Download Genes and Inheritance

DNA and Inheritance
Biology 30
Genes & Heredity

Your biological traits are
controlled by genes, which are
located on the chromosomes
that are found in every cell of
your body. Only one copy of
each gene is on a chromosome.
Genes are like books; they may
or may not be read by the
chemical machinery of the cell.
• Your traits are a result of the interactions of the
genes from both parents.
• Although you contain half the genes from each
parent, your genes and traits are uniquely your
own.
• There are over eight million combinations
possible from the 23 chromosomes you inherit
from your mom and 23 you inherit from your dad.
• The passing of traits from parents to offspring is
called heredity.
Pioneer of Genetics: Gregor Mendel
Gregor Mendel, an Austrian monk, conducted
some simple but significant experiments in
his monastery garden. (1822-1884).
Mendel’s work with garden peas
not only explained the mechanism
of gene inheritance for plants, but
also provided a basis for
understanding heredity.
Why did Mendel work with the garden pea?
1) Garden peas have a number of
different characteristics that can be
expressed in one of two ways.
(ex. yellow vs. green pea)
2) It was easy for Mendel to go about
his experiments, because of the way the
plant reproduces.
Mendel fertilized
plants by
cross-pollination
rather than
self-fertilization.
Mendel’s Characters
The seven characteristics that Mendel studied
in peas
Mendel’s Experiments
Crossed the pollen from a plant that produced
round seeds with the eggs of the one that
produced wrinkled seeds, the offspring were
round.
Crossed the pollen from a plant that produced
wrinkled seeds with the eggs of the one that
produced round seeds, the offspring were
round.
Mendel then crossed the F1 hybrids
together to produce an F2 generation.
The recessive trait “reappeared”.
F1
F2
He found a 3:1 ratio of dominant
to recessive phenotypes.
3 round : 1 wrinkled
1/
4
homozygous dominant
1/
2
heterozygous
1/
4
homozygous recessive
Rr
r
R
1/
4
homozygous dominant
R
RR
Rr
Rr
1/
4
homozygous recessive
r
Rr
rr
He repeated the procedure for other
characteristics (eg. flower colour).
He discovered that one trait dominated
another, whether the sex cell came from
a female or a male part of the plant.
Mendel reasoned that things called
“factors” controlled the traits of a plant.
These factors were later termed genes.
Dominant genes –
Determine the expression of the genetic
trait in the offspring.
(Capitalize this letter).
Recessive genes –
Are “masked” by dominant genes.
(Lower case letter).
Mendel’s Laws of Heredity
First Law (Law of Unit Characters)
Inheritance is governed by genes that exist
in the individual and are passed on to
offspring. These factors (genes) occur in
pairs, one gene comes from the female and
one gene comes from the male. The
alternate forms of the same genes are known
as alleles.
Second Law (Law of Dominance)
One factor, or gene, masks the effect of
another. This process is known as the
principle of dominance. The dominant
expression is seen and the recessive gene is
not seen (remains hidden).
Third Law (Law of Segregation &
Recombination)
A pair of factors (genes) segregate/separate
during the formation of sex cells (meiosis).
As a result, each parent can only contribute
one member (allele) of a pair of genes to their
offspring.
Genetic Terms

Define the following terms:
Genotype –

Phenotype –

Homozygous (“purebred”) –

Heterozygous –

Alleles –

Review Quiz




Draw a diagram that illustrates the difference
between a genotype & a phenotype. (2 marks)
Using the letter “a,” list a genotype that is
homozygous dominant. (1 mark)
Using the letter “t,” list a genotype that is
heterozygous. (1 mark)
What is an allele? (1 mark)
Monohybrid Cross
The combining of single
contrasting traits:
eg. Crossing a tall plant (TT) with a
short plant (tt)


a special chart called a Punnett
square, helps geneticists organize
the results of a cross between the
sex cells of two individuals

