneticists are most interested in pinpointing the exact mechanism
by which a gene is passed on to offspring, especially in the case of genetic
diseases. For example, two parents who do not have cystic fibrosis, a genetic
respiratory disease, may have a child who is born with cystic fibrosis. Where
did the cystic fibrosis gene come from—both parents, or just one? Will all the
children of these parents have cystic fibrosis, or just some? By understanding
how genetic disease genes are passed on, geneticists are able to answer such
questions. Once such questions are answered, researchers can work on treatment and prevention.
In this chapter you will investigate how characteristics are passed on from
parent to child. You will find out how the rules of probability govern the offspring’s characteristics. You will explore the idea of being able to determine
which genes you carry, and the social and ethical consequences of finding
out. Finally, you will see how certain characteristics may be valuable to one
individual and detrimental to another.
starting pOInTS
Answer the following questions using your current
knowledge. You will have a chance to revisit these questions
later, applying concepts and skills from the chapter.
1. How do you think genetic information gets passed on
from generation to generation?
2. How might an understanding of genetic processes and
2
the inheritance of biological characteristics benefit
individuals and society?
3. Why is there so much variation in the human population
with respect to what humans look like?
4. How does society benefit from the screening of harmful
genes?
Chapter 5 • Mendelian Genetics—Patterns of Inheritance
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FPO
All "tasters" detect a quite bitter
flavour. No taster perceives the PTC
as only "slightly" bitter.
Therefore, I have removed the
references to "super tasters" - ok to
mention in the TR but there is really
way for a student to know if they
Mini now
Investigation
are a taster or super taster. Ie how
pTC Tasting—“I Don’t Like my vegetables”
"bad" is "bad"? I for example am a
Skills: taster
Predicting,-Performing,
Analyzing
Evaluating
the PTC
tastes
disgusting to
me - but I have no way of knowing if it
Phenylcarbamide (PTC) is not naturally found in foods, but
even worse
to others
don't
severaltastes
related compounds
are. Whether
or not so
you Ican
taste
know
if
I
am
a
super
taster
or
not.
PTC or any of its related chemicals in food depends on the
genes that you inherited from your parents. If you are able to
taste PTC, there is a good chance that you are a non-smoker
and do not like Brussels sprouts grapefruit juice, or green tea.
Equipment and Materials: PTC test paper
1. Predict if you are able to taste PTC using knowledge of
your dietary habits.
Obtain
2. Take a strip of PTC test paper out of the stock bottle.
3. Place the strip on the tip of your tongue for 5 seconds.
Then remove the strip.
4. Record your taste result: no taste, slightly bitter taste,
extremely bitter taste
5. Pool your data with those of the rest of the class.
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SKILLS
HANDBOOK
A. Report the class data. How many of your classmates taste
PTC? How many of your classmates cannot taste PTC? T/I
B. What percentage of students in the class were nontasters (recorded no taste)? What percentage of students
in the class were super-tasters (recorded an extreme
taste)?
C. Which group is larger: super-tasters or non-tasters?
Is one group much larger than the other? Extrapolate
your results to the general population by comparing the
percentages. T/I
D. Is there a correlation between your dietary habits and
whether or not you could taste PTC? Explain your
thinking. T/I A
E. Share your answer to E with a partner. Do your partner’s
dietary habits correlate with his or her ability to taste
PTC? T/I A
Introduction
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5.1
trait a particular version of a
characteristic that is inherited, such as
hair colour or blood type
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Mendelian Inheritance
You are likely familiar with the notion of resemblance—two siblings who look similar
either
simple
change
to "simple
to each other
or to delete
their parents.
Youor
probably
resemble
one orbut
moreelegant"
of your family
members. This is because you have many genes in common. When we talk about
resemblance,
we are"simple"
usually referring
to atraits.
A trait is a We
particular
of an
I think
alone is
bit insulting.
bred version
many hundreds
inherited characteristic, such as a person’s eye colour or the shape of a leaf. People
of plants and very carefully observed and documented
have always recognized that traits are hereditary, even though they did not underthousands
of individual
- not
really a advances
simple task.
stand the mechanism
of inheritance.
Overtraits
the last
few centuries,
in genetics
have changed the way we understand inheritance.
We owe much of our understanding of genetics to the simple experiments conducted by Gregor Mendel in the nineteenth century (Figure 1). At that time, some
scientists thought that traits from each parent were blended in the offspring, similar
to mixing red and white paints to make pink paint. However, offspring sometimes
exhibited a trait identical to that of one parent rather than being in between those
of both parents. To explore patterns of inheritance, Mendel crossbred thousands of
plants in his garden and carefully recorded the offsprings’ traits.
Mendel’s Pea Plants
Figure 1 Gregor Johann Mendel
(1822–1884), an inquisitive Austrian
monk, is known as the “father of
genetics.”
true-breeding organism an organism
that produces offspring that are genetically
identical for one or more traits when selfpollinated or when crossed with another
true-breeding organism for the same traits
hybrid the offspring of two different truebreeding plants
Mendel conducted experiments with the garden pea, Pisum sativum. He chose the
garden pea because it reproduces quickly and, more importantly, he could control
which parents produced offspring. The sex organs of a plant are in its flowers. Pea
flowers have both male and female reproductive organs. Garden peas are both selffertilizing and cross-fertilizing. In other words, the pea flower can self-pollinate
(mate with itself) or pollinate others.
Some garden peas are true-breeding plants. This means that, when self-pollinated or
crossed with a similar true-breeding plant, they will always produce offspring that have
the same trait. For example, if a true-breeding pea plant with purple flowers is selfpollinated, or crossed with another true-breeding plant with purple flowers, all offspring
plants will have purple flowers. The offself-pollinating
spring of two different true-breeding plants is
called a hybrid. By preventing pea plants from, Mendel was able to cross-breed plants
with specific traits. Mendel removed the male reproductive organs, the anthers, from the
flowers of true-breeding plants. He then pollinated true-breeding plants with pollen from
other true-breeding plants (Figure 2). Since the parent plants were true breeding but had
different traits, the offsprings’ traits would represent the hybrid condition.
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missing here?>
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LEARNING TIP
Characteristic or Trait
Do not confuse the terms
“characteristic” and “trait.” Traits
represent the variation within a
characteristic. For example, height is
a characteristic, while short and tall
are traits; sight is a characteristic,
while normal vision, near-sightedness,
and far-sightedness are traits.
a)
stigma
carpel
not sure if this is too late - but the second
flower
b) shouldnot have c)anthers - these d)
Pollen been removed early on.
would have
pollen
anthers
Figure 2 Pollen grains form in the anthers. The egg cell is foundhad
in the carpel. Mendel brushed
pollen from one plant (a) onto the stigma of a second plant (b). He cut the anthers from the second
plant so it could not self-pollinate. He then planted the resulting seeds (c) in order to observe the
characteristics of the resulting offspring (d).
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Chapter 5 • Mendelian Genetics—Patterns of Inheritance
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Another significant feature of the pea plant is that it has several observable characteristics, each of which is expressed in one of two ways. For example, the shape of the
pea may be smooth or wrinkled; the colour of the seeds may be yellow or green.
Mendel performed his experiments on seven hereditary characteristics of the pea
plant: flower colour, flower position, stem length, seed shape, seed colour, pod shape,
and pod colour (Table 1). Mendel chose characteristics that always occurred in one
of only two ways so that he could distinguish between these traits and thus interpret
his data easily.
Mendel’s Experiments
In genetics, the breeding of two organisms with different traits is called a cross. In
order to track the inheritance of a single trait, Mendel crossed true-breeding plants
that differed in only one characteristic, such as flower colour. These plants were
his parental generation, or P generation. The hybrid offspring of cross were the filial
generation, or F1 generation, (from the Latin word for son, filius). The F1 generation
differed from each other in only one characteristic, making them monohybrids. This
type of cross, which scientists use to study the inheritance of a single trait from two
true-breeding parents, is called a monohybrid cross.
Figure 3 shows an example of one of Mendel’s monohybrid crosses. Mendel
crossed a true-breeding pea plant with purple flowers with a true-breeding pea plant
with white flowers. He wondered whether the hybrid F1 generation would have pink
(or “blended”) flowers, as some scientists might have predicted. Mendel observed,
surprisingly, that all the F1 plants had purple flowers rather than flowers that were a
blend of the two traits in the P generation. It was as though the trait for white flowers
had disappeared!
Flower colour
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F1 generation
P generation
Table 1 Seven Characteristics of Pea
Plants
Characteristic
Traits
flower colour
purple/white
flower position
axial (along
stems)/terminal
(at tips)
stem length
tall/dwarf
seed shape
smooth/wrinkled
seed colour
yellow/green
pod shape
inflated/
constricted
pod colour
green/yellow
cross the successful mating of two
organisms from distinct genetic lines
P generation parent plants used in a
cross
F1 generation offspring of a P-generation
cross
monohybrid the offspring of two different
true-breeding plants that differ in only one
characteristic
monohybrid cross a cross designed to
study the inheritance of only one trait
×
purple
white
purple
Figure 3 All the offspring of a monohybrid cross between purple true-breeding pea plants and
white true-breeding pea plants have purple flowers.
When Mendel allowed the F1 generation of plants to self-pollinate, the resulting
F2 generation included both plants with purple flowers and plants with white flowers.
This meant that the trait for white flowers had not disappeared but had somehow
been masked.
What Mendel did next was fundamentally important in his pursuit of scientific
knowledge. He recorded the numbers of the F2 generation plants according to their
traits. He then calculated the ratios of the traits for each characteristic.
Mendel found a pattern. In each F1 generation, only one of the two traits was
present. In the F2 generation, both traits were present—the missing trait had reappeared. This disproved the “blending” theory. The traits in the F2 generation were
repeatedly expressed in a ratio of approximately 3:1.
F2 generation offspring of an
F1-generation cross
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5.1 Mendelian Inheritance
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The results of Mendel’s careful analysis are summarized in Figure 4.
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Characteristics
P
F1
F2
seed shape
round × wrinkled
all round
5474 round
1850 wrinkled
2.96:1
seed colour
yellow × green
all yellow
6022 yellow
2001 green
3.01:1
299 constricted
2.95:1
pod shape
inflated × constricted
all inflated
pod colour
green × yellow
all green
428 green
152 yellow
2.82:1
flower
colour
purple × white
all purple
705 purple
224 white
3.15:1
flower
position
axial × terminal
all axial
651 axial
207 terminal
3.14:1
tall × dwarf
all tall
787 tall
277 dwarf
2.84:1
stem length
882 inflated
Ratio
Figure 4 Mendel’s crosses with seven different characteristics in peas, including his results and
the calculated ratios of the offspring
web Link
Mendel’s Life
Mendel lived a fascinating life.
To find out more about Mendel’s life
and work,
g o t o n e l so n sci e nce
Mendel’s Conclusions: The First Law
of Mendelian Inheritance
Mendel concluded that traits must be C05-F04-OB11USB
passed on by discrete heredity units, which he
called factors. Although these factors might not be expressed in an individual, they
can still be passed on. Mendel called the factor that was expressed in all the F1 generations the “dominant factor.” The factor that remained hidden but was expressed
in the F2 generation is the “recessive factor.” In addition, once Mendel had compiled
all the data and realized that there was a definite pattern, he recognized that the 3:1
ratio was an important clue. Why would a trait present in the parent generation not
be expressed in the offspring (F1) but then reappear in 25 % of the second generation
(F2)? Mendel had noticed a pattern in the data. Now, he had to try to explain it.
Mendel’s next two conclusions form the law of segregation:
law of segregation a scientific law
stating that (1) organisms inherit two
copies of genes, one from each parent,
Bio 11
and (2) organismsSci
donate
only one copy of
ISBN
each gene to their gametes because the
#0176504311
genes separate during
gamete formation
Figure Number
Artist
Ann
Sandersonof chromosomes. In addition, Mendel recognized that traits are inherited
discovery
Pass
each
2nd
Passparent. Today, these distinct units are called genes, and we know that they are
• For each characteristic (such as flower colour), an organism carries two
factorsI(genes):
from each
parent.
would one
reword
so that
we don't use the word
• Parent gene
organisms
donateimmediately
only one copy before
of each gene
in their
for unit,
saying
wegametes.
call During
meiosis,
the
two
copies
of
each
gene
separate,
or
segregate.
the units genes.
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Using his data, Mendel was able to predict the results of meiosis long before the
in distinct units and that an organism inherits two copies of each gene—one from
Approved
Not Approved
passed on from one generation to the next. Typically, each gene determines a specific
characteristic that will appear in the individual, such as seed colour or pod shape.
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Alleles: Alternate Forms of a Gene
Recall from Chapter 4 that each gene has a locus, or position, on a chromosome.
Most genes exist in at least two forms. For example, in Mendel’s experiments, there
were two different forms of the gene for flower colour, two different forms of the gene
for stem length, and so on. Each form of a gene is called an allele. Your cells have two
alleles for each gene. One allele for the gene is inherited on a chromosome from one
parent, and the other allele is inherited on the homologous chromosome from the
other parent.
Each parent passes on one copy of each chromosome to the offspring via gametes.
Gametes are formed during meiosis. As you learned in Chapter 4, during anaphase
of meiosis I (Section 4.2 Figure 4), homologous chromosomes separate. This ensures
that each gamete receives only one chromosome from the pair and therefore receives
only one allele for each gene. In other words, only one allele from each parent is
passed to the offspring. Which of the two alleles will be passed on is random and
purely a matter of chance.
The two alleles that an individual inherits from its parents for a particular characteristic may be the same, or they may be different. Different allele combinations can
result in different traits for that characteristic. If the two alleles for a particular gene
are the same, the individual is homozygous for that allele. This would be the case, for
example, if both flowers-colour alleles coded for white flowers. If, however, one allele
coded for white flowers while the other coded for purple flowers, the alleles would
be heterozygous. The term heterozygous describes an organism that has two different
alleles for a gene.
of that an individual has is its genotype. An individual’s genotype
The set of alleles
includes all forms an individual’s genes, even if some of these genes remain “hidden.”
In contrast, the traits of an individual make up its phenotype. The alleles that are
expressed determine an individual’s phenotype.
Dominant and Recessive Alleles
In heterozygous individuals, which allele is expressed? As Mendel observed in his experiments, some alleles were expressed while others remained hidden. A dominant allele is
an allele that expresses its phenotypic effect whenever it is present in the individual. A
recessive allele is expressed only when both alleles are of the recessive form.
In Mendel’s experiments, the allele for purple flowers was dominant over the allele
for white flowers. This explains why, when Mendel crossed two true-breeding plants
with different alleles, all the flowers were the same colour. The resulting F1 generation
expressed only one allele.
Geneticists use letters to represent alleles. Uppercase letters represent dominant
alleles; lowercase letters represent recessive alleles. An individual’s genotype is
expressed with one letter for each allele. As an example, the gene for plant height
results in tall plants or dwarf (short) plants. The allele for tall plants is dominant
and represented by the capital letter T. The allele for dwarf plants is recessive and
assigned a lower case t. Possible genotypes for the plant are homozygous dominant
(TT), homozygous recessive (tt), or heterozygous (Tt). However, there are only two
possible phenotypes: tall and short (Figure 5).
allele a specific form of a gene
homozygous describes an individual that
carries two of the same alleles for a given
characteristic
heterozygous describes an individual
that carries two different alleles for a given
characteristic
genotype the genetic makeup of an
individual
phenotype an individual’s outward
appearance with respect to a specific
characteristic
dominant allele the allele that, if present,
is always expressed
recessive allele the allele that is
expressed only if it is not in the presence
of the dominant allele, i.e., if the individual
is homozygous for the recessive allele
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INFLUENCE OF ALLELES ON PHENOTYPE
It is not possible to determine whether the genotype of a tall pea plant is TT or Tt
just by looking at it. Why do two different genotypes result in the same phenotype?
