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Transcript
Biology 11 Enriched
Tuesday, December 4th,2 012

Human have observed changes in population
for many centuries.
 Humans have domesticated animals and plants
and have artificially selected for various traits in
our livestock and crops.


Character – an observable physical feature
Trait – a particular form of a character

In the 19th Century, 2 theories on how
traits are passed on from one generation
to the next:
1. THEORY OF BLENDING INHERITANCE
2. THEORY OF PARTICULATE
INHERITANCE
BLENDING INHERITANCE
PARTICULATE INHERITANCE
That “hereditary
determinants” (genes)
were found in gametes
 Physical traits from
both parents are
blended and the
offspring will exhibit an
intermediate trait
between the two.


That both hereditary
determinants remain in
fertilized zygote.
 Both parental traits
will be exhibited in the
offspring.


He studied biology, physics, and
mathematics
Developed the fundamental laws of
heredity
 Took some ideas from blended model
and particulate model

Mendel chose garden peas (Pisum
sativum) as his subjects as they are
easily grown and their pollination
is easily controlled.
 He controlled pollination by manually
moving pollen between plants
Funfact: Mendel originally
wanted to breed mice, but
wasn't allowed to because it
was considered scandalous

Pea flowers have both
male and female sex
organs so Mendel was
able to developed
“true-breeding” plants
by self-pollination.

Mendel examined varieties of peas for
inheritable characters and traits for his study.
(stem length, pod shape, seed shape, seed
color, etc.)

In 1865, Mendel published his findings in a
paper called Experiments on Plant
Hybridization, which was mostly ignored at the
time due to a number of reasons.
 First, Mendel was not well known in scientific
community.
 Second, his theory ran against the popular model of
blended inheritance.
Mendel’s Three Laws
1. Law of Dominance
2. Law of Segregation
3. Law of Independent Assortment

A monohybrid cross involves one
(mono) character and different
(hybrid) traits.

Mendel crossed a parental
generation with 2 different traits
of the same character
(in this example, flower color)
 The F1 seeds were all purple
 The white flower trait failed to
appear at all.
 There was also no “blending” of
colors

Mendel then took
purple flowers from the
F1 generation and
allowed them to selffertilize.

The flowers in the F2
generation showed a
3:1 ratio of
purple:white flowers

The purple trait
completely masks the
white trait when truebreeding plants are
crossed
 the purple flower trait is
called dominant
 the white flower trait is
called recessive.
 The Law of Dominance

When Mendel repeated
the procedure but for
other characters and
traits, he would
observe a F1
generation with only 1
trait and the 3:1 ratio
in the F2 generation.
The Law of Segregation
 Mendel concluded that each gamete must hold 1
copy of a gene and the zygote will hold 2 copies
(1 from each parent)
 Character = genes
 traits = alleles of the gene.
But why that 3:1 ratio?

An organism has 2 alleles of a gene (1 from
each parent).
 If both alleles are the same  homozygous
 If alleles are different  heterozygous

Let’s assign the each allele a letter/symbol.
P = purple
p = white
BY CONVENTION:

The dominant trait is given a capital letter, the
lowercase of that same letter is the recessive trait. DO
NOT MIX LETTERS. Pick one and stick to it.

Also, some letters are better than others. Capital S
looks a lot like a lowercase (s). Pick a different letter...
Okay
Short hair = SS
Short hair = Ss
Long hair = ss
Better (use H for hair)
HH
Hh
hh
Generation
Parental
Genotype
PP x pp
Phenotype
Purple x white

Both parents are
homozygous.