from a Punnett square, we can
predict the genotypes & phenotypes
of the offspring:
Draw a Punnett square for a cross between a
heterozygous round-seed plant and a wrinkledseed plant.
R
r
r
Rr
rr
r
Rr
rr
In the F1 generation ½ are round ½ are wrinkled
Rr
r
R
1/
4
homozygous dominant
R
RR
Rr
Rr
1/
4
homozygous recessive
r
Rr
rr
Draw a Punnett square for a cross between
two heterozygous round-seed plants.
R
r
R
RR
Rr
r
Rr
rr
In F1 generation ¾ are round ¼ wrinkled
Draw a Punnett square for the cross between a
heterozygous purple flower with a white flower.
W
w
w
Ww
ww
w
Ww
ww
In the F1 generation ½ are purple ½ white
Phenotypic Ratio:
50% of flowers are purple
50% of flowers are white
Genotypic Ratio:
50% of flowers are Ww
50% of flowers are ww
Test Cross
A test cross is often performed to determine the
genotype of a dominant phenotype. When would
we need to know this?
There would
eg. A sheep farmer wants to
be some
ensure that all of his flock
offspring
will have white hair (black
produced
that
wool is brittle & hard to
have
white
dye). He chooses a white
wool and
ram to mate with the flock.
some with
What if the ram is
heterozygous?
black.
To ensure that the ram is not, a test cross can be
performed to ensure that the ram is homozygous
for the white phenotype
A test cross is always
performed between the
unknown genotype and a
homozygous recessive
(known) genotype.
Possible outcomes:
If 50% of the offspring are black and the
other half white, the unknown genotype must
have been heterozygous (Ww).
If 100% of the offspring are white, then the
unknown genotype must be homozygous
dominant (WW).
Mendel showed that by performing a
test cross with the homozygous
recessive, the genotype could be
determined.
YY x yy
Yy
Yy x yy
Y
y
Yy
yy
Yy
yy
yy
y
y
This test cross
produces
1/2
dominant and
1/2
recessive
phenotypes
Gene Interactions
There are three main types of gene
interactions:
1) Multiple Alleles
There are many traits that are controlled by
more than two alleles for example, eye colour
in Drosophila is controlled by four possible
alleles.
The following phenotypes and dominance
hierarchy is possible:
wild type (red) > apricot > honey > white
When dealing with multiple alleles, it is no longer
necessary to use upper & lower case letters  both
letters & upper case numbers are used.
phenotype
wild type
Apricot
honey
white
genotype
E1
E2
E3
E4
Predict the genotypes and phenotypes of the F1
generation from the mating of wild type (E1E4) with
apricot (E2E3).
Predict the genotypes and phenotypes of the F1 generation
from the mating of wild type (E1E4) with apricot (E2E3).
E1
E4
E2
E1E2
E2E4
E3
E1E3
E3E4
In the F1 generation ½ are wild type; ¼ apricot; ¼ honey
Do Now
phenotype
genotype
wild type
E1
Apricot
E2
honey
E3
white
E4
With the above information, make two
heterozygous crosses what phenotypes do you get?
2) Incomplete Dominance
In some heterozygotes, both alleles of a pair are
expressed in the phenotype. These alleles are said
to be equally dominant. This lack of dominance is
known as incomplete dominance.
eg.
P1 Black