Whenever an individual has at least one copy of the dominant allele, that allele is
expressed. All tall plants are either TT or Tt. It is the T allele that makes them tall.
Only plants that have no T allele—only tt plants—show the dwarf phenotype. In
heterozygous plants, the T allele “dominates” the t allele.
TT
Tt
tt
Figure 5 Heterozygous plants inherit
one T allele and one t allele, and are tall.
5.1 Mendelian Inheritance
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LEARNING TIP
Dominance
In genetics, dominance refers only
to which gene is expressed in an
organism. It does not mean that the
allele is stronger, better, or more
common than the recessive allele.
What makes an allele dominant or recessive? One common situation occurs when
the dominant allele codes for a working protein, while the recessive allele does not.
For example, melanin is a pigment responsible for colour in our eyes, skin, and hair.
Humans have two forms of the gene for the production of melanin. One form of
the gene—call it allele M—provides instructions for the production of melanin. The
other form—allele m—is unable to code for the production of melanin. Only a single
copy of the M allele (like a single set of instructions) is needed to produce melanin.
Individuals who produce melanin have normal eye, skin, and hair colour. An Mm
individual can make melanin just as easily as an MM individual, but mm individuals
are unable to produce melanin. These individuals have the condition known as albinism. In this case the M allele is said to be dominant over the m allele, because the
normal colour phenotype is expressed whenever an M allele is present, and the albino
phenotype is expressed only in mm individuals.
You will learn more about the relationships between alleles and their resulting
phenotypes in Section 5.2.
Predicting the Inheritance of Alleles
Punnett square a diagram that
summarizes every possible combination
of each allele from each parent; a tool
for determining the probability of a single
offspring having a particular genotype
Geneticists use monohybrid crosses to study inheritance. They cross two truebreeding parents that differ in a single trait. In other words, the study involves the
inheritance of two alleles for a single characteristic. Mendel’s experiments consisted
of many crosses. As a result of these experiments, he developed a way of mathematically predicting the proportions of phenotypes in the offspring. Biologists now use
Punnett squares to copy Mendel’s analysis. A Punnett square is a diagram used to
predict the proportions of genotypes in the offspring resulting from a cross between
two individuals (Figure 6).
parent 1 alieles
P
p
PP
Pp
Pp
pp
quite difficult to tell P from p
unless next to each other. Do
not italicize the "P" - only the "p"
P
parent 2
alieles
possible
genotypes
of offspring
p
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Figure 6 A Punnett square is a grid system for predicting the possible genotypes of offspring
probability the possibility that an
outcome will occur if it is a matter of
chance
Probability is a measure of the chance that an event will happen. For example, when
you flip a coin to settle a dispute, there is a 50 % chance (50:50 ratio) that the coin
will land on the side you have selected. However, this does not mean that if you flip
the coin 10 times you will get heads 5 times and tails 5 times. Each flip of the coin is
an independent event.
Tutorial 1
8
Predicting Single-Characteristic Inheritance
Chapter 5 • Mendelian Genetics—Patterns of Inheritance
Ontario Biology 11 U SB
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Tutorial 1 Predicting Single-characteristic inheritance
Sample Problem 1: Homozygous Dominant/Homozygous Recessive Cross
Can we not italicize the capital YY
In pea plants, the allele for yellow seed colour, Y, is dominant over that for green seed colour, y. Consider a cross between a pea
and Y alleles? - it is very difficult to
I think
this tutorial
should
in the form of
plant that is homozygous
yellow
seeds yand
a plantunless
that is they one of four possible
combinations
of alleles
that the be
offspring
tellforthe
Y and
apart
Step
1,
Step
2.
homozygous recessive for green seeds. Create a Punnett square to may receive (Figure 8).
are next to each other. Y and y are
determine the possible genotypes and phenotypes of the offspring.
much more distinctive
The first step is establishing the proper
Solution
yellow
seed
gametes, the second
is placing
them in the
The plant that is homozygous
for yellow
seedto
colour
has a - then in C05-F07-OB11USB
If we don't
want
do this
table.
YY
genotype of YY. The plant
is homozygous
green
the that
diagrams
(butfor
not
theseed
text)
colour has a genotypecould
of yy. we use a larger font for the
StepDraw
1: a Punnett square
that shows
the genotypes
of the
Y
Y
Y alleles
so they
are clearly
two parents and the gametes that they can produce. Write the
I think including short thin
distinct?
symbols for the gametes across the top and along the left side
lines is useful but I don't like
of the square (Figure 7). Note that, each parent can supply two
Yy
Yy
y
the long thick black
lines.
possible
gametes, each containing one of two possible alleles.
NO ONE that can find
uses
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anything like these thick
black lines except parent
us ingenotypes
the
old 11U. They are
in
distracting and misleading. alleles
gametes
Most books place gametes
in little circles to show they
y
represent alleles "in sex
cells" not just letters on a
yy
page.
green seed
y
yellow seed
green seed
YY
Y
Y
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Crowle Art Group
2nd pass
y
Yy
Figure 8 A complete Punnett square showing all the possible
genotypes for the offspring.
process - not just show the result. We are
instructing the student on "how to"
Step 2: Now fill in the boxes of the Punnett
square by combining the gametes
corresponding to each row and column. Each
box represents a possible offspring genotype.
parent genotypes
Yy
alleles in
gametes
the possible gametes from each parent across the top and down
Not Approved
the left-hand side. In this case the two possible gametes from
offspring.
each parent are Y and y. Proceed to form all the possible zygotes
(Figure 8).
Yy
In the completed Punnett square, there are 1 YY, 2 Yy,
and 1 yy genotypes. Therefore, the genotype ratio is 1
homozygous dominant plant (YY ) to 2 heterozygous plants
(Yy) to 1 homozygous recessive plant (yy), or 1:2:1. Both YY
and Yy plants have yellow seeds. The phenotype ratio in the
Figure 8
F2 generation is 3 yellow seed plants (YY 1 2Yy) to 1 green seed
plant (yy), or 3:1.
Y
y
Y
YY
Yy
y
Yy
yy
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Yy
The four possible genotypes for the cross between a pea
We don't tell the student where these letter
plant that is homozygous for yellow seed colour and a pea plant
come from. I realize it is "obvious" from the
that is homozygous recessive for green seed colour are Yy, Yy,
figure
is the
a tutorial
Yy,
and Yy.but
The because
offspring willthis
all have
genotype we
Yy, soshould
will all
use
wording
that
describes
each
step
in
the
have the yellow seed phenotype.
I prefer including little
arrows in this first drawing
Figure- 7I think
An incomplete Punnett square showing the possible
which I include here
gametes from each parent
it is visually meaningful as it
shows where the gametes
The letters in each box within the Punnett square represent
Ontario Biology 11 U SB
"come from". So the
students "follow the
arrows"
Sample
Problem0176504311
2: Heterozygous/Heterozygous Cross
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- this models howTwo
they
heterozygous yellow seed plants (Yy) are crossed. Determine
Crowle Art Group
CO
decide what letters
the genotype and phenotype ratios of the F2 generation offspring.
represent the possible
Solution
2nd pass
Pass
gametes from each
and label the parent genotypes. Insert
particular parent.Draw a Punnett square
Approved
11 U SB
yy
posted to 2nd pass folder 9-1-10
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Sample Problem 3: Determining Parent Genotype Using Offspring Phenotype Ratios
Table 2 shows the phenotypes of offspring produced in a cross.
Determine the probable genotype of the parents. Which allele is
dominant? Assume that the trait is influenced by only two alleles
and follows the laws of Mendelian inheritance. Use the letters R
and rsteps
to represent
alleles.
the
andthe
think
we have a
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I like
mismatch
if don't include them above
Table 2
when there clearly "are" steps.
Offspring phenotype
Number of plants
red tomato
1821
yellow tomato
Rr
R
r
R
RR
Rr
r
Rr
rr
Rr
615
Step 1. Determine the whole number ratio of red tomato plants
to yellow tomato plants.
red
1821
2.97
5
5
5 3:1
yellow
615
1
Step 2. Since this is a 3:1 phenotype ratio, it matches a cross
between two heterozygous parents.
Therefore, we predict that red (R) is dominant to yellow (r), and
the parent plants were both Rr.
Check Your Answer: Use a Punnett square like the one shown
in Figure 10 to cross two heterozygous red tomato plants.
Figure 10
The cross between two heterozygous red tomato plants
produces a 3:1 phenotype ratio of red to yellow tomato plants.
Genotypically, one tomato plant is homozygous dominant (RR),
two are heterozygous (Rr), and one is homozygous recessive
(rr). In this case many heterozygous tomato plants were crossed,
3
and of 2436 (1821 1 615) tomato plants, about produced red
4
1
tomatoes and produced yellow tomatoes.
4
Practice
1. A researcher crossed a homozygous yellow seed plant (YY ) and a heterozygous yellow seed
plant (Yy). Determine the genotype and phenotype ratios of the offspring.
2. A researcher crossed a heterozygous yellow seed plant (Yy) and a recessive green seed plant
Ontario Biology 11 U SB
(yy). Determine the genotype and phenotype ratios of the offspring.
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Test Crosses
Crowle Art Group
CO
A test cross is used to determine if an individual exhibiting a dominant trait is
homozygous or heterozygous for that trait. A test cross is always performed between
1stunknown
pass
the
genotype and a homozygous recessive genotype. This is achieved by
Approved
crossing the individual with the dominant trait with an individual that exhibits the
Not Approved recessive trait. The results reveal the genotype of the parent:
• Ifalltheoffspringdisplaythedominantphenotype,thentheindividualin
question is homozygous dominant.
• Iftheoffspringdisplaysbothdominantandrecessivephenotypes,thenthe
individual is heterozygous.
test cross a cross used to determine the
genotype of an individual expressing a
Pass
dominant trait
Test crosses work well with species that reproduce quickly and in large numbers.
For example, test crosses can be used on mice because they produce large litters (7 to
12 mice on average) and have a gestation period of only 18 to 21 days. Also, mice can
become pregnant again while nursing a litter. Therefore, a large sample size of mice
can be studied in a short period of time. Cows, on the other hand, give birth to only
one calf each year and the gestation period of a cow is 9 months. This makes testcrossing cows difficult. Although farmers often attempt breeding cows with beneficial
traits, it takes a long time to improve a herd.
Today, test crosses are rarely performed. Advances in molecular biology techniques allow geneticists to test for specific alleles within the genotype of an organism
directly, rather than having to wait for the production of offspring.
10
Chapter 5 • Mendelian Genetics—Patterns of Inheritance
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Tutorial 2 Determining an Unknown Genotype
Sample Problem 1: Performing a Test Cross
Animal and plant breeders are often interested in whether or not
an individual will consistently produce offspring with a desired
trait. A breed of rooster has a dominant trait (S)—a comb that
resembles a series of fingers—while a breed with a recessive
trait (s) has a flat comb (Figure 11).
A breeder would like to use a true-breeding, homozygous,
Again, asrooster
a tutorial
think we
five-fingered-comb
as a studI don't
in her breeding
program.
should
state
it
in
this
way.
We
should
She has many roosters to choose from but does not know if they
have . (Ss)
. . .or homozygous dominant (SS) for the trait.
are heterozygous
(a) What type of hen should she cross with the roosters in order
Step 3.whether
Draw athe
Punnett
to
to determine
particular
roostersquares
is homozygous
or heterozygous
for
the
fi
ve-fi
ngered
comb?
Explain
your
determine the expected results (Figure
reasoning
12) using Punnett squares.
(b) What are the expected results?
No matter which genotype hen is crossed with a
homozygous dominant rooster (SS), all the offspring will
inherit an S allele from the rooster and have a fivefingered comb.
However,
the heterozygous
roosters could
Unfortunately
again
- the
S and s look
pass
on
either
an
S
or
an
s
allele.
Therefore,
you can tell
quite similar unless side by side. Could
them apart if you can detect this s allele in the offspring.
we try increasing the font size of the "S"
The only way to tell if an offspring receives an s from the
or only
use italics for the "s"?
rooster is if the offspring also receives an s from the hen
and is born with a flat comb. Therefore, to ensure that all
the offspring receive an s allele from the hen, the breeder
should choose a homozygous recessive (ss) hen. The
Punnett squares for the crosses of a homozygous ss
hen with the two genotypes of roosters are shown in
Figure 12.
rooster
Ss
rooster
SS
S
s
S
S
s
Ss
Ss
s
Ss
Ss
s
Ss
ss
s
Ss
Ss
hen ss
(a)
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(b)
(a)
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hen ss
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Figure 11 A rooster could have a five-fingered comb (a) or a flat
comb (b).
Step 1. List the possible genotypes of roosters and hens.
The roosters of interest all exhibit the dominant trait,
so they must be either homozygous dominant (SS) or
heterozygous (Ss).
There are three possible hen genotypes: homozygous
dominant (SS), heterozygous (Ss), and homozygous
recessive (ss).
Step 2. Decide which hen genotype could be used to distinguish
homozygous roosters from the heterozygous roosters.
(b)
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Figure 12 (a) Homozygous recessive hen and homozygous dominant
rooster cross (b) Homozygous recessive hen and heterozygous
rooster cross
Answers:
(a) The breeder should cross the rooster with a hen with a
flat comb. Using a homozygous recessive (ss) hen ensures
that all the eggs will contain a recessive allele from the
hen and none will contain a dominant S allele that would
mask the presence of a recessive s allele in the rooster.
(b) If the rooster is homozygous dominant, all the offspring
will express the five-fingered comb. If the rooster
is heterozygous, we would expect that 50 % of the
offspring will have a five-fingered comb while the
remaining 50 % will have a flat comb.
Ontario Biology 11 U SB
Ontario Biology 11 U SB
0176504311
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FNGroup
Crowle Art
CO
Crowle Art Group
CO
Pass
Approved
Not Approved
2nd pass
Pass
Approved
Not Approved
2nd pass
5.1 Mendelian Inheritance
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Practice
1. The gene for whisker length in seals occurs in two different alleles. The dominant allele (W )
codes for long whiskers, and the recessive allele (w) codes for short whiskers.
youlong-whiskered
expect to and the other parent is short-whiskered,
(a) If one parent is heterozygous
what percent of offspring would have short whiskers?
(b) A male long-whiskered seal is mated in captivity with a number of different females. With
some females all their offspring are long-whiskered, and with some females there are
both long- and short-whiskered offspring.
(i) What is the genotype of the male? How can you be sure?
(ii) Would it be possible to find a female mate that would produce only short-whiskered
offspring? Explain.
2. Mendel found that crossing wrinkle-seeded (rr) plants with homozygous round-seeded (RR )
plants produced only round-seeded plants. What genotype ratio and phenotype ratio can be
expected from a cross of a wrinkle-seed plant and a heterozygous plant for this characteristic?
Mini Investigation
What Are the Chances?
SKILLS
HANDBOOK
Skills: Performing, Analyzing, Evaluating, Communicating
This is ALL F1
When
two heterozygous
are crossed, the probability of
generation.
Theyindividuals
do
producing
each
genotype
is
25
%
homozygous
dominant: 50 %
not perform a second
heterozygous:
25 % homozygous
cross. Change
all F2's recessive. In this investigation
you will model a cross between two heterozygous individuals.
to F1's.
You will then determine the genotype and phenotype ratios of
your model F2 generation. In addition, you will investigate the role
that sample size and probability play in producing a 25:50:25
ratio in the F2 generation.