We can use a Punnett Square to show us the
allele combinations
Male gametes
Female
gametes
Genotype
P
P
p
Pp
Pp
p
Pp
Pp
All offspring in the
F1 generation are
heterozygous
dominant

In his experiment, he then crossed the F1
generation to produce F2 generation
Male gametes
Female
gametes
Genotype
P
p
P
PP
Pp
p
Pp
pp
3 out of 4 are purple
• 1 is homozygous
• 2 are heterozygous
1 out 4 is white
• Homozygous
recessive
1. Polka dots are dominant to stripes.
2. Long sleeves are dominant to short sleeves.
3. Collared shirts are recessive.
4. Buttons are dominant over snaps.
5. Pockets are recessive.
1. A round seeded plant (RR) is crossed with a wrinkle
seeded plant (rr). What are the phenotypes of the
offspring?
2. Two heterozygous purple flowered pea plants are
crossed. What are the phenotypes of their offspring
and in what proportion?
3. A plant with green seeds (yy) is crossed with a
heterozygous plant. What percentage of their
offspring have yellow seeds?
In dragons...wings are a
dominant trait, but some
dragons are born wingless.
1. What are the chances that
two heterozygous dragons
have a whelp that is
wingless?
2. If a wingless dragon is
crossed with one that is
heterozygous, how many
of its offspring will also be
wingless?
Help, help! I don't
know what my
genotype is!!
Am I Dd or DD?
D = winged
d = wingless
I can help
you! Let's have
offspring!
Because we know wingless dragons are homozygous
recessive, we can breed wingless dragons with a winged
dragon.
 By looking at the ratios, we can tell if the winged dragon is
homozygous or heterozygous
If female dragon is PP
then 100%
heterozygous winged
dragons
Genotype
Female
gametes
•
Male gametes
p
p
P
Pp
Pp
P
Pp
Pp
Genotype
Female
gametes
If female dragon is Pp
then a 1:1 ratio is
observed.
Male gametes
p
p
P
Pp
Pp
p
pp
pp
 Mendel's Law of Independent Assortment is illustrated by the
dihybrid cross
 The second law describes the outcome of dihybrid (two
character) crosses, or hybrid crosses involving additional
characters.
▪ A dihybrid is an individual that is a double heterozygote
(e.g., with the genotype RrYy - round seed, yellow seed).
R = round/r = bumpy, Y = yellow/y = green
▪ What are the gametes that can be produced by an
individual that is RrYy?
▪ RY, Ry, rY, ry
RY
Ry
rY
ry
RY
RRYY
RRYy
RrYY
RyYy
round, yellow round, yellow round, yellow round, yellow
Ry
RRYy
round, yellow
RRyy
round, green
RrYy
round, yellow
Rryy
round, green
rY
RrYY
RrYy
round, yellow round, yellow
rrYY
bumpy,
yellow
rrYy
bumpy,
yellow
ry
RrYy
round, yellow
rrYy
bumpy,
yellow
rryy
bumpy, green
Rryy
round, green

The ratio that is seen is a 9:3:3:1 ratio
 a total of 4 phenotypes are observed:
▪ 9 round, yellow
▪ 3 round, green
▪ 3 bumpy, yellow
▪ 1 bumpy, green (double recessive)

You have 30 minutes to complete as much of
this package as you can.

At 2:20, we are moving to notes.

A Punnet Square for a dihybrid cross is pretty
epic!
 a Punnett Square is helpful for 1 or 2 genes but a
little troublesome for more characters
▪ A trihybrid cross needs 64 boxes
▪ A tetrahybrid cross needs 256 boxes
▪ TOO MUCH EPIC!

Mendel’s laws of segregation and
independent assortment reflect the rules of
probability

Remember that the alleles of different (and
unlinked) traits ending up in a gamete is
independent of the chances of other alleles.

To find the probability of an series of
INDEPENDENT events happening together, the
individual probabilities of the events are
multiplied together:
P(A and B) = P(A) x P(B)



If an event is absolutely certain to happen, its
probability is 1.
If it cannot possibly happen, its probability is
0.
All other events have a probability between 0
and 1.
An example of independent
events is the flipping of a coin.
Each flip of a coin is
independent from the previous
or next flips. They don’t
influence each other.

