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Putman’s Biol 160 Lab 9: Genetics Take-Home Lab
LAB 9: Genetics Take-Home Lab
Introduction
The science of genetics touches every aspect of our lives. Agriculture, industry, medicine,
criminology, conservation, materials science and many other fields employ the concepts of
genetics.
The general purpose of this lab is to give you the information and practice necessary to
solve real-life genetics problems.
Laboratory Objectives
After mastery of this laboratory, including doing the assigned readings and required
laboratory work, the student should be able to:
1. If given the dominance/recessiveness data for any alleles, perform monohybrid and dihybrid
crosses, and test crosses, and state the expected genotypic and phenotypic frequency of the
offspring.
2. If given incomplete dominance data for any alleles, perform monohybrid and dihybrid
crosses, and test crosses, and state the expected genotypic and phenotypic frequency of
offspring.
3. If given codominance data for any alleles, perform monohybrid and dihybrid crosses, and test
crosses, and state the expected genotypic and phenotypic frequency of offspring.
4. If given epistasis data for any alleles, perform monohybrid and dihybrid crosses, and test
crosses, and state the expected genotypic and phenotypic frequency of progeny.
5. From genotypic and phenotypic frequency data, determine the expected numbers of offspring
of each genotype and phenotype, if given the total number of offspring.
6. If given appropriate data, perform a pedigree analysis.
7. From memory, state, and apply in problems, the following genetics:
a) black vs white coat color in rabbits and Guinea pigs
b) pelage length in rabbits and Guinea pigs
c) height in pea and tomato plants
d) wire-haired and smooth –haired coats in dogs
e) coat color in cocker spaniels
f) solid vs spotted coat patterns in cocker spaniels
g) yellow, cream, and white coat color in Guinea pigs
h) hemophilia
i) hypertrichosis
j) ABO blood group
k) rhesus factor
l) black vs brown pelage in mice as affected by the epistatic gene for pelage color vs noncolor expression
8. Perform a 2 analysis by hand and computer of given statistical data to test congruence with
hypothetical Mendelian distributions.
9.1
Putman’s Biol 160 Lab 9: Genetics Take-Home Lab
Exercise 9.1: Dominant-Recessive Crosses
Introduction
Working with pea plants, Gregor Mendel found that certain traits were dominant to other
traits, traits he called recessive. For instance, pure-breeding tall pea plants, when crossed with
pure-breeding short pea plants, would yield all tall pea plants. When this generation, the F1, was
cross-pollinated, the subsequent generation, the F2, would be 75% tall and 25% short.
In order to solve genetics problems, we first write down the genotype of the parents, or
"P" generation. In the example above, we could designate the allele for tallness in pea plants as
"T" and the allele for shortness at "t." Note that by convention, dominant alleles are usually
designated by capital letters and recessive alleles by lower-case letters. In Mendel's P generation,
the pure-breeding tall plants were homozygous tall, TT, and the pure-breeding short plants were
homozygous short, tt. All of the individuals in Mendel's F1 generation were heterozygous, or Tt.
The second step in solving genetics problems is to determine all possible gametes. In the
above example, the heterozygous (Tt) pea plants can give copies of either the T allele or t allele
to each gamete, but not both.
The third step is to construct a Punnet square. A Punnet square is a statistical device that
gives you the expected percentages or frequencies of particular outcomes. For our cross above,
we construct a Punnet square (below left), put the possible alleles from one parent on top and the
possible alleles from the other parent to the left (below center), then combine alleles to yield
genotypic frequencies (below right):
T

Frequencies:
t
T
T

t
genotypic
25% TT
50% Tt
25% tt
t
T TT
Tt
Tt
tt
t
phenotypic
75% tall
25% short
Let's say a litter of 4 animals is produced. How many would we expect to be homozygous
black? The answer is 1. How many would we expect to be heterozygous black? Similarly, the
answer is 2. Then, of course, we'd expect the last animal to be white, which must be
homozygous. Remember that these are only expected frequencies, and by chance, all of the litter
could be black, or all of the litter could be white, which would, however, be highly unlikely.