Equipment and Materials: two small pouches containing 40
beads each (20 white beads and 20 red beads)
1. Assign the red bead the dominant allele, R, and the white
bead the recessive allele, r. Label the two pouches “P1” and
“P2.” Each pouch therefore contains beads that represent
the gametes from one heterozygous individual (20 R and
20 r). Together, the two pouches represent the parent
generation.
2. Without looking, draw one bead from Pouch P1 and one bead
from Pouch P2. Place the beads together on a flat surface.
This represents the joining of two gametes to form a new
individual. The colours represent the alleles and the resulting
genotype of the offspring. For example, two red beads would
represent a new RR member of the F1 generation. Return
each bead to the pouch that you drew it from.
3.B., 10.
then tally the total number of homozygous dominant,
heterozygous, and homozygous recessive individuals
produced.
4. Pool your data with your classmates’ data.
A. What was your percentage ratio of homozygous dominant:
heterozygous: homozygous recessive individuals in your
20 F1 offspring? Calculate each percentage by dividing
your genotype counts by the total sample size (20) and
multiplying by 100. For example, if you had 4 homozygous
dominant individuals, the percentage of homozygous
4
individuals would be
3 100 5 20 %. T/I
20
B. Did your ratio approach a 25:50:25 ratio?
T/I
C. Answer A and B with the pooled class data. Remember
to use the total class sample size to calculate percentage
values. T/I
D. Check with other students. Were the percentages using the
pooled data closer to or further from the theoretical value
than the percentages using single-student data. Why is
sample size important? T/I
E. What aspect of Mendel’s own experimental design suggests
he understood the effects of sample size? T/I
3. Repeat Step 2 another 19 times, producing a total of 20
offspring. Record the genotype of each offspring and
12
Chapter 5 • Mendelian Genetics—Patterns of Inheritance
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5.1 Summary
• Gregor Mendel studied heredity in pea plants. He was the first person to
successfully record and quantify heredity data.
• Genes have alternate forms known as alleles. The alleles of a gene are found in
a specific position on a specific chromosome. Individuals have two alleles for
new bullet
each gene.
two alleles
areoffound
at specific
• Each parent passes on toThe
its offspring
only one
its two alleles
for eachmatching
gene.
This is called the law of segregation.
locations on homologous chromosomes.
• Some alleles are dominant, while others are recessive. Dominant alleles are always
expressed in the phenotype, but recessive alleles do not show up unless they are
the only
present
thesame
genotype. This is called the law of dominance.
twoallele
copies
of inthe
• Individuals who carry only one type of allele are homozygous for that gene.
Individuals who carry different alleles are heterozygous for that gene.
• A Punnett square is a tool that can be used to illustrate how alleles are
distributed from parent to offspring and to predict the frequency of
phenotypes and genotypes within a population. set of offspring.
• A cross between two heterozygous individuals produces a genotype ratio of
1:2:1 and a phenotype ratio of 3:1.
could be cut definition
rather than
summary
point.
5.1 QuestionsAlthough unlikely it is possible that
the man
is homozygous
so this 9. Humans who have an abnormally high level of cholesterol
1. Why was the pea plant
an excellent
choice for Mendel’s
question does not have a definitive are said to suffer from familial hypercholesterolemia. The
inquiry into heredity? answer.
Soexperimented
change towith truegene for this disorder is dominant (C ). A man who has
2. Why was it important that Mendel
K/U
breeding-variety plants? K/U
. . a man
who is
heterozygous
3. What were the phenotype
and genotype
ratios
of Mendel’s
familial
hyper
.
.
.
.
F1 crosses? What do the numbers represent? K/U
4. List Mendel’s conclusions from his experiments. How do
the conclusions relate to what is known today in the field of
genetics? A
5. Differentiate between the following: (a) dominant and recessive
(b) gene and allele
(c) homozygous and heterozygous
K/U
6. State the law of segregation. How does the law relate to
meiosis? A
7. Explain why it is important that Mendel had a large sample
size of offspring to count in his experiments. T/I
8. The smooth pea pod allele (S) is dominant, while the
wrinkled pea pod allele (s) is recessive. A heterozygous,
smooth pea pod plant is crossed with a wrinkled pea pod
plant. Use a Punnett square to solve the following: T/I C
(a) Determine the predicted genotype ratio of the offspring.
(b) Determine the predicted phenotype ratio of the
offspring.
(c) If this cross produced 50 plants, how many plants
would you predict would be wrinkled pea pod plants?
for
familial hypercholesterolemia marries a woman who does
not. What is the probability that they will have children that
suffer from this disorder? T/I C
10. Holstein dairy cattle normally have black and white spotted
coats. On occasion calves with a recessive red and white
spotted coat are born. A dairy farmer purchases a prized black
and white spotted bull. To the farmer’s dismay the bull produces
a red and white spotted calf when mated to one of his cows.
(a) What is the genotype of the bull? (Use R and r for the
colour alleles.) T/I C
(b) What phenotype ratio is expected in the offspring if the
bull is mated to a red and white spotted cow?
11. At one time, if a farmer wanted to improve his or her cattle
herd, he or she would have to buy an expensive bull from
another farmer who had a herd proven to show desirable
characteristics. Now, semen from bulls with desirable
characteristics can be shipped all over the world to help
farmers improve their herds. Use the Internet to learn more
about the use of artificial insemination (AI) as a cattle
K/U T/I
A
breeding option. (a) What are the primary advantages and disadvantages of
using AI for cattle breeding?
(b) How popular is AI as a breeding method for the beef
and dairy industry?
See Over matter
5.1 Mendelian Inheritance 13
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12. Plants contain many hormones that determine their
characteristics. Mendel was unaware of the hormones that
T/I
A
resulted in the different traits in his plants. (a) Using the Internet or other sources, research the role
that the hormone gibberellin plays in determining stem
length (plant height).
(b) How could knowledge of gibberellin release in plants
help agriculturists?
I would cut - very tangential to genetics. Better
go to nare
els on s c i en c e
for the plant unit where hormones
discussed in detail anyway. If needed, include
a research question that has a genetics
aspect to the "answer."
Over matter
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5.2
determine the phenotype
complete dominance an allele will
always be expressed, regardless of the
presence of another allele
In codominance an allele (such as
the A blood type allele) is also
always and fully expressed
regardless of the presence or
absence of other alleles.
incomplete dominance neither allele is
dominant; all alleles contribute equally to
the phenotype, to result in a blend of the
traits
Variations in Heredity
Mendel’s experimental work involved the crossing of what he called “typical” plants
(homozygous dominant) with “atypical” plants (homozygous recessive). Mendel had
discovered complete dominance, in which only one of the alleles is expressed, despite
the presence of the other allele.
Not all traits are passed on from parent to offspring in the simple patterns that
Mendel proposed. Variations in the patterns of heredity exist, and dominance is not
always complete.
Incomplete Dominance and Codominance
Mendel’s work provided an explanation of why the traits of parents did not blend in
the offspring. Yet blended inheritance is common in nature. In snapdragons one of
the genes that controls flower colour has one allele for red (R) and one allele for white
(W). A homozygous RR plant will produce red flowers, while a homozygous WW
plant will produce white flowers. However, the heterozygous plants will produce pink
flowers (RW). In this case, the actual flower colour (phenotype) is a result of varying
amounts of red and white pigments. The homozygous (RR) plant produces red pigment, the homozygous (WW) plant produces white pigment, and the heterozygous
(RW) plant produces both red pigment and white pigment. Neither of the alleles is
dominant, because the red pigment cannot mask the white pigment and the white
pigment cannot mask the red pigment. This type of interaction, in which a heterozygous phenotype is a blend of the two homozygous phenotypes, is known as incomplete
dominance. Interestingly, in this case, incomplete dominance still results in the same
Mendelian genotype ratio of 1:2:1 (Figure 1).
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codominance both alleles are expressed
fully to produce offspring with a third
phenotype
CR
CW
CR
C R CR
C R CW
CW
C R CW
CWCW
1 red
C R CR
Use the proper allele
designations
immediately and
have the Learning tip
next to that
paragraph. If the
learning tip comes
late student won't
have a clue where
this Punnett square
notation came from.
I would call this "mixed" because both alleles are
4
Learning Tip
2 pink C R C W
F2
expressed - there is no "new" version. In incomplete
4
Notation of Alleles
dominance you have a "new" phenotype - pink is NOT 1 white C R C W
Notation of alleles for a specific
red or white. But in codominance - like blood type an 4
gene can be represented using
1 Colour
snapdragons
is an example
incomplete dominance. When crossed, redAB
person
is A andFigure
B - the
"A" intrait
is all/fully
A withofno
superscripts. For
example,
consider
flowering
and
white-flowering
snapdragons
produce
different
in phenotype from an AA or AO individual. thepink-flowering offspring. A cross between
the alleles for colour
in snapdragons
these pink F1 individuals produces an F2 generation with a ratio of 1 red to 2 pink to 1 white (1:2:1).
shown in FigureB1.is
Theall
gene
B.is C for
colour. The alleles are red (R) and
white (W ). When you combine the
notations for genes and alleles, the
result is C R for the red allele and C w
for the white allele.
Another type of interaction between alleles occurs when both allele products apmixed
pear in the offspring at the same time. In this case, a third
phenotype is generated.
This type of interaction is called codominance. A classic example of codominance
pure
purecow will produce a roan
appears in shorthorn cattle. A red bull crossed
with a white
calf (Figure 2). Roan calves have intermingled white and red hair.
14 Chapter 5 • Mendelian Genetics—Patterns of Inheritance
7380_Ch05_pp002-045.indd 14
Ontario Biology 11 U SB
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F1 generation
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roan cow
roan bull
roan bull
red bull
P generation
white
cow
Hr
Hr
Hw
H rH w
H rH w
Hw
H rH w
H rH w
×
roan
calf
red
H rH r
roan
H rH w
roan
H rH w
white
H wH w
F2 generation
Figure 2 In codominance, one allele does not mask the other allele. Both alleles influence the final
phenotype. In shorthorn cattle, roan calves have intermingled red and white hair.
I would replace with.
Codominance and Dominance: ABO Blood Types
If i is paired
IA or
B, then the individual expresses the dominant allele
Human
bloodwith
type
is Iboth
a codominant and dominant genetic trait. There are
(I
A
or
I
B
)
and
is
either
type
A
or type
four major blood types: A, B,
AB, B.
and O. The blood type gene has three posA
B
sible alleles. They are I , I , and i. Each allele codes for a different enzyme that
places different types of sugars on the surface of a red blood cell. If you are
I A I A (type A), an enzyme places, one type of sugar on the surface of the cell. If you are I B
I B (type B), another enzyme places, a different sugar on the cell surface. If you are I A I B
(type AB), both sugars are placed on the cell surface. Type AB blood is an example of
codominance. The allele i codes for an enzyme that makes a simpler surface molecule
that lacks the extra sugars of the A, B, or AB blood types. If an individual is ii, he has type
O blood. If i is paired with I A or I B, then the individual is heterozygous for the respective
allele (I A or I B). Type I Ai blood and type I Bi blood are examples of dominant inheritance.
Table 1 shows the distribution and expression of the blood type alleles. One of the gametes
is provided by the father and the other is provided by the mother.
io Biology 11 U SB
Table 1 The Distribution and Expression of the Blood Type Alleles
These two columns
04311
are notC05-F13-OB11USB
really
Gamete 1
Gamete 2
Genotype
neededCrowle
- the Art Group A
I
student knows what
the gametes
were if I A
2nd pass
ovedthey are given the
IB
genotype.
We
could
pproved
IB
use the space for
the column - "Able
IA
to
i
Receive Blood
from ." . .
Blood type
IA
I AI A
A
i
I Ai
A
IB
I BI B
B
i
I Bi
B
IB
I AI B
AB
i
ii
O
don't split
genotype I^BI^B
We don't actually say that A and B
are dominant over i here where we
are discussing the details. We
don't have to tell them they are
heterozygous - this is not new info
and not relevant.
New<rom>
column . .
Able to recieve
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blood from
A, O
A, O
B, O
B, O
A, B, AB, O
O
Seems only fair to also include type O and AB
An individual with type A blood produces an immune response against type B and
individuals However, this is long way to present this
type AB blood. An individual with type B blood produces an immune response against
typeAB blood and type AB blood. A blood transfusion can take place onlyinformation
be-tween twoclearly. This is the info that is missing . . .
An can
individual
people who have compatible types of blood. Individuals with any blood type
receive with type O blood produces an immune
response
against type A,B and AB blood. An
type O blood, because it does not have an identifying A or B sugar on the
surface of the
red blood cells, so the cells are not targeted as foreign by the recipient’s immune
system.
individual
with type AB blood shows no immune
If an incompatible blood type is transfused, the patient’s life may be put
at
risk.
In
an
response to theFigure
other
blood types.
3 Blood banks are always in
emergency situation when there is no time to test the patient’s blood type, or if a certain
need of blood donations. Type O blood is
blood type is in short supply, type O blood may be used (Figure 3).
valuable
because
it is compatible
HOWEVER
this
could
be included
inwith
one column in
The frequency of the blood type alleles varies throughout the world’s population.
all blood types.
(my addition above). Then this paragraph
Genetically isolated populations sometimes have very high frequenciesthe
for table
particular
could
just
have a simple "example" and note that
alleles. For example, about 80 % of the Native Americans of the Blackfeet Nation
NEL
7380_Ch05_pp002-045.indd 15
type O is the "universal
donor"
and type "AB"
the
5.2 Variations
in Heredity 15
universal recipient.
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web LInk
To learn more about blood types,
g o t o n e l so n sci e nce
Investigation
5.2.1
Gummy Baear Genetics
In this investigation, you will use
the type and number of “offspring”
you have to predict the possible
genotypes of the parents. Make sure
you review the different Mendelian
monohybrid crosses.
Pikuni Indians in Montana have type A blood because the frequency of the I A allele
is very high in this population.
Codominance can provide an even greater variation in the population: there are
genes that have many more alleles than just three or four. For example, there is a gene
that plays a role in the acceptance or rejection of a transplant. This gene has more
than 200 different types of alleles.
5.2 Summary
determine the phenotype
• Allelesthatareexpressedregardlessofthepresenceofotherallelesfollowa
pattern of inheritance called complete dominance.
• Aheterozygousindividualwithanintermediatephenotypebetween
the phenotypes of the two homozygous individuals follows a pattern of
inheritance called incomplete dominance.
• Manygeneshavemorethantwoalleles.Bloodtypeisanexampleofagene
with multiple alleles. The three blood type alleles are IA, IB, and i. Different
combinations of the three alleles produce type A, type B, type AB, and type
O blood.
• Codominanceoccurswhenbothallelesareactive.Th
eheterozygotehas
the phenotypes of both homozygotes. Type AB blood is an example of
codominance.
5.2 Questions
1. Explain in your own words the meaning of dominance,
codominance, and incomplete dominance. k/U
2. In some chickens, the gene for feather colour is controlled
by codominance. The allele for black is FB and the allele
for white is FW. The heterozygous phenotype is known as
erminette. T/I A
(a) What is the genotype for black chickens?
(b) What is the genotype for white chickens?
(c) What is the genotype for erminette chickens?
(d) If two erminette chickens were crossed, what is the
probability that they would have a black chick? A white
chick?