P(A and B) = P(A) x P(B)
What is the probability of getting 5 “tails” in a row?
If P(tails) = 0.5 (or ½)
P(5 tails) = ½ x ½ x ½ x ½ x ½
= 1/32 or 0.03125 (never use %)

Rr
Segregation of
alleles into eggs
Rr
Segregation of
alleles into sperm
Sperm
1/
R
2
1/
R
1/
2
R
R
R
1/
Eggs
2
r
1/
4
r
1/
r
2
4
r
r
R
r
1/
4
1/
4
Heads
¼+¼+¼=¾
Tails
1/4



We used the rule of multiplication to find the
probability of a certain genotype.
We used the rule of addition to find the
probability of a certain phenotype.
In calculating the chances for various
genotypes, each character is considered
separately, and then the individual
probabilities are multiplied together
Calculate the probability that an F2 seed will be spherical
and yellow.
Remember, calculate each trait separately.
(create 2 monohybrid squares rather than 1 dihybrid
square)
P(yellow,round) =
P(yellow, wrinkled) =
P(green, round) =
P(green,wrinkled) =
¾ yellow x ¾ round
¾ yellow x ¼ wrinkled
¼ green x ¾ round
¼ green x ¼ wrinkled
A 9:3:3:1 ratio!!
= 9/16
= 3/16
= 3/16
= 1/16
For any gene with a dominant allele A and recessive
allele a, what proportions of the offspring from an
AA x Aa cross are expected to be homozygous
dominant, homozygous recessive, and
heterozygous?
Two organisms, with genotypes BbDD and BBDd
are mated. Assuming independent assortment of
the B/b and D/d genes, write the genotypes of all
possible offspring from this cross and use the
rules of probability to calculate the chance of
each genotype occurring.
Three characters (flower color, seed color,
and pod shape) are considered in a cross
between two pea plants (PpYyIi x ppYyii).
What fraction of offspring are predicted to
be homozygous recessive for at least two
of the three characters?
Three characters (flower color, seed color, and pod shape) are
considered in a cross between two pea plants (PpYyIi x ppYyii).
What fraction of offspring are predicted to be homozygous recessive for
at least two of the three characters?
p
p
P
p
Y
y
I
i
Pp
(1/4)
Pp
(1/4)
pp
(1/4)
pp
(1/4)
YY
(1/4)
Yy
(1/4)
Yy
(1/4)
Yy
(1/4)
Ii
(1/4)
Ii
(1/4)
ii
(1/4)
ii
(1/4)
Pp = ¼ + ¼ = ½
pp = ¼ + ¼ = ½
Y
y
YY = ¼
Yy = ¼ + ¼ = ½
yy = ¼
i
i
Ii = ¼ + ¼ = ½
ii = ¼ + ¼ = ½

The relationship between genotype and
phenotype is rarely as simple as in the pea plant
characters Mendel studied
 Many heritable characters are not determined by
only 1 gene with 2 alleles.
 However, the basic principles of segregation and
independent assortment apply even to more
complex patterns of inheritance

Inheritance of characters by a single gene may
deviate from simple Mendelian patterns in the
following situations:
 When alleles are not completely dominant or
recessive
 When a gene has more than two alleles
 When a gene produces multiple phenotypes
1.
2.
3.
Complete dominance occurs when
phenotypes of the heterozygote and
dominant homozygote are identical
In incomplete dominance, the phenotype of
F1 hybrids is somewhere between the
phenotypes of the two parental varieties
In codominance, two dominant alleles affect
the phenotype in separate, distinguishable
ways
Fig. 14-10-1
P Generation
Red
CRCR
Gametes
White
CWCW
CR
CW
Fig. 14-10-2
P Generation
Red
CRCR
Gametes
White
CWCW
CR
CW
Pink
CRCW
F1 Generation
Gametes 1/2 CR
1/
2
CW
Fig. 14-10-3
P Generation
Red
CRCR
White
CWCW
CR
Gametes
CW
Pink
CRCW
F1 Generation
Gametes 1/2 CR
1/
CW
2
Sperm
1/
2
CR
1/
2
CW
F2 Generation
1/
2
CR
Eggs
1/
2
CRCR
CRCW
CRCW
CWCW
CW