White
F1
P2 Grey
F2
Grey
×
Grey
Ex. Snapdragons
An interaction between the alleles in the
heterozygote shows an intermediate
phenotype.
WW
Ww
ww
3) Codominance
A form of incomplete dominance where
two alleles are expressed in such a way
that the effect of each is noticed
separately in the phenotype.
Both parental phenotypes can be
distinguished in the heterozygote
offspring.
The expression of one allele does not
mask the expression of another.
Ex. A red bull crossed with a white cow =
roan calf (calf has intermingled white & red hair)
Roan cattle and horses have both
coloured and white hair.
Blue Roan
Strawberry Roan
The ABO blood group system is another
example.
Inheritance of blood groups is determined by
the gene “I” which has three different alleles,
only two of which can occur at the locus at
once. The alleles are responsible for producing
antigens on the surface of the red blood cells,
which determines the blood group.
Alleles A and B are co-dominant so that when
they are both present, both A and B antigens are
produced. Both A and B are dominant to O.
Blood Types (A, B, O, AB)
Allele IA – formation of blood factor A (antigen A)
Allele IB – formation of blood factor B (antigen B)
Allele I – no factors result
Genotype
IAIA or IAi
IBIB or IBi
ii
IAIB
Blood Type
A
B
O
AB
Type A
Type B
Type AB
Type O
IA an IB are codominant and i is
recessive.
Rh antigens are straight dominant
vs. recessive.
Blood types of
North America
Blood types of
North America
with RH.
Example:
A mother with blood type A has a child with Blood type
O. The father is blood type B. Indicate the genotypes
of the parents.
Father can be IBIB or IBi and mother can be IAIA or IAi.
To have a child that is type O (ii) both mother and father
must contribute the i allele, therefore the father must be
IBi and the mother is IAi
Sex Chromosomes
So far, what do you know about sex
chromosomes?
In addition to their role in determining sex, the sex
chromosomes, especially X chromosomes, have genes
for many characters unrelated to sex.
We call these sex-linked alleles.
Female cells can differ from male cells in two ways:
1. Female cells show dark spots of chromatin
(called Barr Bodies) during interphase, male
cells do not.
2. Female cells contain 2 X chromosomes and
males contain only one X chromosome.
The Y chromosome carries few genes.
There are very few genes on the Y
chromosome that are common on the X
chromosome, and because of that, little
crossing over may occur between an X
and a Y.
eg.) Calico cats
Male cats tend to be black (XBY) or orange
(X0Y). Female cats can be black (XBXB),
orange (X0X0) or calico (XBX0) – a mixture
between black and orange.
Very few male cats can be calico, why?
Those who do, carry a hidden X
chromosome, and are likely sterile.
• A male embryo does not differ from a female fetus until
the 6th/7th week of pregnancy.
• At this point, the “testes determining factor” (TDF)
gene on the Y chromosome is activated.
• The TDF gene initiates the production of a protein that
stimulates the testes to begin secreting male hormones.
Examples of sex linked traits.
a.
Hemophilia - lack or deformity of
blood clotting factor VII or IX.
b.
Red Green colorblindness
c.
Pattern baldness - sex influenced
not sex-linked.
i.
Humans carry two alleles for
baldness.
ii.
In females the allele for baldness is
recessive but in males, due to
testosterone, it is dominant.
HB - baldness allele
HN - normal hair
Male
Female
HNHN
Normal
Normal
HBHN
Bald
Normal
HBHB
Bald
Bald
So heterozygote females avoid baldness.
We can also perform monohybrid crosses between
sex chromosomes.
For example:
Brown eye color (B) is dominant to blue (b).
Eye color is carried on the X chromosome.
Homozygous dominant female
Heterozygous female
Homozygous recessive female
Dominant male
Recessive male
XBXB (brown)
XBXb (brown)
XbXb (blue)
XBY (brown)
XbY (blue)
Draw a Punnett square for a cross between a heterozygous
female with a recessive male. Calculate the phenotypic &
genotypic ratios.
XB
Xb
Xb
XBXb
XbXb
Y
XBY
In the F1 generation:
XbY
Phenotypic ratios:
1 brown eyed girl: 1 brown eyed boy: 1 blue
eyed girl: 1 blue eyed boy
Genotypic ratios:
1XBXb: 1XbXb: 1XBY: 1XbY
Example #2
Is it possible to get a blue eyed female from crossing a blue
eyed female with a brown eyed male? Explain.
Xb
Xb
XB
XBXb
XBXb
Y
XbY
XbY
No it is not possible, all females would be browned eyed
Pedigrees
• A chart or register showing a line of ancestors
• Circles represent females, squares represent males,
solid circles & squares represent those who have the