3. A geneticist notes that crossing a round radish with a long
radish produces oval radishes. If oval radishes are crossed
with oval radishes, the following phenotypes are noted
in the F2 generation: 100 long, 200 oval, and 100 round
radishes. Use symbols to explain the results obtained for
the F1 and F2 generations. T/I C
4. Describe how Mendel’s conclusions may have differed if
he had worked with plants whose alleles were incomplete
dominant. k/U T/I
5. Thalassemia is an inherited anemic disorder in humans.
Individuals can exhibit major anemia, minor anemia, or be
completely normal. Assume only one gene is involved with
two alleles in the inheritance of this condition. What type
of inheritance is thalassemia governed by? What are the
corresponding genotypes to the three scenarios? k/U T/I
6. An individual has type A blood. List the possible genotypes
this individual may have. k/U T/I A
16
7. Suppose a father of blood type A and a mother of blood
type B have a child of type O. What are the possible blood
types of the mother and father? k/U T/I A
8. Suppose a father of blood type B and a mother of blood
type O have a child of type O. What are the chances that
their next child will be blood type O? Type B? Type A? Type
AB? k/U T/I A
9. Explain why blood type inheritance is an example of both
codominance and complete dominance. k/U
10. An additional characteristic of human blood is the presence
or absence of a blood protein referred to as the Rh factor.
People with the protein are Rh+ and those without it are
Rh–. Research this characteristic to answer the following
k/U T/I
A
questions:
(a) What are the genotypes of individuals who are Rh–
and Rh+? Is this an example of complete dominance,
incomplete dominance, or codominance?
(b) How can the Rh blood type of two parents be of
concern during a pregnancy? How can possible harmful
complications be avoided?
11. Tay–Sachs disease is a fatal lipid storage disease and is an
example of incomplete dominance. Using the Internet and
k/U T/I
A
other resources, research the following:
(a) What is Tay–Sachs disease and how does it affect an
affected individual’s health?
(b) What is the prognosis for an individual with Tay–Sachs?
(c) Why is Tay–Sachs disease considered an example of
incomplete dominance?
Chapter 5 • Mendelian Genetics—Patterns of Inheritance
7380_Ch05_pp002-045.indd 16
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8/31/10 4:34:40 PM
5.3
Pedigrees—Tracking Inheritance
Thanks to the laws of heredity, revealed by Mendel, scientists can now do genetic
analyses of heritable traits. Human genetics follow the same patterns of heredity
seen in organisms such as the garden pea. For example, if we know that a child
is born with a trait that neither parent has, then we can infer that the trait must
not be controlled by a dominant allele and that the child must have inherited two
recessive alleles.
Scientists are especially interested in determining the patterns of inheritance of
genes that are beneficial or detrimental to human health. For obvious reasons, experimental genetic crosses cannot be conducted on humans. However, we can use what
we know about heredity to investigate individuals and track the inheritance of a trait
from generation to generation within a family.
pedigree Charts
The simplest way to visually follow the inheritance of a gene is to construct a family
tree known as a pedigree. A pedigree is a chart
that traces between
the inheritance of a certain
connections
trait among members of a family. It shows the phenotype for all parents and offspring,
the sex of individuals in each generation, and the presence or absence of the trait
being tracked. The chart is composed of symbols that identify gender and relationships between individuals (Figure 1).
C05-F14-OB11USB
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redundant
- the ofpresence
pedigree a diagram
an individual’sor
absence
of the
traitgenetics
"is" the
ancestors used
in human
to
analyze the Mendelian
inheritancefirst
of
phenotype.
So replace
a certain trait; also used for selective
portion
as indicated.
breeding of plants and animals
C05-F17-OB11USB
OK - lots of items here - myopia is quite
complex and a number of references I
found said it was
or affected male
normalrecessive
male
normal female
affected female
predominantly so. Also there is an
environmental component in many
cases. So I think we should change the
mating
siblings
identical twins
fraternal twins
trait entirely. Regardless
of the trait
(female)
however - theC05-F18-OB11USB
caption and figure
were
C05-F19-OB11USB
C05-F20-OB11USB
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both incorrectFigure
for a1 dominant
trait.
Squares represent
males, and circles represent females. Individuals who express a trait
are shown in a shaded circle or square. Mating between two individuals is shown by a horizontal
line, and children are connected to their parents with vertical lines.
So - recommend we switch to a different
dominant trait - Pedigree
freckles,
curly
charts
arehair,
very ordered within the constraints of the available informaIf this 2isthat
trueeach
then
the pedigree
tion
about
the
family.
You
will notice in Figure
generation
is identified
double-jointed.
is notsymbolize
valid. IIIindividuals
2 and IIIwithin
3 have
by Roman numerals and that Arabic numbers
a given
inherited
a
dominant
allele
generation. The birth order within each group
of offspring
is drawn from
leftfrom
to right,
Also - as is - if we were using this
from oldest to youngest. Figure 2 shows a two
pedigree
for a family
with the trait of nearrecessive
parents.
myopia info we would also have to alter
sightedness.
a later tutorial question
where we
ask and analyze pedigrees to help trace the genotypes
Genetic counsellors
construct
Also this caption only mentions
students to determine
whether
myopia
and phenotypes
in a family.
Theyiscan determine if and how any particular trait runs
individuals
when
in a family. For
example,
expectant
might
want to know
how afive
recessive allele
dominant or recessive.
They
have
been parentsfour
exhibit
the
trait.
for
hemophilia
(a
blood
clotting
disorder)
has
been
inherited
in
past
generations.
told the answer AND the pedigree chart
A
genetic
counsellor
could
help
predict
how
that
gene
will
be
passed
on
to future
is again wrong "if" we were sticking with
generations.
dominant myopia.
C05-F22-OB11USB
I
1
2
2
3
II
1
4
5
III
1
2
3
Figure 2 An example of a pedigree
chart spanning three generations. In this
pedigree, the grandmother (I-1), one
of her daughters (II-2), one of her sons
(II-3), and her grandson (III-3) are nearsighted. The allele for near-sightedness
(S) is dominant over the allele for normal
vision (s).
New Fig 2 caption:
An example of a pedigree chart spanning three
generations. In this pedigree, the grandmother
(I-1), one of her sons (II-3), one of her daughters
(II-4), and her grandson (III-3) are nearsighted.
The allele for near-sightedness (S) is dominant
over the allele for normal vision (s).
Ontario Physics 11 SB nel
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FN
CO
5.3 Pedigrees—Tracking Inheritance
17
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Ontario
Physics
11folder
SB
posted
to 2nd
pass
9-1-10
8/31/10 4:34:41 PM
Tutorial 1
Interpreting Pedigree Charts
Figuring out genotypes from phenotypes on a pedigree chart requires you to use a process of
elimination. You can often determine which genotypes are possible, and which are not.
Sample Problem 1: Determining Genotypes of Individuals
Marfan syndrome is a genetic disorder that affect’s the body’s
connective tissue. When the dominant allele (M ) is expressed,
an individual will have Marfan syndrome. People with no defect
in the Marfan allele are homozygous recessive (mm). Individuals
with the syndrome are typically very tall, with disproportionately
long limbs and fingers, and sometimes have problems with their
hearts and eyes. Use the pedigree chart (Figure 3) to determine
the genotypes of all individuals, if possible. What information in
the pedigree confirms that Marfan syndrome is a dominant trait?
2
II
1
2
The shapes for the mother (I-2) and the two sons (II-1
and II-3) are not shaded, indicating that they do not have
Marfan syndrome. Therefore, they are all homozygous
recessive (mm) (Figure 5).
I
NO information confirms it
is dominant. This identical
chart is possible for a
recessive trait.
I
1
Step 2. Determine which individuals do not carry a dominant
Marfan allele.
3
Figure 5
C05-F23-OB11USB
Figure 3 A family’s pedigree showing the inheritance of Marfan
syndrome
Step 1. Determine which individuals carry a dominant Marfan
allele.
The shapes for the father (I-1) and the daughter
(II-2) are shaded, indicating that they have Marfan
syndrome. The Marfan allele is dominant, so all
individuals expressing this trait must be either MM or
Mm. Therefore, the father and daughter must be either
heterozygous (Mm) or homozygous dominant (MM ). I-1
and II-2 have a dominant allele (M ) and either another
dominant allele (M ) or a recessive one (m) (Figure 4).
1
M_
2
mm
II
1
mm
2
M_
C05-F25-OB11USB
3
mm
Step 3. Determine which individuals are heterozygous or
homozygous for the Marfan allele.
The mother is mm, so she can pass on only a
normal allele to her offspring. The daughter must
be heterozygous (Mm). The two sons do not have
Marfan syndrome, so they must both have inherited a
normal allele (m) from the father. The father must be
heterozygous (Mm). The completed pedigree chart is
shown in Figure 6.
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I
1
Mm
2
mm
I
1
M_
II
2
1
mm
II
1
Figure 4
11 SB
-g1
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2
3
Ontario Physics 11 SB
M_
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FN
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CO
Allan Moon
Pass
1st pass
2
Mm
3
mm
Figure 6 Individuals with Marfan syndrome must have at least one
M, and recessive individuals must be mm.
Approved
Allan Moon
Not Approved
1st pass
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Chapter 5 • Mendelian Genetics—Patterns of Inheritance
7380_Ch05_pp002-045.indd 18
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Sample Problem 2: Determining the Mode of Inheritance for an Allele
Individuals with albinism have a defect in an enzyme that is involved in the production of melanin,
a pigment normally found in the skin. These individuals have little or no pigment in their skin, hair,
and eyes (Figure 7). The characteristic is governed by only two alleles: the normal allele and the
albinism allele. Analyze the pedigree chart below (Figure 8) to determine whether the albinism
allele is a dominant or recessive allele. Then determine the genotypes of each individual. Use P
and p to represent the dominant and recessive alleles, respectively.
This is ok but we
did tell them that
albinism was
recessive
previously (page
8).
I
1
3
2
4
II
1
2
3
4
5
6
III
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Figure 7 The lack of melanin makes a person
with albinism much more susceptible to sun
damage.
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1
2
3
Figure 8 A family’s pedigree chart for albinism
I
P_
Step 1. Determine which individuals carry the recessive albinism allele.
P_
pp
P_
PP_
pp
pp
P_
[ARTPp
ALTs:Pp
- add labels for I, II,
III, as in Figure 8
- make uppercase P
P?
pp
italP?x10] pp
PP_
PP_
Pp
C05-F28-OB11USB
Pp
Figure 9 All individuals can be labelled P or pp
based on their expressed traits. Both parents of pp
individuals,
andart
all offspring
of awider
pp parent,
must
This
is 6 picas
than
the C
have at least one p allele.
I
recessive individuals are labelled pp, and all P_
dominant
P_ individualsP_are P_
labelled with one P (Figure 9).
width of 18p6 ...
but this is the standard size for pedigree charts
Can the charts have different sizes without confusion?
Pp
Pp
Pp
Pp
II
P?
P_
Some
butPhysics
not all11
of SB
the missing alleles can be filled in by looking at the
Ontario
0-17-6504311-g1
parents
and offspring of recessive individuals (Figure 10).
P_
P_
III
you
of individuals
you already "know" - so either
Step 2. Determine which individuals carry one copy of the dominant normal
use my cuts
OR change to "Identify those
allele.
individualsIndividuals
that are
homozygous . . . .
who do not have albinism must have at least one P allele. All
P_
P_
II
Individuals II-2, II-4, and II-5 have albinism, but none of their parents
exhibit this trait. It is not possible to inherit a dominant trait from a
parent who is not also dominant. Therefore, the trait must be caused
by a recessive allele. The F1 offspring who have albinism (II-2, II-4, II-5)
have inherited two copies of the p allele, making them homozygous
recessive (pp) for the characteristic. The genotypes of II-2, II-4, and II-5
don't are
need
to determine
labelled
pp (Figure 9). the genotypes
Step 3. Determine the genotypes of homozygous non-albino individuals and
heterozygous non-albino individuals.
P_
pp
P_
pp
pp
P_
P_
pp
P?
pp
pp
III
PPp
PPp
P?
[ART ALTs:
- add labels for I, II,
III, as in Figure 8
- add hair space
between uppercase
P and ? x 3
- make uppercase P
ital x10 ]
PPp
FN explanation
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Add
- make this a thorough tutorial not leaving
things to interpretation.
Figure 10
CO
Moon
"Every
parentAllan
of an
albino child must have at least one "p" and every albino parent
Pass on a "p"
1st pass
passes
allele to each ofThis
their
art children."
is 6 picas wider than
the C width of 18p6 ...
Practice
I
Approved
but
this
is
the
standard
size
for pedigree charts
1. Phenylketonuria (PKU) is a genetic disorder caused by a dominant allele.
1
2
Not Approved
Can therecessive
charts
have
different sizes without confusion?
Individuals
with phenylketonuria accumulate phenylalanine
in their
body.
High amounts of phenylalanine lead to delayed mental development.
II
Figure 11 is a pedigree chart that shows the inheritance of the defective
1
2
3
4
5
6
7
Ontario Physics 11 SB
PKU allele in one family.
p NOT P
(a) How many generations are shown in the0-17-6504311-g1
pedigree chart?
III
(b) Determine the genotypes of the individuals in Figure 11. Let P represent
FN
C05-F28 and 29-OB11USB1
2
3
4
5
the dominant phenylketonuria
Not theallele.
greatest
pedigree chart - I would note that it is actually
impossible IF PKU was
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CO
recessive
nel
7380_Ch05_pp002-045.indd 19
Allan Moon
11
a dominant trait
but since that is not theFigure
case
this chart is "possible" but highly
Pass
1st pass
improbable. I would be tempted to at least change
II4 to a blank
and perhaps
5.3 Pedigrees—Tracking
Inheritance
19 II5 and
Approved
II7 and III5 as well. We have two carriers that have passed a higher than predicted
Not Approved
number of recessive alleles on and both their children that marry, happen to marry
people with PKU - quite a high frequency in the pedigree for a RARE disorder.
posted to 2nd pass folder 9-1-10
8/31/10 4:34:45 PM
"IF" myopia was dominant like we told them
previously this chart would not be "possible".
shows those individuals that are
sensitively to poison-ivy.
2. The following pedigree is for myopia (near-sightedness) in humans.
(a) Analyze the pedigree chart (Figure 12) and determine whether the
disorder is inherited as a result of a dominant or recessive trait.
(b) Determine the genotype for each individual if possible.
However - lets just switch to a recessive trait
and then it will be fine.
I
II
Not all humans react strongly to poisonivy and this trait is thought to be
controlled by a single allele.
III
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Figure 12 A family’s pedigree chart for myopia
poison-ivy sensitivity
Sex Linkage—Following the X and Y Chromosomes
The
sentences
underlined in
Different organisms have different numbers of chromosomes.
Humans
have 23
green arewhile
all saying
the same
pairs of chromosomes. One set of chromosomes is the sex chromosomes,
the
thingis- found
and already
other 22 sets are autosomes, the non-sex chromosomes. If an allele
on an known to
autosome, it is said to be under the control of autosomal inheritance
. With autosomal
autosomal inheritance inheritance of
students
- nothing new.
alleles located on autosomal (non-sex)
inheritance, both males and females are affected equally, since there is no difference
chromosomes.
I really think this should be rearranged. between the autosomes of males and the autosomes of females.
However, some alleles that cause genetic disorders are found on the X chromosome.
sex-linked an allele that is found on one
Females
(XX) may have up to two copies of the gene, but males
only(XY) with only one X
of the
sex chromosomes,
X or Y,like
and when
Why not just
state
the obvious
we did chromosome have only a single copy. If a female has inherited one copy of a defective
passed on to offspring is expressed
and then talk
about the male first - the
recessive allele, the other copy of the gene, on the other X chromosome, being dominant,
expression
of an they and will mask the effect of the recessive allele. A female who carries the recessive genetic
male is the X-linked
oddity.phenotypic
Both the
fact that
allele that the
is found
on the from
X chromosome
disorder allele on only one X chromosome will not express the disorder. So, the female
"always" express
allele
their
mother - whether recessive or dominant is heterozygous and a carrier of the recessive allele. A female must inherit two copies of
LInk
and that it is web
impossible
for them to pass the recessive gene—one on each X chromosome—in order to express the disorder.