Dominant alleles are not necessarily more
common in populations than recessive alleles
 For example, one baby out of 400 in the United States
is born with extra fingers or toes
▪ The allele for this unusual trait is dominant to the allele for
the more common trait of five digits per appendage
▪ In this example, the recessive allele is far more prevalent
than the population’s dominant allele

Most genes exist in
populations in more than 2
allelic forms

For example, blood types
The 4 blood types are:




Type A
Type B
Type AB
Type O

the 4 phenotypes of the ABO
blood group in humans are
determined by 3 alleles for the
enzyme (I) that attaches A or B
carbohydrates to red blood cells:
IA, IB, and i.

A alleles = Carbohydrate A

B alleles = Carbohydrate B

i alleles = no carbohydrates
Allele
IA
IB
Carbohydrate
A
B
i
none
(a) The three alleles for the ABO blood groups
and their associated carbohydrates
Genotype
IAIA
Red blood cell
appearance
Phenotype
(blood group)
IA i
A
IBIB or IB i
B
IAIB
AB
ii
O
or
(b) Blood group genotypes and phenotypes
Blood types is
also an example
of codominance.

Most genes have multiple phenotypic effects
 a property called pleiotropy
 For example, pleiotropic alleles are responsible for
the multiple symptoms of certain hereditary
diseases, such as cystic fibrosis and sickle-cell
disease

Some traits may be determined by two or
more genes

In epistasis, a gene at one locus alters the
phenotypic expression of a gene at a 2nd locus
 For example, in mice and many other mammals,
coat color depends on two genes
▪ One gene determines the pigment color
(BB/Bb = black & bb = brown)
▪ The other gene determines whether the pigment will be
deposited in the hair
(CC/Cc = color & cc = no color)
Fig. 14-12
BbCc

BbCc
Sperm
1/
4 BC
1/
4 bC
1/
4 Bc
1/
4 bc
Eggs
1/
1/
1/
1/
4 BC
BBCC
BbCC
BBCc
BbCc
BbCC
bbCC
BbCc
bbCc
BBCc
BbCc
BBcc
Bbcc
BbCc
bbCc
Bbcc
bbcc
4 bC
4 Bc
4 bc
9
: 3
: 4

Quantitative characters are those that vary in
the population along a continuum/spectrum
 Quantitative variation usually indicates polygenic
inheritance, an additive effect of two or more genes
on a single phenotype
Fig. 14-13

Skin color in humans
is an example of
polygenic inheritance
Eggs
AaBbCc
AaBbCc
Sperm
1/
1/
8
1/
8
1/
8
1/
8
1/
1/
8
1/
1/
8
8
1/
8
1/
64
15/
8
1/
1/
8
8
8
1/
8
1/
8
1/
8
8
1/
Phenotypes:
64
Number of
dark-skin alleles: 0
6/
64
1
15/
64
2
20/
3
64
4
6/
64
5
1/
64
6

Another departure from Mendelian genetics
arises when the phenotype for a character
depends on environment AND genotype
 The norm of reaction is the phenotypic range of a
genotype influenced by the environment
Fig. 14-14
For example, hydrangea flowers of the same genotype
range from blue-violet to pink, depending on soil acidity

An organism’s phenotype includes its:
 physical appearance
 internal anatomy
 physiology
 behavior

An organism’s phenotype reflects its overall
genotype and unique environmental history

Some genes do not sort independently
 They are often inherited together because they are
on the same chromosome.