trait being studied.
•Horizontal lines between circles and squares represent
mating between male and female.
• A vertical line joins parents and children.
• Pedigrees are often used to study sex-linked traits
such as color-blindness and hemophilia
PEDIGREE SUMMARY
This will help you determine a trait in a pedigree as
with autosomal or X-linked:
Autosomal Dominant
 must be in each generation
 affected individuals transmit to minimum ½ of their
offspring
 males and females are equally affected
 cannot have carriers, they will all be affected!
Autosomal Recessive
 may skip a generation
 affected offspring generally have normal
(but heterozygous) parents
 male and female are equally affected
X-linked Dominant
 very likely to be observed in each generation
 females pass on to half of either sex
 no transmission from father to son (only
daughters)
X-linked Recessive
 affect males more than females
 no transmission from father to son
 daughters of males are carriers
 females pass onto ½ sons
 affected females have affected fathers and carrier
mothers
Pedigree analysis
Probability
• Probability is the likelihood of an event
happening.
• Probability can be expressed by the
following formula:
Probability = # of chances for an event
# of possible combinations
Therefore, when a coin is tossed, there are
two possibilities – heads or tails.
What are the chances of getting heads?
1 head/2 head/tail =
50 % chance
What is the probability of two coins being
tossed and getting heads?
½ = 50 % chance
Coin 1 –
Coin 2 –
½ = 50 % chance
The Rule of Independent Events –
Chance has no memory. This means that previous
events will not affect future events. Ex. If you
tossed two heads in a row, the probability of
tossing heads again will still be ½.
Product of the probabilities of two separate
events:
½ × ½ = ¼ therefore a 25% chance
The Product Rule –
The probability of two or more independent
events occurring together is the product of the
individual probabilities if each individual event
occurs separately.
Dihybrid Crosses
Mendel also studied
two separate traits
with a single cross by
using the same
procedure he had
used for studying
single traits.
Mendel crossed a purebred yellow round pea
with a purebred green wrinkled pea.
Pure breeding round = RR
Pure breeding wrinkled = rr
Pure breeding yellow = YY
Pure breeding green = yy
Genotype for the yellow, round parent is RRYY
Genotype for the green, wrinkled parent is rryy
P1
YYRR x yyrr
Purebred yellow round × Purebred green wrinkled
The entire F1
Male gametes
generation has the
YR
YR
YR
YR
genotype: YyRr
and is
yr YyRr YyRr YyRr YyRr
phenotypically
yellow
round!
Female
gametes
yr
YyRr YyRr YyRr YyRr
yr
YyRr YyRr YyRr YyRr
yr
YyRr YyRr YyRr YyRr
F1
Now let’s cross the F1 generations with one
another and see what we get…
9/16 Yellow Round
3/16 Yellow wrinkled
3/16 green Round
1/16 green wrinkled
The
phenotypic
ratio 9:3:3:1 is
the ratio you
will find in all
heterozygous
dihybrid
crosses!
Example:
In summer squash, white fruit color is dominant
“W” and yellow fruit color is recessive “w”.
Another allele produces disc shaped fruit “S” while
its recessive allele “s” yields sphere-shaped fruit. If
a pure breeding white disc variety is crossed with a
homozygous yellow sphere variety, the F1 are all
white disc hybrids. If the F1 generation is allowed
to mate, what would be the expected phenotypic
ratio in the F2 generation?
Try this!
P1
WWSS x wwss
F1 All offspring will be WwSs
P2
WwSs x WwSs
WS
Ws
wS
ws
WS WWSS WWSs WwSS WwSs
Ws WWSs WWss WwSs Wwss
wS
WwSS
WwSs wwSS wwSs
ws
WwSs
Wwss wwSs
wwss
F2
Rhesus Factor & Birth
P1 : Female Rh- ×
Male Rh+
Baby is Rh+ because father is. Mother’s
blood produces antibodies upon birth, (since
blood mixes at birth). First baby is okay.
Second pregnancy- mom’s antibodies can
now move across the placenta and cause
baby’s RBC’s to clump (agglutinate) if second
baby is also Rh+. This decreases oxygen
delivery in the baby – “blue baby.”
What can be done?
 Mom can be given an injection of a drug that
inhibits antibody production immediately after
delivery.
What happens if this is undetected?
 Baby could be given a blood transfusion
while in the womb. Fairly uncommon.
Blood Types & Rhesus Factor Question
R – dominant allele (Rh+)
r – recessive allele (Rh-)
Example: A woman homozygous for blood
type A and heterozygous for the rhesus
allele, Rh+, has a child with a man with
type O blood who is Rh-. What is the
probability that their child will have blood
There
will
be
a
type A, Rh+?
50% chance.
Do Now