A mother who is a carrier has a 50 % chance of passing on the recessive allele to
the trait on toRoyal
their
sons.
Genes
her children. Since the allele with the disorder is found on the X chromosome and is
Queen Victoria and her descendants
recessive, this type of inheritance is called sex-linked and, more specifically, X-linked. If
constitute
one ofis
theinherited
most famous and
For the female
the trait
a male
X chromosome from a mother who carries the recessive allele, he
Ontarioinherits
Physics the
11 SB
charts. To explore this
expressed inpedigree
an entirely
normal way. So itwill
express the disorder because the Y chromosome cannot mask the effects of that
0-17-6504311-g1
pedigree in a case study and gain
seems odd tofurther
be talking
about
"carriers of allele. The male cannot inherit an X-linked disorder from his father, since a father
understanding
of X-linked
to a son.
FN on a Y chromosome
C05-F31-OB11USB
disorders" in inheritance
females before mentioning passes
Some
examples
of
X-linked
inheritance are red–green colour blindness, hemoCO
Allan Moon
the situation in males.
g o t o n e l so n sci e nce
philia A, and male-pattern baldness. Individuals who have hemophilia
A are
(Xh )
Pass
1st pass
not able to form a clot when they are cut and may bleed for a lengthy period of
Approved
time.
In Figure 13 the mother is a carrier of the hemophilia allele, and the father
Not
Approved
does not have hemophilia. The probability of this couple producing a son who
has hemophilia (XhY) is 25 %, and the probability of producing a daughter who
is a carrier (XHXh) is 25 %. There is a 50 % chance that the couple will produce
a daughter or son who does not inherit the hemophilia allele (XHXH and XHY).
XHXh
XH
Xh
XH
XHXH
XHXh
Y
XHY
XhY
XHY
C05-F32-OB11USB
Figure 13 Hemophilia A is X-linked. A female carrier can pass on the hemophilia allele to her sons
and daughters. Males cannot pass on hemophilia to their sons.
20
Chapter 5 • Mendelian Genetics—Patterns of Inheritance
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posted to 2nd pass folder 9-1-10
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8/31/10 4:34:46 PM
It might be worth mentioning that "of course" there are many traits on the Y
chromosome that are inherited (they determine male gender!) but are not disorders.
It just seems like we are discussing the sex chromosomes only in the context of
disorders.
Y-linked disorders also exist and are passed on from father to son. Very few Y-linked
disorders exist, since the Y chromosome is small and does not carry as much genetic
information as the X-chromosome. Male infertility can be caused by a Y-linked disorder. Males who possess this disorder can have children using medical intervention.
Y-linked phenotypic expression of an
allele that is found on the Y chromosome
seems odd - maybe "reduced
fertility" but not "infertility" - at
least not unless this is recent
mutation for which ALL fathers
have had medical intervention.
The main reason there are few disorders is because evolution selects
5.3 Summary
against them and since EVERY male in the affected lineage would have
the disorder
the selective pressure would be quite unrelenting and it is
• Pedigreechartsarevisualrepresentationsofafamilytreethatcanbeusedto
the inheritance
of a trait.
unlikely forfollow
the disorder
to persist.
X linked are not the same because they
• Ifanalleleislocatedonanautosome,oranon-sexchromosome,itis
are not selected
against in female carriers. So when the frequency of the
The way it is written every male
through
autosomal
inheritance.
allele is lowtransmitted
there is no
lineage
that gets
"targeted" by selection. All Y linked with the disorder must have had
• Sex-linkedinheritanceoccurswhenarecessivealleleisfoundontheXorY
"disorders"
must be quite mild (or be take effect later in life).
a father that underwent this
chromosome and that chromosome is passed on to the offspring.
intervention.
• InX-linkedinheritance,thesexesexhibitdiff
erentphenotypicratios.More
I know we
can't get into the evo here - so please
males than females will express the recessive phenotype, but more females are
a) include my
above note in the TR and
carriers of the recessive X-linked allele.
b) change wording to "one of the factors responsible for the fewer number
of Y linked
. ."
5.3 . .Questions
become
1. Sickle-cell anemia iscan
a condition
in which the red blood why not have one that is
cells of an individual are shaped like the letter “C.” This dominant? Here both 14 I
1
2
shape prevents the red blood cells from moving easily
and 15 show a recessive
through blood vessels. It can result in the cells clumping,
changed
because the sickle
trait. Reverse
every
II
I
blocking blood flow and causing pain, infection, and organ
shape
only
happens
when
colour
in
fig
15
and
it
will
1
2
1
2
3
damage. The allele that causes sickle-cell anemia is
oxygen
levels drop
a allele can be be converted to a
autosomal recessive
(s), andbelow
the dominant
certain
level
the cell.
represented
by S. within
For the following
families, determine dominantII trait and
III
the genotypes of the parents and offspring. When it is not remain a solvable
1
2
3
2
1
3
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possible to decide which genotype an individual is, list question.Figure
14
Figure 15
both. T/I C
(b) Label the genotype of each individual in the pedigree
(a) Two normal parents have four normal children and one
chart. Assume that the dominant allele is A and the
with sickle-cell anemia.
recessive allele is a.
(b) A normal male and a female with sickle-cell anemia
have six children, all normal.
5. Hairy ears is a rare condition that is sex-linked. Let H be
(c) A normal male and a female with sickle-cell anemia
unknowns in this
the dominant allele (non-hairy ears) and h many
be the recessive
have six children; three are normal, and three have
T/I
allele (hairy ears).
pedigree - do we want
sickle-cell anemia.
(a) Examine the pedigree chart in Figure 16.
Determine if
this?
(d) Construct a pedigree chart for the families in (b) and (c).
the condition is X-linked or Y-linked.
2. Distinguish between autosomal inheritance and sex-linked
inheritance. k/U
3. A male with hemophilia (X hY ) marries a woman who does
not carry the hemophiliac gene (X HX H ). Use a Punnett
square to answer (a) and (b). T/I
(a) What is the probability of producing sons or daughters
who have hemophilia?
(b) What is the probability of producing daughters who are
carriers of the hemophiliac allele?
(b) Label all possible genotypes.
I
1
nel
2
II
1
2
3
4
5
6
1
2
3
III
4. Examine the pedigree charts in Figures 14 and 15. T/I
1116
SB
(a) Determine whether the mode of inheritance for theOntario Physics
Figure
0-17-6504311-g1
affected individual is autosomal dominant or autosomal
recessive.Ontario Physics 11 SB
0-17-6504311-g1
I would prefer to switch
colours for II2 and III4.
Forces them to carry
through that extra
generation.
C05-F35-OB11USB
FN
C05-F34-OB11USB
CO
Allan Moon
FN
C05-F33-OB11USB
Pass
1st pass
CO
Allan Moon
Approved
Pass
1st pass
Not Approved
Also - bit of a shame the
affected males have no
4children to produce
carrier daughters.
5.3 Pedigrees—Tracking Inheritance
21
Approved
Not Approved
7380_Ch05_pp002-045.indd 21
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Biology JOURNAL
5.4
The Gene Hunters
Abstract
The road to scientific achievement is a challenging one that requires scientists
to have determination, perseverance, and innovation. This is well illustrated
by the history of our studies of Huntington’s disease. This deadly genetic
disorder strikes people in midlife and does not yet have any effective cure
or treatment. Motivated by personal tragedy, and supported by advances in
DNA technology, Dr. Nancy Wexler was a pioneer in the hunt for the faulty
gene. These efforts not only led to the discovery of the Huntington’s disease
gene (and many others), but also opened up avenues of future research that
offer hope for possible treatments and a cure.
Introduction
Huntington’s disease (HD) is a devastating neurological
genetic disorder. Inherited as a dominant autosomal trait,
the symptoms of this late-onset disease do not usually
appear until individuals are between 30 and 50 years of age.
Symptoms include uncontrollable movements, intellectual
and emotional deterioration, and other health complications that may lead to death.
First described as early as the sixteenth century, the unusual
symptoms were thought by some to be evidence of demonic
possession. Tragically, ignorance of genetics and such beliefs
likely led to the execution of many HD sufferers during the
witch trials of the Middle Ages. It wasn’t until 1872 that
American physician George Huntington (Figure 1) provided
the first detailed description of the disease and established it as
an inherited disorder.
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Venezuelan physician Americo Negrette unknowingly
made a major contribution to unravelling the mystery. He
studied two villages on Lake Maracaibo, Venezuela, that
had a very high incidence of a neurological disease known
locally as el mal (the bad). His findings, published in 1955
in Spanish, went largely unnoticed for more than a decade,
until a young researcher discovered them while searching
for answers to her own questions.
Personal Motivation
In 1968, at the age of 23, Nancy Wexler became very
interested in genetics—her mother had started to show
the symptoms of HD. Nancy’s maternal grandfather and
three uncles had already died of the disease. Fighting
the disease became the primary focus for Nancy and
her family. With great determination, she graduated
from university at the top of her class and became a
researcher for the National Institutes of Health in the
United States.
The Search for Answers
Figure 1 George Huntington (1850–1916) published his landmark
paper on HD when he was only 22 years old.
By the early twentieth century, Mendelian laws of inheritance
had become widely accepted in science, and researchers had
learned that HD causes parts of the brain to degenerate.
Unfortunately, with no understanding of the molecular
basis of inheritance, the genetic cause of HD remained a
mystery to scientists.
Wexler believed that the first step to finding a treatment or
cure for HD would be to discover the gene responsible for
the disease. By 1981, after obtaining vital federal funding,
she headed to the shores of Lake Maracaibo.
Wexler studied family histories and prepared pedigrees
of thousands of individuals in the Lake Maracaibo communities (Figure 2). In 1983, using pioneering advances in
DNA technology, researchers were able to identify a section
of DNA near the tip of chromosome #4 that is a marker
for the HD gene. A marker for a genetic disease is a DNA
sequence that is associated with a specific gene and is found
in the same position on a chromosome in people who have
or are predisposed to having that disease. The discovery of
this marker meant that, for the first time, there was a conclusive test for people at risk of inheriting the disorder.
22 Chapter 5 • Mendelian Genetics—Patterns of Inheritance
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living organisms. Genetically modified mice were created
that contained the actual human HD gene. These “model
organisms” could be used to conduct experiments and
study the activity of the gene. The mice were also useful for
preliminary testing of new drugs without putting human
patients at risk.
The Challenges Ahead
Figure 2 Since the early 1980s, Nancy Wexler has compiled a
community pedigree of over 18 000 individuals.
Innovations and Advances
In 1983, modern genetics and the ability to analyze DNA
was still in its infancy. It took the efforts of many more
scientists more than a decade before the actual gene for
HD was identified. The discovery of the HD gene in 1993
was monumental. The gene is abnormally long and prone
to repeated mutation events (changes in the DNA code as
it replicates). These mutations result in large numbers of
copies of a very short portion of the gene. This discovery led
to an understanding of an entire family of genetic diseases
caused by similar abnormal “repeat sequences” of DNA.
By 1996, advances in genetic technology had enabled
scientists to begin conducting research on the HD gene in
Even with tremendous strides in understanding, scientists
still do not know exactly how the protein produced by the
Huntington’s gene actually causes cell deterioration and
death. There are not yet any truly effective treatments or a
cure.
Still, researchers and patients remain hopeful as advances
in science continue. Perhaps soon an effective drug will be
found. Perhaps someday a patient’s own stem cells will be
used to regrow and replace defective or lost brain cells.
Perhaps someday we will have the ability to actually correct
or replace the defective gene using gene therapy.
Meanwhile, the villagers on the shore of Lake Maracaibo,
and other people with Huntington’s around the world,
remain the subject of intense scientific investigation as they
continue to suffer from this horrific disease.
Further Reading
[to come]
5.4 Questions
1. Prepare a brief chronology of events leading to our current
understanding of HD. C
2. Nancy Wexler could not have accomplished her goals on
her own. Use examples from this article to describe how
the discipline of science builds up a knowledge base over
time. A
3. Use the example of Huntington’s disease to illustrate how
advances in one area of science can be applied to others. A
4. Why would Dr. Nancy Wexler look at family pedigree charts
of the people of Lake Maracaibo as a starting point for her
search for the Huntington gene? T/I
5. Nancy Wexler has a strong affinity for the villagers of Lake
Maracaibo. She said, “The Venezuelan families have given
us many gifts. . . . It is important that the world understand
how much they have given. It would be fitting if they could
be the first to reap the benefits of all future therapies.” T/I
C
(a) Use the Internet to investigate the latest therapies that
are being researched for future implementation. Report
back on two that you think sound promising.
(b) Do you agree with Dr. Wexler that the people of Lake
Maracaibo should be the first to receive a successful
treatment? Why or why not?
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5.5
Genetic Disorders
cystic fibrosis (CF) a potentially fatal
genetic disease that affects the respiratory
and digestive tract of an individual
Many human disorders have a genetic component, but the onset of a disorder may
vary depending on life conditions. Some disorders can be detected at birth, while
others do not manifest themselves until an individual has reached a certain age range.
Some disorders can be treated and managed, while others lead to debilitating symptoms and premature death. Research geneticists spend a great deal of time and money
to better understand the genetic components of such disorders. Their intent is to one
day be able to detect, prevent, or fix the genetic component of the disorder.
Cystic Fibrosis
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Figure 1 A child undergoing physical
therapy for cystic fibrosis. The child’s
chest or back is tapped repeatedly to
loosen the mucus in the lungs. This
makes it easier for the child to rid his
body of the mucus.
grammar ok?
mutation a change in the genetic code of
an allele; the change may have a positive,
negative, or no effect
carrier testing a genetic test that
determines whether an individual is
heterozygous for a given gene that results
in a genetic disorder
genetic screening tests used to identify
the presence of a defective allele that
leads to a genetic disorder
Cystic fibrosis (CF) is the most common fatal genetic disease in Canada. The life expectancy for individuals with CF is less than 40 years, though this is increasing every
year—some individuals with CF have lived into their seventies. The disease causes the
body to produce thick, sticky mucus that clogs the lungs, leads to infections, and blocks
the release of enzymes from the pancreas. The pancreas produces digestive enzymes
that help break down protein, fats, and carbohydrates during digestion, so children
and adults who have CF must take a large number of replacement enzymes daily in
pill form. In addition, individuals with cystic fibrosis must undergo physical therapy or
such
the day
result
of loosen accumulated mucus in the lungsfirst
other treatments every
to help
(Figure
1).
Cystic fibrosis is caused by a mutation, or a change in the genetic code. The defective gene was isolated and identified in 1989 by research geneticist Dr. LapChee Tsui
and his team at the Hospital for Sick Children in Toronto. The mutated copy of the
gene is recessive, so a child must inherit both copies of the defective allele from his
or her parents in order to express CF. In the past, parents realized that they were both
carriers only when their child was born with CF.
Today, carrier testing is used to identify individuals who carry disorder-causing
recessive genes that may be inherited by their children. In fact, the Canadian College
of Medical Geneticists recommends that carrier testing for cystic fibrosis be available
to anyone who has a family history of CF. Genetic counsellors work with couples that
have an increased risk of conceiving a child with CF.
Testing for the presence of the mutated gene in the genome is known as genetic
screening. Currently, we are aware of approximately 200 mutations that lead to cystic
fibrosis. The severity of symptoms depends on which DNA mutation an individual
has. The genetic test is based on a blood sample that is sent to a molecular diagnostic
laboratory. In Canada the test is able to detect approximately 85 % of CF mutations.