New allele combinations
can be produced if
chromosomes cross-over
during meiosis
 Recombinant chromosomes
and gametes are produced

Females: XX
Males: XY

Some genes are found only on the X
chromosome.
 Because males are heterogametic (X and Y) and
hemizygous (only 1 X), whatever genes are on the
X chromosome will be expressed

The gene for seeing color has 2 alleles:
B and b. The loci for this gene is on the X
chromosome
Genotype
Phenotype
Genotype
XBXB
Color
XBXb
Color / carrier
for color
blindness
XBY
X bY
XbXb
Color-blind
Phenotype
Color
Color-blind

Humans are not good subjects for genetic
research
 Generation time is too long
 Parents produce relatively few offspring
 Breeding experiments are unacceptable

However, basic Mendelian genetics endures
as the foundation of human genetics
Key
Male
Female
1st generation
(grandparents)


A pedigree is a family tree
that describes the
interrelationships of
parents and children across
generations
Pedigree analysis allows us
to figure out whether an
allele controlling a
particular phenotype is
dominant or recessive.
Affected
male
Affected
female
Ww
Mating
Offspring, in
birth order
(first-born on left)
ww
2nd generation
(parents, aunts,
Ww ww ww Ww
and uncles)
ww
Ww
Ww
ww
3rd generation
(two sisters)
WW
or
Ww
ww
No widow’s peak
Widow’s peak
(a) Is a widow’s peak a dominant or recessive trait?
1st generation
(grandparents)
Ff
2nd generation
(parents, aunts,
FF or Ff ff
and uncles)
Ff
ff
ff
Ff
Ff
Ff
ff
ff
FF
or
Ff
3rd generation
(two sisters)
Attached earlobe
Free earlobe
(b) Is an attached earlobe a dominant or recessive trait?
Dominant trait:
 Every affected individual (G2) has an affected
parent (G1)
 About ½ of offspring (G3) are affected
Fig. 14-15a
Key
Male
Female
Affected
male
Affected
female
Mating
Offspring, in
birth order
(first-born on left)
Fig. 14-15b
1st generation
(grandparents)
2nd generation
(parents, aunts,
and uncles)
Ww
ww
ww
Ww ww ww Ww
Ww
Ww
ww
3rd generation
(two sisters)
WW
or
Ww
Widow’s peak
ww
No widow’s peak
(a) Is a widow’s peak a dominant or recessive trait?
Recessive trait:
 Affected people have parents that are not
affected (skip generations)
Fig. 14-15c
1st generation
(grandparents)
Ff
2nd generation
(parents, aunts,
and uncles)
FF or Ff ff
Ff
ff
ff
Ff
Ff
Ff
ff
ff
FF
or
Ff
3rd generation
(two sisters)
Attached earlobe
Free earlobe
(b) Is an attached earlobe a dominant or recessive trait?


Pedigrees can also be used to make predictions
about future offspring
We can use the multiplication and addition rules
to predict the probability of specific phenotypes
Biology 11 Enriched
Tuesday, December 18th, 2012

The Hardy-Weinberg Equilibrium is a
mathematical theory that describes, in detail,
the conditions that must be met for evolution to
not occur (for allele frequencies to remain the
same)
 Thus, it is a null hypothesis with which natural
populations (that are NEVER at H-W equilibrium) can
be compared to.
 Useful model to measure if forces are acting on a
population
▪ Measuring evolutionary change
For a population to NOT evolve, the following
conditions MUST be met:
1. Mating is random
2. Large population
3. No movement in to or out of population
4. No mutation
5. No natural selection
If any of the 5 conditions for maintaining a Hardy-Weinberg
equilibrium are not met, then evolution must be occurring.
Of course, none of these conditions is ever permanently met in
any known natural population of organisms:
 Mutations occur at a slow but steady rate in all known
populations.
 Many organisms, especially animals, enter (immigration) and
leave (emigration) populations.
 Most populations are not large enough to be unaffected by
random changes in allele frequencies.
 Survival is virtually never random.
 Reproduction in organisms that can choose their mates is also
virtually never random.
Therefore, according to the Hardy-Weinberg
Equilibrium Law, evolution (defined as changes
in allele frequencies over time) must be
occurring in virtually every population of living
organisms.
In other words, “Evolution is as ubiquitous and
inescapable as gravity.”