Now, let's say we have a litter of 5 Guinea pigs or rabbits. How many of them would be
expect to be black? The answer is 4. Statistically speaking, we'd multiply the expected
phenotypic frequency by the size of the litter: 0.75 x 5 = 3.75; thus it's most likely we'd get 4
black offspring. How many would we expect to be white? Multiply the expected phenotypic
frequency by the size of the litter to get 0.25 x 5 = 1.25, which we'd round down to 1.
The above example is that of a monohybrid, or single trait cross. In a dihybrid cross, two
traits are crossed at the same time.
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Putman’s Biol 160 Lab 9: Genetics Take-Home Lab
For instance, in Guinea pigs and rabbits, black pelage is determined by a dominant allele,
B, and white pelage by a recessive allele, b; further, short pelage is determined by a dominant
allele, L, while long pelage is determined by a recessive allele, l. Let's determine the expected
genotypic and phenotypic frequency of a cross between two Guinea pigs (or rabbits) that are
heterozygous for both of these conditions:
P generation:
genotype
phenotype
gametes produced:
BbLl
x
black, short fur
BbLl
black, short fur
BL, Bl, bL, bl
BL, Bl, bL, bl
Note that in determining which copies of which alleles go into the gametes, we use
Mendel's law of independent assortment--half of the gametes will get a B, half a b, half an L, and
the other half an l.
We next construct a Punnet square large enough to accommodate all of our gametes, then
fill the square in:
BL
Bl bL
bl
BL BBLL BBLl BbLL BbLl
Bl BBLl BBll BbLl Bbll
bL BbLL BBLl bbLL bbLl
bl BbLl Bbll bbLl bbll
Frequencies:
genotypic
1/16 (6.25%) BBLL
2/16 (12.5%) BBLl
2/16 (12.5%) BbLL
4/16 (25%) BbLl
1/16 (6.25%) BBll
2/16 (12.5%) Bbll
1/16 (6.25%) bbLL
2/16 (12.5%) bbLl
1/16 (6.25%) bbll
phenotypic
9/16 black, short fur
3/16 black, long fur
3/16 white, short fur
1/16 white, long fur
Note that the typical outcome of a dihybrid cross, heterozygous for both traits, is the
classical ratio 9:3:3:1, which is a good way to check to see if your “book keeping” is in order!
Using the above dihybrid, heterozygous cross as our example, the expected number of
offspring of a particular genotype or phenotype is determined as above. For example, the
expected number of offspring with black, short pelage would be the litter size x (9/16). In a litter
of 7 offspring, this would be 3.9, which rounds up to 4. The expected number of white, short
pelage individuals that are homozygous for pelage length in a litter of 7 would be 7 x (1/16) 0.44. In this case, we would not be surprised if we didn't have any offspring that matched this
description. How many of our 7 offspring would we expect to be black and long-furred?
Multiply 7 by (3/16) to get 1.3, which rounds down to 1.
9.3
Putman’s Biol 160 Lab 9: Genetics Take-Home Lab
Problems 9.1:
1. Black pelage color in Guinea pigs is a dominant trait while white pelage color is recessive.
When a pure black Guinea pig (BB) is crossed to a pure white one (bb), what percentage of
the resulting litter would we expect to be homozygous black, heterozygous black,
homozygous white, and heterozygous white? Show all work including a Punnet square!
2. Heterozygous black Guinea pigs are crossed among themselves. They have a litter of 7. What
are the expected genotypes and phenotypes of the 7 baby Guinea pigs? Show all work
including a Punnet square!
3. Heterozygous black Guinea pigs are mated to white Guinea pigs. What are the expected
genotypic and phenotypic ratios of the F1 generation? If 5 babies are born, how many of them
would we expect to be black? White? Show all work including a Punnet square!
4. Tallness and shortness are simple dominant and recessive traits in sweet pea plants. You
cross a homozygous tall plant with a short plant to produce the F1 generation. You then cross
two individuals from the F1 generation to produce the F2 generation. What is the expected
genotype and phenotype of the F2 generation? If the crossing produced 187 seeds that were
then sewn to produce 187 plants (assuming that all of the seeds germinated), how many of
these plants would you expect to correspond to each genotype and phenotype? Show all
work!