Table 1 shows the probability of conceiving a child with CF according to different
test result scenarios.
Table 1 Risk of Having a Child with Cystic Fibrosis before and after Carrier Testing In Canada
Ff
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F
F
FF
f
Ff
Test status of parents
Risk of having a child with CF
No test performed
1 in 2500
Both partners tested: results show one positive, one negative
1 in 600
Both partners tested: results show both positive
1 in 4
Source: Canadian Cystic Fibrosis Foundation,Carrier Testing For Cystic Fibrosis
Ff
f
Ff
ff
Figure 2 The probability of these two
parents conceiving a child who does not
express cystic fibrosis is 75 %.
From Mendel’s Punnett square for a recessive autosomal trait, we know that the
probability of two heterozygous parents (Ff and Ff ) conceiving a child with CF (ff ) is
25 %, a child who is a carrier (Ff ) is 50 %, and a child who does not carry the mutation
(FF) is 25 % (Figure 2). Therefore, if both parents have tested positive for the mutation,
1
the probability of giving birth to a child with CF is , as indicated in Table 1.
4
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research This
Life with a Genetic Condition
not sure there is an
"issue" here.
Skills: Researching, Analyzing the Issue, Communicating,
Evaluating
SKILLS
HANDBOOK
T/k
Cystic fibrosis is a genetic condition that can be managed by a variety of therapies and
lifestyles. In the past people with CF barely lived past childhood. Today, with treatment
and healthy lifestyle choices, many people with CF live well into adulthood. They go to
school, pursue careers, play sports, and can have families. Some individuals with CF even
live to be senior citizens. With advances in genetics, it is quite reasonable to think that a
child born with CF today may live to see a cure for the disorder.
1. Use the Internet or other sources to find stories about people who have cystic fibrosis.
2. Make a list of strategies that individuals with CF use to manage their condition. Research
and record the ways each person’s strategy helps them, as stated by that individual.
A. What similarities do the strategies share?
A
B. Think about how each person’s strategies would fit into your daily life, if you had CF.
Describe in paragraph form how your daily schedule would change. Do you think
some CF strategies could fit easily into your own lifestyle? What changes would you
have to make? What would stay the same? C A
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Breast Cancer
Not all genetic disorders appear right from the onset of life. As you learned in
Section 5.4, Huntington’s disease does not usually show up until an individual is at
least middle-aged. At this point, the potentially harmful allele may have been passed
on to offspring. Cancer genes also fall into this category. Cancer is uncontrolled cell
division.
BRCA1 and BRCA2 are human genes that belong to a class of genes known as
tumour suppressors. Tumour suppressors produce chemicals that inhibit the growth of
tumours in the body. If a woman carries a defective copy of either of these genes, she
has an increased chance of developing breast or ovarian cancer at an early age.
Not all breast cancers are caused by a mutation in the BRCA1 or BRCA2 gene. A
woman has a 12 % chance of developing breast cancer if she does not carry a copy of
one of the defective genes, but a chance of 60 % if she is a carrier of even a single copy
shouldn't
say "some"
of either the defective BRCA1 or BRCA2
genes. Thwe
e environment
plays a significant
lawn
role in determining whether cancer genes
are pesticides?
turned on or off. For example, a woman
who is exposed to estrogen-mimicking chemicals found in lawn pesticides and certain types of plastics has a greater chance of developing breast cancer.
A woman who has a family history of breast cancer may choose to undergo genetic
screening for the mutated BRCA1 or BRCA2 gene. If she does have a mutation, she
may decide to have a mastectomy, a procedure that involves the removal of part or
all of the breast tissue. Alternatively, she may choose to increase surveillance of her
breast and ovarian health.
cancer a disease in which cells start to
divide uncontrollably, resulting in tumours
tumour suppressors a class of human
genes that produce chemicals that inhibit
the growth of tumours
web LInk
For more information on estrogenleaching plastics,
go t o n els on s c i en c e
phenylketonuria
Breast cancer and cystic fibrosis are potentially fatal conditions caused by a mutation
in a gene. In contrast, individuals with phenylketonuria (PKU) are able to live normal
lives with modifications to their lifestyle.
Phenylketonuria is a genetic disorder in which the affected individual is unable to
utilize the amino acid phenylalanine, which the body needs in order to make proteins and essential brain chemicals. In Canada, about 1 in 20 000 people have PKU.
Individuals with PKU do not possess the enzyme that breaks down phenylalanine
5.5 Genetic Disorders
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phenylketonuria (PKU) an autosomal,
recessive, inherited genetic disorder
that results in the accumulation of
phenylalanine in the tissues and blood
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25
8/31/10 4:34:51 PM
(found in most foods), and it accumulates in the blood and tissue. The brain function
of a child with PKU does not develop normally.
Phenylketonuria is a recessive trait. Both parents must be carriers of the defective
allele in order to pass it down to the child. Unlike for cystic fibrosis, researchers have
not isolated the phenylketonuria gene. Hence prospective parents cannot be counselled or tested for the probability of producing a PKU child. Instead, every newborn
is screened for the presence of PKU.
When a baby is about three days old, a few drops of its blood are tested for elevated
amounts of phenylalanine and other compounds that are associated with genetic disorders. If the test is positive, more tests are performed to confirm or rule out PKU. If the
infant does suffer from PKU, he or she must follow a diet that severely limits intake of
high protein foods such as meat, fish, eggs, and milk. If the diet is followed, the child
will be able to lead a normal life. If PKU is left untreated, the child will suffer arrested
mental development, decreased body growth, and poor development of tooth enamel.
Other Disorders
Table 2 summarizes some common genetic disorders. Some of them are discussed
elsewhere in this unit. Researchers are hunting for the genetic mutations that cause
these disorders and are working on developing tests to screen for them.
Table 2 Genetic Disorders and Their Mechanism of Inheritance
and pancreatic duct
Adverse health effects
alkaptonuria
the accumulation of alkapton in the body
kidney stones, damage to
cartilage
recessive allele
cystic fibrosis
causes the body to produce thick, sticky
mucus that clogs the lungs
infections; blocks the
release of enzymes from
the pancreas
recessive allele
galactosemia
the inability to digest galactose
infants experience
jaundice, failure to thrive,
vomiting, and diarrhea;
may lead to death if
undiagnosed
recessive allele
hemophilia
body cannot form blood clots
excessive bleeding
recessive allele
Lesch–Nyhan syndrome
buildup of uric acid in the body
gout, kidney problems,
self-injuring behaviour
recessive allele
phenylketonuria
accumulation of phenylalanine in blood
poor mental development
and growth, weak tooth
enamel
recessive allele
Tay–Sachs disease
nerve cells in the brain are affected by
the accumulation of gangliosides
deterioration of muscle
and physical abilities
recessive allele
Huntington’s disease
progressive, irreversible degeneration
of nervous system
loss of muscle control and
cognitive abilities
dominant allele
hypercholesterolemia
high levels of cholesterol accumulate in
the blood
premature heart disease
dominant allele
neurofibromatosis
nerve cells grow tumours
tumours may be harmless
or may cause damage by
pushing on other nerves
dominant allele
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Mechanism of
Inheritance
Disorder
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Ontario’s Newborn Screening Program
In Ontario, newborns are currently screened for at least 28 genetic disorders. These
disorders include cystic fibrosis, a number of metabolic disorders such as PKU, blood
disorders including sickle-cell anemia, and disorders of the endocrine system. Even
babies from families with no known history of these rare disorders are at risk.
Each year more than 140 000 newborns are tested, and of these, approximately
150 test positive for one of these rare genetic disorders. A positive result indicates the
baby may have a genetic disorder. Only further genetic testing will confirm the presence or absence of a disease. Babies with one of these disorders often appear normal
at birth but without early diagnosis may suffer irreversible damage and even death.
Early diagnosis enables doctors to begin treatments that may eliminate or reduce
serious health conse-quences and even prevent infant death.
5.5 Summary
• Cystic fibrosis is a common, fatal disease passed on through a recessive
mutated gene.
typo? I don't know what a genetic
• Genetic counsellors construct pedigree charts and interpret results of genetic
background is.
tests for individuals who do not have a genetic background.
• Genetic screening is a means by which the presence of a mutated copy of a
gene can be detected.
• Individuals who have the BRCA1 or BRCA2 gene have a higher chance of
developing breast cancer.
• Ontario’s newborn screening program is used to test for many genetic
disorders, including phenylketonuria and cystic fibrosis.
5.5 Questions
1. What factors may shorten the lives of people who have
cystic fibrosis? K/U
2. Some people with cystic fibrosis find that athletic exercise
helps their condition and keeps them in better health.
Research and compare the stories of some people with CF
C
A
who have become notable athletes. 3. Why is it important for infants to be genetically screened
for PKU (phenylketonuria) at birth? K/U
4. Choose one of the disorders listed in Table 2. Use the
Internet to research the following aspects of the disorder: A
(a) symptoms
(b) frequency of the disorder in the general population or a
specific population
(c) research currently being conducted
(d) existence of any genetic tests or screening mechanisms
available for the disorder
(e) treatment of the disorder
5. Does the presence of a defective BRCA1 or BRCA2 gene in
a woman’s genes guarantee that she will develop breast
cancer? Why or why not? K/U
6. Using your guidance department, the Internet and other
sources, research the educational background an individual
C
must have in order to become a genetic counsellor. A
7. Conduct research online to learn more about the Newborn
K/U
Screening Ontario (NSO) program. (a) Where in Ontario are the actual genetic tests conducted?
(b) In addition to helping the newborn, for what other purposes
can NSO results be used?
(c) How old is the baby when the blood sample is taken?
(d) The NSO program is mandatory for all newborns. Why do
you think this policy was put in place?
5.5 Genetic Disorders 27
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5.6
Explore an Issue in Genetic Screening
Skills Menu
•Defining the
Issue
•Researching
•Identifying
Alternatives
•Analyzing the
Issue
•Defending a
Decision
•Communicating
•Evaluating
Who Decides? Who Wants to Know?
Anyone could be a carrier of a genetic disorder. Every person has alleles that produce
defective proteins, but most of these defective proteins do not harm us. Depending
on their family history, some people might want to use gene testing to find out if
there is anything in their DNA that might turn up later in life as a genetic disorder.
The results of genetic tests might cause people to make life-altering changes, so the
decision to have a predictive gene test can be very complicated.
A negative gene test may bring relief. For example, consider individuals at risk
of developing Huntington’s disease. If they find out that they do not carry the
Huntington gene, they will be free of the worry of developing the disease or passing
it on to their children. However, if a test for Huntington’s disease turns out positive,
they may be in a better position to plan for the remaining years of their life financially, socially, and emotionally.
In our society now, the decision to have a genetic test is personal. But what if genetic testing was prescribed and law enforcement agencies, employers, schools, and
adoption agencies all had access to medical files on a “need to know” basis? What if
notinsurance,
sure weoruse
scenarios
- we
just offer
you could be turned down forI'm
a job,
adoption
based on
a genetic
test?
How much information should
readily
available Maybe
about an individual’s
genetic
proa be
few
examples.
replace with
issues
file, and who should have access
to
it?
Does
the
average
person
understand
enough
or "factors" or "situations"
about basic genetic principles to know how to interpret and apply information about
someone’s DNA? In this activity, you will explore different scenarios involving privacy and access to genetic information issues.
At present in Canada, no regulation exists on the use of genetic information for
purposes other than criminal investigations. Yet, in the future it is plausible that an
individual’s genetic information could be used for other purposes. What if DNA
testing was required for a marriage licence? If one of the partners is carrying a lethal
allele such as the Huntington gene, should the other partner be told? The questions
posed here may seem farfetched in today’s society. Yet society will be asked to consider such privacy questions in the future.
The Issue
The ethics of legal access to and ownership of DNA information needs to be considered carefully as technology for collecting and testing DNA for genetic disorders
increases. Who decides? Who wants to know? Who should know? And who should
be required by law to tell?
You are part of a committee that has been asked to compile a list of guidelines for
a government task force that will eventually use your guidelines to compose “Access
to Genetic Information” legislation. The committee consists of the following people:
• a person whose parent has Huntington’s disease but does not want to be tested
himself
• a person whose future mother-in-law has Parkinson’s disease
• an airline executive who is advocating for the genetic testing of all current and
future pilots
• the president of a firm that conducts genetic tests
• a physician who advocates for genetic testing on the basis of preventive
medicine
• half of a couple who wants to adopt a child
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Goal
To create a set of guidelines and recommendations for access to genetic information
Research
Work in pairs or in small groups to learn more about genetic screening in Canada.
Research these questions:
• Which genetic disorders can be treated early if detected by genetic screening?
web Link
• Compile a list of companies in Canada that offer genetic testing. What tests
are offered and at what cost? Does medical insurance cover the costs of those
To start your research into these
tests?
questions,
typo?
Why to
would you test parents or a
• What guidelines or laws currently exist in Canada with respect
to access
go t o n els on s c i en c e
pregnant woman? Did we
mean children,
genetic information? Identify Solutions
You may wish to consider the following issues:
fetuses? We already said that the newborn
testing program "is" the law in Ontario so
perhaps we should cut this?
• Parenting: Should parents be able to test their children for any genetic
disorder regardless of evidence of risk? Should parents, or pregnant women,
be required by law to be tested for genetic disorders?
• Relationships: Should people have access to a potential partner’s genetic
information? What if your partner is going to develop Huntington’s disease?
Should you have the legal right to know? If you are carrying a lethal gene,
should the law require you to disclose this information to your prospective
spouse?
• Employment: Should prospective employers have access to genetic
information about an applicant? If so, how much of this information should
be used to determine eligibility for employment? For example, should an
airline be able to screen its pilots for a gene that may cause heart attacks?
What if the test is later proven to be less accurate than originally thought?
Make a Decision
Create a list of recommendations or guidelines that your committee recommends with
respect to the issue. Provide rationale for each guideline and identify any stakeholders.
Communicate
In an oral presentation, present your committee’s recommendations to the government task force. Your recommendations should be supported by your rationale as
well as research. Remember that your audience is made up of individuals who may
Learning Tip
Making Decisions as a Group
Before you come to any decisions as
a group, make a list on chart paper
of pros and cons for the question at
hand. Place the chart paper where
all group members can see and have
access to it. This can be helpful in
forming new thoughts and ideas
about the discussion at hand. During
a brainstorming session like this one,
take an accepting attitude of all the
ideas presented to the group. Wait
until the page is full to critique the
ideas.
Plan for Action
Tests for screening for prostate cancer and breast cancer are
considered one of the triumphs of genetic testing. The Prostate
Cancer Research Foundation of Canada and the Canadian
Breast Cancer Foundation both recommend early screening.
• Screening for breast cancer and prostate cancer can
result in fewer deaths. Find out what each foundation
recommends with respect to age and frequency of
screening.
Choose a type of cancer. Design and produce a pamphlet that
could be used to educate people on the importance of screening
for that specific cancer. Consider making a number of copies
of the pamphlet and distributing them to your parents, aunts,
uncles, and other people in your life who would benefit from the
information you have compiled.
• What tests are conducted to screen for each of these
cancers? How do the tests work? How effective are they
in preventing deaths?
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5.7
Multi-trait Inheritance
As you have learned, Mendel’s monohybrid crosses were based on one characteristic
controlled by one gene such as height (tall or dwarf) or seed shape (round or wrinkled) in a pea plant. As far as we currently know, genetic disorders such as cystic
fibrosis and Huntington’s disease each involve only one gene. What happens when
more than one characteristic is involved in a cross? Gregor Mendel asked such a question and then conducted ex-periments to determine the answer.