Hardy-Weinberg Theorem:
 Assumes 2 alleles (B,b)
▪ Frequency of the dominant allele (B) = p
▪ Frequency of the recessive allele (b) = q
Frequencies in a gene pool must add up to 1
so:
p+q=1
B
b
B
b
BB
Bb
Bb
bb
p = B allele
q = b allele
p+q=1

If we break down frequencies of individual
genotypes, then:
 Frequency of homozygous dominant = p x p = p2
 Frequency of homozygous recessive = q x q = q2
 Frequency of heterozygous = (p + q) + (q + p) = 2pq

Frequencies of individuals must add up to 1:
2
p
+ 2pq +
2
q
=1
2
p
+ 2pq +
Homozygous
dominant
2
q
=1
Homozygous
recessive
Heterozygous

p+q=1
Alleles in a gene pool:
B

Individuals:
p2 + 2pq + q2 = 1
BB
BB
b
Bb
Bb
bb
bb
population:
100 cats
84 black, 16 white
How many of each
allele?
p2=.36
BB
q2 (bb): 𝟏𝟔/𝟏𝟎𝟎 = 𝟎. 𝟏𝟔
q (b): 𝟎. 𝟏𝟔 = 𝟎. 𝟒
p (B): 𝟏 − 𝟎. 𝟒 = 𝟎. 𝟔
2pq=.48
Bb
q2=.16
bb
Whatassume
Must
are the genotype
populationfrequencies?
is in H-W equilibrium!

The following 4 examples will allow you to
practice solving HW Equilibrium questions.

In a population of pigs color is determined by one
gene. If the black allele (b) is recessive and the
white allele (B) is dominant, what is the frequency
of the black allele in this population?
q2
q
p
p2
2pq
p+q=1
p2 + 2pq + q2 = 1

In a population of 1000 fruit flies, 640 have red
eyes and the remainder have sepia eyes. The sepia
eye trait is recessive to red eyes. How many
individuals would you expect to be homozygous for
red eye color?
q2
q
p
p2
2pq
p+q=1
q2 + 2pq + q2 = 1

In a population of squirrels, the allele that causes
bushy tail (B) is dominant, while the allele that
causes bald tail (b) is recessive. If 91% of the
squirrels have a bushy tail, what is the frequency
of the dominant allele?
q2
q
p
p2
2pq
p+q=1
p2 + 2pq + q2 = 1

In the U.S. 1 out of 10,000 babies are born with
Phenylketonuria, a recessive disorder that results
in mental retardation if untreated. Approximately
what percent of the population are heterozygous
carriers of the recessive PKU allele?
q2
q
p
p2
2pq
p+q=1
p2 + 2pq + q2 = 1
p2=.36
Assuming
H-W equilibrium
2pq=.48
q2=.16
BB
Bb
bb
p2=.20
=.74
BB
2pq=.64
2pq=.10
Bb
q2=.16
bb
Null hypothesis
Sampled data
How do you
explain the data?

Sickle cell anemia
 inherit a mutation in gene coding for hemoglobin
▪ oxygen-carrying blood protein
▪ recessive allele = HsHs / normal allele = Hb

low oxygen levels causes RBC to sickle
 breakdown of RBC
 clogging small blood vessels
 damage to organs

often lethal

High frequency of heterozygotes
 1 in 5 in Central Africans = HbHs
 unusual for allele with severe
detrimental effects in homozygotes
▪ 1 in 100 = HsHs
▪ usually die before reproductive age
Why is the Hs allele maintained at such high levels
in African populations?
Suggests some selective advantage of being
heterozygous…
Single-celled eukaryote parasite (Plasmodium)
spends part of its life cycle in red blood cells
1
2
3

In tropical Africa, where malaria is common:
 homozygous dominant (normal) : HbHb
▪ die or reduced reproduction from malaria
 homozygous recessive: HsHs
▪ die or reduced reproduction from sickle cell anemia
 heterozygote carriers are relatively free of both: HbHs
▪ survive & reproduce more, more common in population
Hypothesis:
In malaria-infected cells,
the O2 level is lowered
enough to cause sickling
which kills the cell &
destroys the parasite.
Frequency of sickle cell allele &
distribution of malaria