5. Tall tomato plants are produced by a dominant allele D and short plants by its recessive allele
d. Hairy stems are produced by a dominant allele H and hairless stems by its recessive allele
h. A plant that is homozygous tall and heterozygous hairy is crossed with a plant that is
heterozygous tall and homozygous hairless. What is the expected genotype and phenotype of
the F1 generation? Show all work!
6. You can get $50 for every long-haired, white rabbit that your ranch, Buns-R-Us, can
produce. You know that short hair in rabbits is governed by a dominant gene, L, and long
hair by its recessive allele, l. Black hair is coded for by a dominant gene, B, and white hair by
its recessive allele, b. You mate two rabbits that are heterozygous for both traits and they
produce 6 offspring. What is your gross profit? Show all work!
Exercise 9.2: Incomplete Dominant Crosses
Introduction
Traits whose phenotypic expression blends follow the laws of incomplete dominance. For
instance, in Japanese 4 o'clock flowers, the alleles RR code for red flowers, WW code for white
flowers and RW code for pink flowers. If we were to cross a red Japanese 4 o'clock with a white
one, our expected frequencies of the F1 generation would be as follows:
W
R
R
RW RW
W
9.4
RW RW
Putman’s Biol 160 Lab 9: Genetics Take-Home Lab
Frequencies:
genotypic
100% RW
phenotypic
100% pink
Crossing two members of the F1 generation would result in the following:
Frequencies:
R
R
RR
W
RW
W
RW WW
genotypic
25% RR
50% RW
25% WW
phenotypic
25% red
50% pink
25% white
Problems 9.2:
1. Color in Andalusian chickens obeys the laws of incomplete dominance, with BB producing a
plack phenotype, Bb a blue phenotype and bb a splashed white phenotype. You cross a blue
rooster with a blue hen. What are the expected genotypic and phenotypic frequencies of the
progeny? Show all work!
2. In Japanese 4 o'clock flowers, red petals are coded for by RR, white petals by WW, and pink
petals by RW, since they obey the laws of incomplete dominance. You cross a white
Japanese 4 o'clock with a pink one. What is the expected genotypic and phenotypic
frequency of the F1 generation? Show all work!
Exercise 9.3: Codominant Crosses
Introduction
Codominant alleles are both expressed without being blended; they are both fully
dominant. For instance, in cattle, the alleles RR code with a red coat, WW for a white coat and
RW for a light red coat called roan. An examination of the coat of the roan phenotype reveals
both white and red hairs, not a blending of the color. The expected genotypic and phenotypic
frequency of a red x roan cross (RR x RW) would be as follows:
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Putman’s Biol 160 Lab 9: Genetics Take-Home Lab
R
R
W
RW RW
W
RW WW
Frequencies:
genotypic
25% RR
50% RW
25% WW
phenotypic
25% red
50% roan
25% white
Problems 9.3:
1. In shorthorn cattle, red is governed by the genotype RR, roan by RW and white by WW.
When roan shorthorns are crossed among themselves, what genotypic and phenotypic
frequencies are expected in the F1 generation? Show all work!
2. The palomino horse is a hybrid exhibiting a golden color with a lighter mane and tail. The
palomino condition is coded for by a pair of codominant alleles, CW. A horse with the alleles
CC is chestnut in color (reddish) and one with the alleles WW is called a cremello (nearly
white). You're a horse breeder. Palominos get you the most money. If you breed two
palominos, what is the expected genotypic and phenotypic frequency of the foals? If your
mare produces a foal, what are the chances that it will be a palomino? Show all work!
3. The shape of radishes may be long (LL), oval (LR) or round (RR), and obey the laws of
codominance. What would be the expected genotypic and phenotypic frequencies if an oval
radish plant were crossed to one that was round? Show all work!
Exercise 9.4: Test Crosses
Introduction
It is oftentimes important for plant and animal breeders to determine whether their stock
is homozygous or heterozygous for a particular trait. For instance, in cattle, a recessive gene
produces calves with amputated legs (no legs) in the homozygous condition. These calves have
to be destroyed--an expensive loss of livestock. If you have a bull that you suspect is a carrier of
this deleterious gene, you perform a test cross by mating it with a cow that you know carries the
gene. If the test cross produces a calf with amputated legs, you can then decide if you want to
keep mating the bull, which is very irresponsible, ship the bull off to be made into hamburger, or
put the old bull out to pasture. The test cross gives you the information you can use to make your
decision.