Dihybrid Crosses and the Law
of Independent Assortment
dihybrid cross a cross that involves two
genes, each consisting of heterozygous
alleles
Mendel focused on two characteristics: seed shape and seed colour. Recall that the
dominant allele for seed shape is round (R) while the recessive allele is wrinkled (r). The
dominant allele for seed colour is yellow (Y), while the recessive allele is green (y). Mendel
crossed two individuals that were heterozygous for seed shape (Rr) and seed colour (Yy).
A cross between two individuals for two pairs of heterozygous alleles is called a dihybrid
cross. In this case the individuals were both RrYy. Figure 1 shows how Mendel used
true-breeding parent plants to produce plants (the F1 generation) that were heterozygous for these two traits. He then used these F1 plants to perform a dihybrid cross.
C05-F37-OB11USB
round, yellow
RRYY
parent genotypes
alleles in
gametes
Since this is all about dihybrid
crosses I think the caption should
be about the fact that this is how
Mendel produced the dihybrids for
the cross. This is in the text but not
mentioned at all in the very long
caption.
RY
RY
ry
RrYy
RrYy
ry
RrYy
RrYy
rryy
wrinkle,
green
Figure 1 Mendel’s parent phenotypes were true-breeding varieties. One parent plant was
homozygous dominant round and yellow seed (RRYY ). The other parent plant was homozygous
recessive wrinkled and green seed (rryy ). The gametes of one parent are RY,OUCH
and the gametes
- not ofclose on this
the other parent are ry. The genotypes of the offspring are heterozygous for both
traits
(RrYy
), so
one! Where
did this come
the F1 phenotypes are all round yellow seeds.
Why does the insertion
say "on the next page"
when it is right here?
from?
on the next page
Here is a crude replacement
version. It is missing the "arrow
heads" which are needed.
law of independent assortment if genes
are located on separate chromosomes,
It might work
to colour
they willbetter
be inherited
independently of
code each
letter
and
keep all
each other
A heterozygous individual for two characteristics will produce four possible gametes. A parent that is RrYy can generate the possible gametes RY,Not
Ry, rY,
andwhere
ry. You to begin sure
will see in Figure 2 that the alleles of the two genes—R and r, and Y and y—separate
there is ONE parent independently during the formation of the gametes. This is called Mendel’s law of indecontaining all four alleles in
pendent assortment. Each allele is independent of the other, and no two alleles are linked
everyfour
cell.
to each other. The alleles are found on different chromosomes, creating
diffNO
erentneed for four
spheres
of
different
colours.
gametes in different combinations of the four alleles (R, r, Y, y).
R
the arrows black. Ontario Biology 11 U SB
0176504311
FN
CO
C05-F37-OB11USB
Crowle Art Group
R
Y
R
r
y
Y
r
VERY misleading.
y
parent
Y
There are FOUR possible
gametes - each with two
letters - one of each type.
gametes
r
y
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30
Figure 2 A pea plant that is heterozygous for both seed shape and seed colour
can produce
four is drawn
They
way this
2nd pass
Pass
different possible gametes. The four gametes encompass all the possible combinations.
There
students willarethink that each
Approved
four possible combinations (RY, Ry, rY, and ry), and all are equally likely.
lower letter represents a
Not
Chapter
5 •Approved
Mendelian Genetics—Patterns of Inheritance
nel
7380_Ch05_pp002-045.indd 30
gamete!
posted to 2nd pass folder 9-1-10
8/31/10 4:34:52 PM
Mendel crossed numerous heterozygous F1 generation plants. He then counted
the F2 offspring and noticed that the characteristics were not linked. He noted that
although the original parents (or “grandparents” of the F2 generation) were either
round yellow or wrinkled green, the F2 offspring characteristics were of every combination possible, including round green and wrinkled yellow. It appeared that the
inheritance of seed shape had no influence over the inheritance of seed colour. Four
distinct combinations of seeds were produced in a 9:3:3:1 ratio (Figure 3). The alleles
that control these two characteristics assort themselves independently.
C05-F39-OB11USB
Cross: Rr Yy Rr Yy
Note that if we stick with the horrid heavy
bold lines then we would MUST add two
more lines at top and side since these
lines are separating gametes.
Gametes (pollen)
RY
Ry
rY
ry
RR YY
RR Yy
Rr YY
Rr Yy
RY
The second "usual" genotype
(matching this punnett square
pattern) would be
GGRR, GGRr . . .
Ry
Gametes
(eggs)
RR Yy
RR yy
Rr Yy
Rr yy
Rr YY
Rr Yy
rr YY
rr Yy
Rr Yy
Rr yy
rr Yy
rr yy
rY
LEARNING TIP
But I think I would only list the first
Gene. Order
one
ry
When completing the Punnett square
for a dihybrid cross, keep the genes
in alphabetical order and write the
dominant alleles (capital letters) first,
for example GGRR, GgR.
Figure 3 A summary of the dihybrid cross—9 round yellow seeds, 3 round green, 3 wrinkled
yellow, and 1 wrinkled green. A dihybrid cross results in a phenotypic ratio of 9:3:3:1.
Tutorial 1 Solving Dihybrid Cross Problems
use version based on my sample
A dihybrid cross is an extension of a monohybrid cross. It involves two genes and up to four alleles.
Sample Problem 1: Determining the Phenotypic Ratio
In watermelons, the green colour gene (G) is dominant over
the striped colour gene (g), and round shape (R) is dominant
over long shape (r). A heterozygous round green colour (GgRr)
watermelon plant is crossed with another heterozygous round
green colour (GgRr) plant. Determine the expected phenotypic
ratio of the F1 generation.
Ontario Biology 11 U SB
Step 1. Determine the possible gametes from each parent.
0176504311
Each parent is heterozygous for green colour and round
C05-F39-OB11USB
FN
shape (GgRr). The alleles assort independently of each
other. Four different gametes are produced (Figure 4).
Crowle Art Group
CO
Pass
Approved
Not Approved
G
R
G
g
R
r
G
r
g
R
Figure 4
parent
g
r
gametes
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How is a student
expected to see
FOUR gametes
here??
2nd pass
5.7 Multi-trait Inheritance
NEL
7380_Ch05_pp002-045.indd 31
posted to 2nd pass folder 9-1-10
31
9/1/10 11:09:28 AM
include a "sampling" of
arrows to help guide students
Step 2. Draw a Punnett square of the dihybrid cross.
Since there are four possible gametes from each parent,
draw a 4 3 4 grid, with the possible gametes from one
parent across the top and the possible gametes from the
other parent down the left-hand side (Figure 5).
GR
Gr
gR
gr
GR
Gr
gR
gr
GR
GGRR
GGRr
GgRR
GgRr
Gr
GGRr
GGrr
GgRr
Ggrr
gR
GgRR
GgRr
ggRR
ggRr
gr
GgRr
Ggrr
ggRr
ggrr
GR
Gr
gR
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Figure 6
Step 4. Compile a list of plants that have the same phenotypes.
gr
Start with plants that will exhibit both dominant
characteristics, followed by plants that are dominant
for only one of the characteristics, and then finally with
plants that are recessive for both characteristics.
C05-F41-OB11USB
Figure 5
green and round: GGRR, GGRr, GgRR, GgRr, GGRr, GgRr,
GgRR, GgRr, GgRr (underlined in Figure 6)
Step 3. Execute the cross.
Be careful when combining alleles. Work systematically
across a row while keeping track of which four alleles
will go into each square. Pair up the alleles for the same
gene, as shown in Figure 6.
total 5 9
green colour and long: GGrr, Ggrr, Ggrr
total 5 3
striped and round: ggRR, ggRr, ggRr
Ontario Biology 11 U SB
0176504311
C05-F42-OB11USB
FN
Crowle Art Group
CO
Pass
Approved
Not Approved
U SB
-F41-OB11USB
wle Art Group
pass
total 5 3
striped and long: ggrr
total 5 1
Ditto - ratio
these
all must
be
Therefore, the phenotypic
is 9:3:3:1
for a dihybrid
replaced. Same idea as
cross.
2nd pass
before.
Sample Problem 2: Examining a Homozygous Cross
Assume that in guinea pigs, black fur (B) is dominant over white
fur (b), and a rough coat (R) is dominant over a smooth coat (r). If
a black, rough-furred guinea pig that is homozygous dominant for
both traits (BBRR) is crossed with a white, smooth-furred guinea
pig (bbrr), what are the expected phenotypes in a large litter?
Step 1. Determine what the gametes will be from each parent.
Parent One is homozygous dominant for black and rough
fur (BBRR). The only gamete that this individual can
produce is BR (Figure 7(a)). Parent Two is homozygous
recessive for white colour and smooth fur (rrbb). The
only gamete that this individual can produce is br
(Figure 7(b)).
32
B
B
(a)
R
B
b
Chapter 5 • Mendelian Genetics—Patterns of Inheritance
7380_Ch05_pp002-045.indd 32
R
R
R
B
R
parent 1
B
R
gametes
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b
b
r
(b)
Figure 7
B
posted to 2nd pass folder 9-1-10
b
r
r
r
b
r
parent 2
b
r
gametes
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Step 2. Cross the gametes using a Punnett square.
All of the offspring will be heterozygous for black and
rough fur in any size litter.
BR
br
BbRr
Figure 8
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Practice
1. In some breeds of dogs, a dominant allele controls the characteristic of barking (B) while on a
scent trail. The allele for non-barking trailing dogs is (b). In these dogs an independent gene
(E ) produces erect ears that is dominant over drooping ears (e). For each of the following
mating situations, calculate the phenotypic ratio of the offspring:
(a)A non-barking trailer with erect ears (heterozygous) is mated with a heterozygous barking
trailer with drooping ears (bbEe 3 Bbee).
(b)A non-barking trailer with drooping ears is mated with a heterozygous barking trailer with
drooping ears (bbee 3 Bbee).
(c)A heterozygous barking trailer with heterozygous erect ears is mated with a heterozygous
barking trailer with heterozygous erect ears (BbEe 3 BbEe).
(d)A heterozygous barking trailer with heterozygous erect ears is mated with a non-barking
trailer with drooping ears (BbEe 3 bbee).
Probability
Genetic ratios are probabilities. Recall that if a cross occurs between two heterozygous
individuals where one of the alleles is dominant, such as widow’s peak (Ww 3 Ww),
the phenotypic ratio is 3:1. The total number of possible events is four, and the prob1
ability of producing an individual without a widow’s peak (ww) is , or 25 %. Similarly
3 4
the probability of producing an individual with a widow’s peak is , or 75 %.
4
U SB
Many
THE PRODUCT LAW
5-F44-OB11USB
owle Art Group
d pass
Since only one type of gamete is supplied by each
parent, only a 1 3 1 grid is needed (Figure 8).
Practice Makes Perfect
To take a dihybrid cross quiz,
go t o n els on s c i en c e
Genetic events occur independently of one another. A couple has a 50 % chance of
producing a boy for their first child. The probability of having a boy as a second child is
also 50 %, because the sex of the second offspring is not affected by the sex of the first.
When two events are independent of each other, the probability that both events
will occur can be calculated using the product law. The product law states that the
probability of two or more outcomes occurring is equal to the product of their indi1
1
1
vidual probabilities. The probability of giving birth to two boys is 3 5 . Therefore,
2
2
4
there is a 25 % (50 % 3 50 % 5 25 %) chance of having two boys or, conversely, two
girls.
Recall that in a dihybrid cross the phenotypic ratio is 9:3:3:1. Consider Mendel’s
work with heterozygous yellow seeds crossed with heterozygous yellow seeds
9
(RrYy 3 RrYy). The probability of producing a round yellow seed is
(about 56 %).
16
You know that the alleles of seed colour and seed shape are independent of each other.
If a cross is conducted between two heterozygous yellow seed plants (Yy), the prob3
ability of producing a yellow seed plant is (75%). If a cross is conducted between
4
two heterozygous round seed plants, the probability of producing a round plant is
product law the probability of two
random events occurring simultaneously is
the product of the individual probabilities
of each event
5.7 Multi-trait Inheritance 33
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7380_Ch05_pp002-045.indd 33
web Link
posted to 2nd pass folder 9-1-10
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3
(75%).Using the product rule, the probability of producing a round yellow seed plant
4
3
3
9
from a cross between two heterozygous round yellow seed plants is 3 5 .
4
4
16
The probability predicted by the product law is the same probability that is predicted by a dihybrid Punnett square. The product law also holds true for the other
three probabilities in a dihybrid cross (Table 1).
Table 1 Using the Product Law to Determine Probabilities of a Dihybrid Cross When Both Genes are
Heterozygous
Learning Tip
Sample Size
Note that Mendel observed these
ratios only because of his large
sample size. He crossed hundreds
of plants, producing thousands of
offspring. As sample size increases,
the closer the experimental (actual)
probability gets to theoretical
(expected) probability.
discontinous variation when the
expression of the products of one gene
has no bearing on the expression of the
product of a second gene
continuous variation when the product
of one gene is affected by the product
of another gene, the gene products may
be additive, or one product may negate
another product
additive allele the product of the allele
comprises one part of the total for a
phenotype
Round or wrinkled
seed: probability from
monohybrid cross
Yellow or green seed:
probability from
monohybrid cross
Round seed (Rr 3 Rr )
3
5 75 %
4
Yellow seed (Yy 3 Yy )
3
5 75 %
4
round, yellow seed (RrYy 3 RrYy )
3
3
9
3 5
4
4
16
75 % 3 75 % 5 56.25 %
round seed (Rr 3 Rr)
3
5 75 %
4
green seed (Yy 3 Yy)
1
5 25 %
4
round, green seed (RrYy 3 RrYy)
3
1
3
3 5
4
4
16
75 % 3 25 % 5 18.75 %
wrinkled seed (Rr 3 Rr)
1
5 25 %
4
yellow seed (Yy 3 Yy)
3
5 75 %
4
wrinkled, yellow seed (RrYy 3 RrYy)
1
3
3
3 5
4
4
16
25 % 3 75 % 5 18.75 %
wrinkled seed (Rr 3 Rr)
1
5 25 %
4
green seed (Yy 3 Yy)
1
5 25 %
4
round, green seed (RrYy 3 RrYy)
1
1
1
3 5
4
4
16
25 % 3 25 % 5 6.25 %
Discontinuous versus Continuous Variation
In Mendel’s work with pea plants, the genes that control two characteristics did not
interact with each other. This is the result of discontinuous variation. Pea plants were tall
or short, and seeds were yellow or green. There were no in-between values.
As you no doubt have noticed, there are many examples in nature where this is
not the case. Continuous variation is when the phenotypic variation is not clear cut. For
example, in the general population there are many variations of skin colour, from
pale white to dark black. This is because skin colour is not controlled by one gene, but
rather by three or more separately inherited genes from the father and mother. The
genes are on different autosomal chromosomes and their interaction is additive. An
additive allele contributes a set amount to a phenotype and is an example of continuous variation. The colour of a person’s skin varies depending on which combination
of six alleles he or she inherit from parents. Each allele makes its own contribution
and exhibits incomplete dominance.
Other alleles may also play a role. For example, alleles for freckles and red hair
have a role in determining skin colour. Other traits that are under the control of additive alleles are height, hair colour, and eye colour. Continuous variation explains the
substantial variation of phenotypes found in nature.
34 Chapter 5 • Mendelian Genetics—Patterns of Inheritance
7380_Ch05_pp002-045.indd 34
Dihybrid cross probability
See Over matter
posted to 2nd pass folder 9-1-10
NEL
8/31/10 4:34:56 PM
5.7 Summary
• Dihybrid crosses are crosses between individuals who differ in two pairs of
alleles; if the individuals are heterozygous for both alleles, the phenotypic
ratio of the offspring is 9:3:3:1.