Biology creatures have many strange traits.
Blinking eyes “B” is dominant and burning
eyes are recessive “b”. Another allele
produces pleasant smelling pheromones “P”
while its recessive allele “p” yields rotting fruit
pheromones. If a pure breed blinking pleasant
smelling biology creature is crossed with a
heterozygous blinking pleasant smelling
biology creature, what would be the expected
phenotypic ratio?
Techniques used in order to produce the
genotype or phenotype that you want:
Selective Breeding: the crossing of desired
traits from plants or animals to produce
offspring with both characteristics
Inbreeding: the process by which
breeding stock is drawn from a limited
number of individuals possessing
desirable phenotypes.
Hybridization: Blending of desirable
but different traits.
Gene Interaction
1) Many traits studied by Mendel were controlled
by one gene.
2) Some traits are regulated by more than one
gene; many of your characteristics are determined
by several pairs of independent genes –
polygenic.
eg.) skin color, eye color and height, feather colour in parakeets
3) One trait controlled by more than one allele
(multiple alleles).
Eg. Blood types, Drosophilia eye colour
4) Genes that interfere with the
expression of other genes are called
epistatic.
Example:
-allele B produces a black coat color in dog
- b produces a brown coat color
-A second gene, W prevents the formation
of pigment
- w does not prevent color
- genotype wwBb would be black
- genotype WwBb would appear white
-W allele masks the effect of the B color gene
- If wwBb is crossed with a WwBb, state the
phenotypes produced.
8/16 = white
6/16 = black
2/16 = brown
5) Complementary Interaction occurs when two
different genotypes interact to produce a phenotype
that neither is capable of producing by itself.
Example:
-Allele R produces a rose comb in chickens
-Allele P (on a different chromosome) produces a
pea comb.
- R and P alleles both present = walnut comb
- The absence of rose and pea alleles results in an
individual with a single comb
Chromosomal Theory
The chromosomal theory is as follows:
i) Chromosomes carry genes, the units of hereditary
ii) Paired chromosomes segregate during meiosis. Each
sex cell or gamete has half the number of
chromosomes found in a somatic cell
iii) Chromosomes sort independently during meiosis.
Each gamete receives one of the pairs and that one
chromosome has no influence on the movement of a
member of another pair
iv) Each chromosome contains many different genes
Chromosome Mapping and Gene Linkage
• A single chromosome contains many genes
linked together and so does the other
chromosome in the homologous pair.
• The sequence of genes on each chromosome
pair should match each other exactly.
• Gene linkage reduces the chance for genetic
recombination and variety among the offspring.
• Parts of a chromosome holding many genes,
may separate and switch places with the
matching part of the other chromosome =
crossing over.
• The closer genes are to each other, the less
likely they will separate during crossing over =
linked genes.
• Scientists use crossover frequencies on genes to
determine their positions on chromosomes
eg.) if the crossover frequency of a gene is
5%, then the two genes are 5 map units
apart.
• Crossover frequency is determined by the
following formula:
crossover % = number of recombinations x 100
total number of offspring
• Gene markers are usually recessive genes that
are easily observed in offspring and can be used
to identify other genes found on the same
chromosome.
• By using crossover frequencies, we can
determine gene maps.
• Gene maps show the relative positions of
genes on a chromosome (loci).
• Gene maps are constructed by:
- ordering fragments of DNA
- studying chromosomal alterations
- performing crosses to see how frequently
crossing over occurs between fragments.
Problem 1: 3 genes A, B, C
AB – 12%
CB – 7%
A
12 map units
C
5 map units
AC – 5%
B
7 map units
Problem 2: AB - 3%
BC - 28%
31 map units
A
B
28 map units
3 map units
AC - 31%
C
Problem 3:
Genes
X
Y
Z
X
X
10
15
Y
10
5
Z
15
5
-
15 map units
10 map units
Z
Y
5 map units
Crossover Frequency of Some Genes on
Chromosome #6
Genes
Cross-over Frequency
Diabetes(1) and Ovarian cancer (2)
21%
Diabetes (1) and RH blood group(3)
12%
Ragweed allergy (4) and RH blood group(3)
10.5
RH blood group(3) and ovarian cancer (2)
9%
Ragweed allergy (4) and ovarian cancer (2)
19.5
Hint: Start here
Transposons (see handout)
Gene Therapy: when defective genes are replaced
with normal genes in order to cure genetic diseases
Human Genome Project: to determine the complete
sequence of the 3 billion DNA subunits (bases),
identify all human genes, and make them accessible
for further biological study.