Problems 9.4
1. Your friend has a black Guinea pig of unknown genotype. She wishes to know what the
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Putman’s Biol 160 Lab 9: Genetics Take-Home Lab
genotype is. You know that fur color in Guinea pigs obeys the laws of
dominance/recessiveness, that B codes for black fur and that b codes for white fur. How
could you determine the genotype of the unknown Guinea pig?
2. In foxes, red fur is coded for by a dominant allele B and silver-black fur by a recessive allele
b. You are a fox breeder. You mate two red foxes together and their union produces a litter of
5 red foxes (one you name "Fred"). Two of these red foxes you mate, and they produce 6
foxes: 5 red and 1 silver-black. How could you determine where the silver-black gene came
from? Could Fred be carrying it?
Exercise 9.5: Epistatic Alleles
Introduction
Alleles that affect the phenotypic expression of other alleles are called epistatic alleles.
For instance, in mice, a dominant allele, B, codes for black pelage, whereas a recessive allele, b,
codes for brown pelage. A codominant allele, C, allows the black and brown alleles to be
expressed, whereas the recessive allele, c, inhibits color expression and produces a mouse with
white pelage.
As an example, we would expect that a cross between a white mouse with the genotype
Bbcc and a brown mouse with the genotype bbCc would produce a litter containing 50% white
mice, 25% black with 25% brown mice. This was determined as follows:
P generation:
genotype
phenotype
Bbcc
white
gametes produced
x
Bb, bc
bC, bc
Punnet square:
bC bc
Bc BbCc Bbcc
bc bbCc bbcc
Frequencies:
genotypic
25% BbCc
25% bbCc
25% Bbcc
25% bbcc
9.7
bbCc
brown
phenotypic
25% black
25% brown
50% white
Putman’s Biol 160 Lab 9: Genetics Take-Home Lab
Now, let's say we have a litter of 5 baby mice. We would expect that 2 or 3 of them
would be white, 1 or 2 black, and 1 or 2 brown. It is, however, possible, although unlikely, for
the entire litter to be black, brown or white!
Problems 9.5:
1. In mice, the allele for black pelage, B, is dominant to the allele for brown pelage, b; further, a
dominant allele, C, at a different locus, allows pelage color to be expressed, whereas a
recessive allele, c, inhibits pelage color expression. What is the expected genotypic and
phenotypic frequency of a cross between BbCc and BbCc? What is the phenotype of the
parents? If a litter of 7 mice are produced, how many would we expect to be black? Brown?
White?
2. Using the information from problem 1 above, what is the expected genotypic and phenotypic
frequency of a cross between a brown male mouse that carries the recessive color expression
gene and a black female mouse that is heterozygous for both alleles? If a litter of only 3 mice
are produced, how many would we expect to be black? Brown? White?
3. A male brown mouse mates with a female brown mouse. The subsequent litter contains 3
white mice and 3 brown mice. What is the genotype of the parents?
4. Black coat color in cocker spaniels is governed by a dominant allele, B, and red coat color by
its recessive allele, b; solid pattern is governed by the dominant allele of an independently
assorting, epistatic locus, S, and spotted pattern by its recessive allele, s. A spotted black
male is mated to a solid red female and they produce a litter of six puppies. At least one of
the puppies is red and spotted. What is the genotype of the male and female dog and what is
the expected genotype and phenotype of the puppies?
Exercise 9.6: Pedigree Analysis
Introduction
Genes for blood clotting in humans are, at least in part, carried on the X chromosome; a
person whose blood does not clot normally is called a hemophiliac and had the disease
hemophilia. Females have two X chromosomes and males have one X chromosome. Normal
blood clotting is governed by the gene XH whereas hemophilia is governed by the gene Xh. If a
female has one hemophilia gene (XHXh), her blood clots normally; if she, however, has a
hemophilia gene on both of her X chromosomes (XhXh), she is a hemophiliac. Since males have
only one X chromosome, a nonhemophiliac male (XHY) does not carry the gene whereas a
hemophiliac male (XhY) carries the gene on his X chromosome.