• Mendel’s law of independent assortment states that alleles of different genes
separate into gametes independently of each other.
• Probability is a measure of chance. As sample size increases, the actual
probability approaches the expected probability.
• The probability of two independent events occurring at the same time may be
calculated using the product law.
• Punnett square ratios are one way to show probability.
• Discontinous variation occurs when a trait is either expressed or is not. There
is no in-between trait.
• Continuous variation occurs in nature when the expression of a characteristic
is a sum of the expression of all alleles involved. Skin colour and height are
examples of continuous variation.
ok - but best to avoid asking section
based on info from an
5.7 questions
Questions
investigation that might not have been
1. State Mendel’s law of independent assortment. Provide an
completed.
example that illustrates the law. K/U
2. List the possible gametes from the following individuals: (a) a widow-peaked (heterozygous), attached-earlobe individual
(b) a free-lobed (heterozygous), albino individual
(c) a dwarf, yellow seed colour (homozygous) pea plant
K/U
3. Assume that curly hair (C ) is dominant to straight hair.
Albinism (P ) is recessive to normal skin pigmentation. A
woman who is heterozygous for curly hair and albinism
has a child. The father is homozygous dominant for curly
hair and has albinism. Determine the possible phenotypes
for their child. What is the probability that the child will be
male for each phenotype? T/I
typo 4.
- reword.
sure in
the
intent.
You
What is thenot
probability
Question
3 that
the are
child not
will
a male express
"for" aalbinism
phenotype.
and have curly hair like his father? T/I
5. In your own words, differentiate between continuous and
Wording
if very difficult
discontinuous
variation.here
K/U - to be perfectly
clear.
Do
you
mean?
6. Assume that in horses, black coat colour is dominant (B) to
chestnut colour (b). The trotting gait is due to a dominant
Calculate
different
of a(t). T/I
allelethe
(T ), four
and the
pacing gateprobabilities
is a recessive allele
child being
both
a trotting
male horses
and of
(a) If two
black
areeach
mated and have four
offspring, all of which are black and pacing, what does
phenotype?
the literacy folks might not like this.
An ELL student might not know the
"horse" meaning of trot, pace, stud,
mare,
or even
colt.
that reveal about the probable genotypes
of the parent
horses?
(b) A stud is black and trotting. He is mated with a mare
who is also black and trotting. Their colt is black and
pacing. The same stud is mated to a second mare
who is chestnut and pacing. Their colt is chestnut and
trotting. What is the probable genotype(s) of the stud,
Mare 1, Colt 1, Mare 2, and Colt 2?
7. Design an experiment in which you could determine
whether height in tomato plants is governed by continuous
variation or discontinuous variation. T/I
8. The alleles for human blood types A and B are codominant,
but both are dominant over the type O allele. The Rh factor
is separate from the ABO blood group and is located on a
separate chromosome. The Rh+ allele is dominant over the
Rh–. Indicate the possible phenotypes from mating a woman
with type O, Rh–, with a man with type A, Rh+. The man is
homozygous for type A and heterozygous for Rh factor. T/I
Don't they have to be if they are
9. Why is sample part
size important
in a geneticcross?
investigation? A
of a trihybrid
10. Is it possible to have a trihybrid cross? If so, how many
possible gametes would be formed if the parents were
heterozygous for each of the three characteristics? T/I
This is quite challenging given the examples
we have provided. I think we need to at least
begin with a much easier question.
OM34 Over matter
7380_Ch05_pp002-045.indd 34
posted to 2nd pass folder 9-1-10
8/31/10 4:34:56 PM
CHAPTER
5
Investigations
Investigation 5.2.1
OBSERVATIONAL STUDY
Gummy Bear Genetics
Mendel derived his laws of inheritance by counting the
number of different types of offspring that resulted from
different crosses of peas, and then analyzing the results.
Using a large sample size provided Mendel with enough
data to see trends. The trends in the data led to the
development of Mendel’s law of inheritance.
In this investigation you will be given a sample of
“offspring” from a Mendelian monohybrid cross. By
sorting and counting the number of offspring, you will
be able to determine the genotypes of the parents that
produced the offspring.
purpose
To determine the genotypes of parents by counting and
sorting the phenotype ratios of the offspring
equipment and Materials
• alabelledbagcontaininggummybears
Caution: This is a laboratory procedure. Food is not to be
eaten in the laboratory.
procedure
SKILLS
HANDBOOK
T/k
1. Copy Table 1 into your notebook.
Table 1 Phenotype Ratios of Gummy Bears in Individual Bags
Total
number
Phenotypes
Monohybrid of
Phenotype
Parental
gummy and
cross bag
numbers
ratio
Genotypes cross
bears
label
A
B
•Questioning
•Researching
•Hypothesizing
•Predicting
•Planning
•Controlling
Variables
•Performing
•Observing
•Analyzing
•Evaluating
•Communicating
Analyze and evaluate
(a) For each bag, calculate the phenotype ratio of bears
in the bag. Record this ratio in Table 1. T/I
(b) Determine which bear colour is dominant and which
bear colour is recessive. Use T to represent the
dominant allele and t to represent the recessive allele.
Hint: look at the phenotype ratio of Bag C. T/I
(c) Use the ratios and your knowledge of Mendelian
genetics to determine the likely genotype(s) of
each type of bear found in the bags. Record these
genotypes in Table 1. T/I
(d) Determine the genotypes of the parents that
produced the offspring in each of the bags. Record
this information in Table 1. T/I
(e) For each bag, draw a Punnett square to show that the
cross of the genotypes you predict for the parents will
produce the ratio that you observed. T/I
Apply and extend
(f) Why is having a large number of gummy bears in
each bag important? T/I
(g) Pool your class data in Table 2. T/I
Table 2 Phenotype Ratios of Gummy Bears—Class Results
Total
number
Phenotypes
Monohybrid of
Phenotype
Parental
gummy and
cross bag
numbers
ratio
Genotypes cross
bears
label
A
C
B
2. Obtain a bag of gummy bears from your teacher and
record the letter label of the bag.
3. Empty the contents of the bag onto your desk. Count
the gummy bears and record the total in Table 1.
4. Sort the gummy bears into groups based on
phenotype differences that can be easily observed
and quantified, such as colour.
5. Count and record the number of bears of each
phenotype in Table 1.
6. Repeat Steps 2 to 5 for each of the different (A, B, C,
. . .) bags available from your teacher.
C
(h) Why are the ratios not exactly the same as Mendel’s
ratios? T/I
(i) What type of inheritance does each of the three
monohybrid crosses represent? T/I
(j) Design a set of bags that would represent the
phenotype ratios found in crosses that involve alleles
that exhibit incomplete dominance. What colour of
gummy bears would you use? How many of each?
What type of cross would you execute? T/I A
5.7
Chapter
Multi-trait
5 Investigations
Inheritance
nel
7380_Ch05_pp002-045.indd 35
SkILLS MenU
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35
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Investigation 5.4.1 OBSERVATIONAL STUDY
Skills Menu
•Planning
•Observing
?? something wrong here - there •Questioning
•Researching
•Controlling
•Analyzing
My Genetic Profile
is NO place to fill in anything
•Hypothesizing
Variables
•Evaluating
•Performing
•Communicating
about genotype in table 1. There •Predicting
The individuals of every population have a variety of
is no provided dominant
phenotypes. These variations are a result of genotypic
genotype.ofI different
would simply
cut #2
variance caused by the inheritance
alleles. The
2. Fill in the missing genotypes for each characteristic
entirely
not
needed.
presence or absence of an allele inherited according to the
in Table 1. The homozygous dominant genotype has
Mendelian laws of inheritance can be followed through
been provided already.
many generations with aOddly
pedigree
chart. as well since
worded
3. Using the information in Table 1 as a guide, draw
In this activity, you will
identify your phenotype
for
characteristics
do not have
and complete a data table similar to Table 2 that lists
six characteristics and use
those
observations
to
try
to
genotypes - individuals with
all 8 characteristics. If, for a particular trait, you do
determine the associated genotypes. Then, you and a
not know whether you are homozygous dominant
specific traits/phenotypes have
partner will determine the phenotypes and genotypes
or heterozygous, simply record your genotype as
genotypes. offspring and
for possible first- and second-generation
heterozygous.
construct a pedigree chart.
Table 2 Determining Your Genotype
Purpose
To determine your phenotype and genotype for a
characteristic
To track allele inheritance through three generations
Characteristic
Phenotype
Possible
genotype(s)
Genotype
dimples
earlobes
Equipment and Materials
• scissors
• beakers
• coloured pencils (green and red)
Characteristic
Possible
Dominant
phenotypes
and
phenotype
associated allele
dimples
present (D)
absent (d )
earlobes
freely hanging (F )
attached (f )
hair on mid-finger
hair present (H )
hair absent (h)
2nd toe longer than big toe
longer (L)
shorter (l )
hairline
widow’s peak (W )
straight (w )
Part B: The F1 Generation
Now you will work with a partner to simulate the passing
on of genetic information to a single offspring. For this
activity, ignore gender.
4. Copy Table 3 into your notebook, with rows for all
8 characteristics.
5. To determine the genetic makeup of the offspring,
you and your partner (Parent 1 and Parent 2) must
each contribute a single allele for each characteristic.
To choose each allele, you must make 8 coin
flips—one for each characteristic. If you flip “heads,”
you contribute the first allele of your genotype; if
you flip “tails,” you contribute the second allele. For
example, if you flip heads for “thumb flexibility”
and you are Tt, then you enter T under your parent
column in Table 3.
6. After you have determined the alleles for each
characteristic, complete the genotype and phenotype
columns in your table.
tongue roll
roller (R)
non-roller (r )
Table 3 Genotype and Phenotype of F1 Generation Offspring
thumb flexibility
bent back (T )
straight (t )
thumb placement with
hands clasped
left thumb on
top (P )
right thumb
on top (p)
Procedure
Part A: Determining Your Genotype
1. Copy Table 1 into your notebook.
simply headings
Table 1 Genotypes for Some Common Human Phenotypes
Recessive
phenotype
Allele
from
Characteristic Parent 1
Allele
from
Parent 2
Genotype
Phenotype
dimples
earlobes
36 Chapter 5 • Mendelian Genetics—Patterns of Inheritance
7380_Ch05_pp002-045.indd 36
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NEL
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Part C: The F2 Generation
Now you will simulate the mating of your offspring with
the offspring of another pair of students.
7. Repeat Step 5 to model the contribution of alleles
from the two F1 offspring and determine the
genotype and phenotype of their F2 offspring—your
“grandchild” in this simulation. Record your data in a
new copy of Table 3 titled “Genotype and Phenotype
of F2 Generation Offspring.”
8. Choose any single trait in your grandchild and draw
a pedigree chart to track its inheritance over the three
generations. Be sure to include all four grandparents
and both parents in the chart.
(c) Look at your table of genotypes and phenotypes. How
many of your 8 traits are recessive? T/I
(d) Look at your partner’s table. How many of his or her
8 traits are recessive? T/I
(e) How did the number of recessive traits in your
offspring compare with that of yours as “parents.” Did
this surprise you? What factors might account for
these differences? T/I
(f) Write a birth announcement for one of your
offspring. List the baby’s characteristics. T/I C
Apply and Extend
(g) Tally the class results and calculate the percentage of
traits that were recessive for the class as a whole. T/I
Analyze and Evaluate represents an event in
(h) Do the same tally for the F1 and F2 generations. How
(a) In this activity, which step(s) represent meiosis? Is
did the percentage of recessive traits compare across
this a true representation? Why or why not? T/I
all three generations. Is there any convincing evidence
(b) Use examples from the activity to explain your
that recessive traits will become more or less common
understanding of the terms “genotype,” “phenotype,”
with time? Explain your reasoning. T/I C A
“recessive,” and “dominant.” T/I C
may be just replace
with . . .
I think we need to delete (or replace) H.
Based on the class data,
do you think dominant
phenotypes are always
more or always less
common than their
corresponding recessive
phenotypes? Use
examples to support your
answer.
Students will ALWAYS get skewed data
because every time a student has a
dominant trait they will have recorded their
genotype as heterozygous. This is
necessary for the activity but creates a data
bias which will result in more recessives in
the F1 than would otherwise be expected.
The F2 data will also have more recessives
than the P1, but will be valid based on the
skewed F1.
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CHAPTER
5
SUMMArY
Summary Questions
2. Look back at the Starting Points questions at the
beginning of the chapter, on page xxx. Use examples
from the chapter to answer the questions again.
Compare your latest answers to your initial answers.
Note how your answers have changed.
1. Look back at the Key Concepts listed at the beginning
of the chapter, on page xxx. Create a graphic organizer
with the word “inheritance” in the middle to organize
the main ideas and concepts presented in this chapter.
vocabulary
trait (p. xx)
homozygous (p. xxx)
codominance (p. xxx)
tumour supressors (p. xxx)
true-breeding organism (p. xxx)
heterozygous (p. xxx)
pedigree (p. xxx)
pheynylketonuria (p. xxx)
hybrid (p. xx)
genotype (p. xxx)
autosomal inheritance (p. xxx)
dihybrid cross (p. xxx)
cross (p. xxx)
phenotype (p. xxx)
sex-linked (p. xxx)
P generation (p. xxx)
dominant allele (p. xxx)
X-linked (p. xxx)
law of independent assortment
(p. xxx)
F1 generation (p. xxx)
recessive allele (p. xxx)
Y-linked (p. xxx)
product law (p. xxx)
monohybrid (p. xxx)
Punnett square (p. xxx)
cystic fibrosis (p. xxx)
discontinuous variation (p. xxx)
monohybrid cross (p. xxx)
probability (p. xxx)
mutation (p. xxx)
continuous variation (p. xxx)
F2 generation (p. xxx)
test cross (p. xxx)
carrier testing (p. xxx)
additive allele (p. xxx)
law of segregation (p. xxx)
complete dominance (p. xxx)
genetic screening (p. xxx)
allele (p. xxx)
incomplete dominance (p. xxx)
cancer (p. xxx)
career pATHWAYS
Grade 11 Biology can lead to a wide range of careers. Some require a college diploma or a B.Sc.
degree. Others require specialized or postgraduate degrees. This graphic organizer shows a few
pathways to careers mentioned in this chapter.
SKILLS
HANDBOOK
T/k
more
careers
1. Select two careers, related to Genetic Processes that youwe
find need
interesting.
Research
the here.
educational pathways that you would need to follow to pursue these careers. What is
breeder,
involved in the required educational programs? Prepare aplant
brief report
of your fihorse
ndings. trainer, market
2. For one of the two careers that you chose above, describegardener,
the career, main
duties and oncologist,
pediatrician,
responsibilities, working conditions, and setting. Also outline
how
the
career
benefits society
insurance adjustor,
demographer
and the environment.
laboratory technician
M.Sc.
B.Sc.
12U Biology
Ph.D.
research geneticist
Canadian Association of Genetic Counsellors Acrcreditation
B.A.
OSSD
11U Biology
registered nurse
B.Sc.N.
college diploma
laboratory technician
dog breeder
38
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is this a
program?
If so it still
needs to
go t o n els on s c i en c e
link to a
career!
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CHAPTER
5 SELF-Quiz
K/U
Knowledge/Understanding T/I
Thinking/Investigation C
Communication A
Application
[QUESTIONS TO COME]
To do an online self-quiz,
go to n els on s c i en c e
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CHAPTER
5 Review
K/U
Knowledge/Understanding T/I
Thinking/Investigation C
Communication A
Application
[QUESTIONS TO COME]
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Chapter 5 Review 41
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