In the multiple allele system of the ABO blood group, the autosomal A allele codes for
the blood type A oligosaccharide antigen found on the surface of erythrocytes, the B codes for
9.8
Putman’s Biol 160 Lab 9: Genetics Take-Home Lab
the B oligosaccharide, and O codes for the absence of an oligosaccharide antigen. If a person
receives a foreign blood type during a transfusion, the immune system attacks the unfamiliar
antigen and agglutination occurs, which may lead to death. The genotypes AA and AO produce
the phenotype (blood type) A, the genotypes BB and BO produce the phenotype B, the genotype
AB produces the AB phenotype, and the genotype OO produces the O phenotype.
The rhesus factor is a second important blood group. If at least one autosomal gene for
the rhesus factor is present (Rh), then the rhesus antigen is produced. A person who does not
produce the rhesus factor has the genotype rhrh and is said to be Rh- (Rh negative), whereas a
person who produces the rhesus factor may be homozygous (RhRh) or heterozygous (Rhrh) for
the rhesus trait and said to be Rh+ (Rh positive)
When reporting a blood type, one often uses the phrase, "AB-" or "B+" which includes
the Rh values.
For medical or criminological reasons, we may wish to know the possible or probable
genotypes and phenotypes of certain individuals. One way to do this is through the use of
pedigree analysis.
In a pedigree analysis, one first writes down what is known of the family of the
individual in question. For instance, if a hemophiliac man (XhY) has four children (two boys and
two girls) with a normal woman, all of his daughters will at least be carriers of the hemophiliac
gene. This is because dad gives copies of his only X chromosome, with the hemophilia gene, to
each of his daughters, thereby determining their sex; if he gave a copy of his Y chromosome,
they’d be boys! If the woman’s mother was also a hemophiliac (XhXh), and her dad was normal
(XHY), we would know for sure that she was a carrier (XHXh), and we’d expect that 50% of her
daughters would have hemophilia; we’d also expect 50% of the sons to have hemophilia as well.
This may be an unacceptable risk, and the couple may decide to adopt.
The above scenario may be abbreviated as follows, by convention, squares indicating
males and circles females:
H
h h
X Y
X X
Xh Y
H
X Y
H h
X X
X HXh
h
X Y
h h
X X
In an example of pedigree analysis involving paternity, let’s say a wealthy celebrity has
been accused of being the father of a girl who is suing the celebrity for ten million dollars. The
mother of the woman has the blood type A- and the girl has the blood type O+. The celebrity has
the blood type AB-. The girl looks a lot like the celebrity. Should he settle out of court?
Let’s see what we know for sure. The mother’s genotype is A?rhrh, the ? referring to the
fact that this allele could be either A or O. Since the girl’s phenotype is O+, the ? must be an O,
9.9
Putman’s Biol 160 Lab 9: Genetics Take-Home Lab
hence the mother’s genotype is AOrhrh, The girl’s genotype is OO?Rh?, since she could be
either homozygous or heterozygous for the rhesus factor and still be Rh+. The celebrity’s
genotype has to be ABrhrh. Could he possibly have been her father? No, because he carries
neither the O nor the rh allele.
Problems 9.6:
1. Copy and determine the genotypes for all of the individuals in the following pedigree for
hemophilia; note that individuals with hemophilia are indicated by filled-in figures.
2. You have A+ blood and your spouse has B-. Your parents were AB+ and O+, while your
spouse’s parents were both B-. If you have a child, is there any chance that he or she could be
O+? Draw a pedigree chart, and show and explain all work!
3. You have been called in as an expert witness for the defense in a paternity suit. The alleged
child of your client has the blood type O-. The child’s mother has the blood type A+ and your
client’s blood type is B-. Could your client have been the father of the child? How would you
advise your client?
4. Green color blindness is linked to the X chromosome. An individual with at least one gene
for normal color vision will have normal color vision so that, in women, XCXC and XCXc
individuals will have normal color vision but XcXc individuals will have green color
blindness. A normal man would be XCY whereas a man with green color blindness would
have the genotype XcY. A woman marries a man with normal vision. They have a male child
with green color blindness. The woman's father has normal color vision. What is the woman's
genotype and phenotype? Show all work, including a pedigree!
5. Hypertrichosis of the pinna of the ear (hairy ears) is a Y-linked trait and was originally
described in Italians. An Italian man with hypertrichosis marries and fathers five children:
three girls and two boys. How many of the children will develop hypertrichosis?
9.10
Putman’s Biol 160 Lab 9: Genetics Take-Home Lab
Exercise 9.7: The
2
Test
Introduction
The 2 (chi-square) test is a statistical test that measures the significance in differences
between expected and actual values. The 2 test takes into consideration sample size and the
possibility that any observed differences may be due to chance. For instance, if you were to flip a
coin 2 times, you’d think nothing of it if both times were to come up heads. Flip that same coin
20 times, and if every result were heads, you’d strongly suspect that the coin was “fixed.” Flip it
200 times and if all you got were heads, you’d be confident that the coin was not a fair coin! The
point here is that as your sample size increases there is a higher confidence level that the results
of a particular experiment, in this case flipping a coin, are not to be dismissed.
To perform a 2 test we do the following:
1. Make a data table summarizing your data. For instance, from a cross of two pea plants we
might obtain 1112 peas, 281 yellow, 570 yellow-green, 261 green:
yellow
Observed 281
Expected 278
yellow-green
570
556
green
261
278
1112
1112
2. Find the expected values by setting the lowest value equal to 1 and finding the closest ratio.
In the above example, we’ll let 261 equal 1. Dividing 1112 by 261 yields about 4. Then
dividing 1112 by 4 yields 278. This gives us a ratio of 1:2:1. Another way of doing this is to
examine the data for clues, do a Punnet square, and calculate the expected frequencies. This
suggests that we have incomplete dominance occurring here. If this is the case, then let YY =
yellow, YG = yellow-green, and GG = green. Since it appears as though we have twice as
many yellow-greens as either of the other, this suggests that the expected number of YY =
GG = (1112)(1/4) = 278, and YG = (1112)(2/4) = 556.
3. We must state a hypothesis that the statistical test is seeking to disprove. The standard
hypothesis is called the null hypothesis, or H0, which is “There is no difference between the
expected value and the observed value.”
4. Calculate 2:
2
=
(observed value – expected value)2
(expected value)
2
=
(281 – 278)2 + (570 – 556)2 +
278
556
= 0.0324 + 0.353 + 1.04
= 1.42
9.11
(261 – 278)2
278
Putman’s Biol 160 Lab 9: Genetics Take-Home Lab
5. Calculate the degrees of freedom. Since we have 3 alleles here, and since the degrees of
freedom for the 2 test are n – 1, we have 2 degrees of freedom.
6. Look at a 2 table (Table 9.1 below). Notice that our value is < 0.05. Hence we do not reject
our H0, meaning that there is no significant difference between the observed and expected
values, further meaning that our experimental data is consistent with the hypothesis of a 1:2:1
ratio.
Table 9.1.
DEGREES
OF
FREEDOM
1
2
3
4
5
6
2
table.
PROBABILITY THAT DEVIATIONS ARE DUE TO CHANCE
0.95
0.004
0.10
0.35
0.71
1.14
1.63
0.90
0.02
0.21
0.58
1.06
1.61
2.20
0.80
0.06
0.45
1.01
1.65
2.34
3.07
0.70
0.50
0.30
0.15
0.46
1.07
0.71
1.39
2.41
1.42
2.37
3.66
2.20
3.36
4.88
3.00
4.35
6.06
3.83
5.35
7.23
NONSIGNIFICANT
7. Various computer programs are also available that run
0.20
1.64
3.22
4.64
5.99
7.29
8.56
2
0.10
2.71
4.60
6.25
7.78
9.24
10.64
0.05
0.01
0.001
3.84
6.64
10.83
5.99
9.21
13.82
7.82 11.34 16.27
9.49 13.28 18.47
11.07 15.09 20.52
12.59 16.81 22.46
SIGNIFICANT
tests.
Problem 9.7:
1. In a sampling of snails, 87 carry pathogenic parasites while 269 do not. Our research group
has a hunch that genetics might be involved in whether or not the parasite in question can
infect the studied snail species. Propose a genetic model and test the hypothesis that there is
no difference between the observed frequency and the genetic model you propose! Do this
problem by hand, showing all work, and then use a computer program (if one is available).
Show all work!
9.12