Exam 3 Review
Supplemental Instruction
Iowa State University
Leader:
Course:
Instructor:
Date:
Lilli Howard
BIOL/GEN 313
Dr. Vollbrecht
03/11/14
Chapter 5
 5.1 Additional Factors at a Single Locus Can Affect the Results of Genetic Crosses
   5.2 Gene Interaction Takes Place When Genes at Multiple Loci Determine a Single
Phenotype
 5.3 Sex Influences the Inheritance and Expression of Genes in a Variety of Ways
Chapter 6
 6.1 The Study of Genetics in Humans Is Constrained by Special Features of Human
Biology and Culture
 6.2 Geneticists Often Use Pedigrees to Study the Inheritance of Characteristics in
Humans
 6.4 Genetic Counseling and Genetic Testing Provide Information to Those Concerned
about Genetic Diseases and Traits
Chapter 7
 Not covered: Fig. 7.10, 7.16, 7.17, 7.18, chi-squared test of independence, multiple
strand crossover, linkage mapping in pedigree analysis, coefficient of coincidence, gene
interference
 7.1 Linked Genes Do Not Assort Independently
 7.2 Linked Genes Segregate Together and Crossing Over Produces Recombination
Between Them
 7.3 A Three-Point Testcross Can Be Used to Map Three Linked Genes
 7.4 Physical-Mapping Methods Are Used to Determine the Physical Positions of Genes
on Particular Chromosomes
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1. When a Chinese hamster with white spots is crossed with another hamster that has no
spots, approximately ½ of the offspring have white spots and ½ have no spots. When two
hamsters with white spots are crossed, 2/3 of the offspring possess white spots and 1/3
have no spots.
a. What is the genetic basis of white spotting in Chinese hamsters?
b. How might you go about producing Chinese hamsters that breed true for white
spotting?
a. The 2:1 ratio in the progeny of two spotted hamsters suggests lethality and the 1:1 ratio in
the progeny of a spotted hamster and a hamster without spots indicates that spotted is a
heterozygous phenotype. If S and s represent alleles at the locus for white spotting,
spotted hamsters are Ss, and solid hamsters are ss. One quarter of the zygotes expected
from a mating of two spotted hamsters are SS, lethal, and missing from the progeny.
b. Impossible
2. Genes at two loci (B, b and R, r) determine the color of fish: fish with a dominant allele
at both loci (B_R_) are brown, fish with a dominant allele at the B locus only are blue,
fish with a dominant allele at the R locus only are red and fish with recessive alleles at
both loci are white. A homozygous brown fish is crossed with a homozygous white fish.
The F1 is then backcrossed with the homozygous white parent and obtained 228 brown
fish, 230 blue fish, 237 red fish, and 222 white fish.
a. Give the genotypes of the backcross progeny.
Brown: BbRr Blue: Bbrr
Red: bbRr
White: bbrr
b. Use a chi-square test to compare the observed numbers of backcross progeny with
the number expected. What conclusion can you make from your chi-square
results?
Expected ratio: 1:1:1:1
X2 = 0.5
df = 3
0.9 < P < 0.975
We cannot reject the hypothesis
c. What results would you expect for a cross between a homozygous red fish and a
white fish?
All progeny would be bbRr (red)
d. What results would you expect if you crossed a homozygous red fish with a
homozygous blue fish and then backcrossed the F1 with a homozygous red
parental fish?
F1 will be all brown (BbRR)
F2 will be ¼ brown (BbRR) ¼ brown (BbRr) ¼ red (bbRR) ¼ red (bbRr)
1060 Hixson-Lied Student Success Center  515-294-6624  sistaff@iastate.edu  http://www.si.iastate.edu
3. In some goats, the presence of horns is produced by an autosomal gene that is dominant
in males and recessive in females. A horned female is crossed with a hornless male. The
F1 offspring are intercrossed to produce the F2. What proportion of the F2 females will
have horns?
(HH) x (hh)  Hh x Hh  HH, Hh, Hh, hh
¼ of the females will have horns
What is the chance that the first offspring will have horns?
½ (3/4) + ½ (1/4) = ½
4. For each of the following modes of inheritance, describe the features that will be
exhibited in a pedigree in which the trait is present: autosomal recessive, autosomal
dominant, X-linked recessive, X-linked dominant, and Y-linked inheritance.
Autosomal Recessive: skips generations, shows up in males and females equally
Autosomal Dominant: shows in every generation, shows up in males and females equally
X-linked Recessive: skips generations, passed from fathers to all daughters and no sons, passed
from mothers to both sons and daughters, always shows in males but not females
X-linked Dominant: shows in every generation, passed from fathers to all daughters and no sons,
passed from mothers to both sons and daughters, always shows in males and females
Y-linked Inheritance: Only passed from father to son, shows in every generations
5. Joe is color blind. Both his mother and his father have normal vision, but his mother’s
father (Joe’s maternal grandfather) is color-blind. All Joe’s other grandparents have
normal color vision. Joe has three sisters (Patty, Betsy, and Lora) all with normal color
vision. Joe’s oldest sister, Patty, is married to a man with normal color vision; they have
two children, a 9 year old color blind boy and a 4 year old girl with normal color vision.
a. Using standard symbols and labels, draw a pedigree of Joe’s family.
b. What is the most likely mode of inheritance for color blindness in Joe’s family?
X-linked recessive
c. If Joe marries a woman who has no family history of color-blindness, what is the
probability that their first child will be a color-blind boy?
No chance
Girl? No chance
d. If Joe marries a women who is a carrier of the color-blind allele, what is the
probability that their first child will be a color-blind boy?
½ (½) = ¼
Girl? ½ (½) = ¼
e. If Patty and her husband have another child, what is the probability that the child
will be a color-blind boy?
½ (½) = ¼
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6. A geneticist discovers a new mutation in Drosophila melanogaster that causes the flies to
shake and quiver. She calls this mutation spastic (sps) and determines that spastic is due
to an autosomal recessive gene. She wants to determine if the gene encoding spastic is
linked to the recessive gene for vestigial wings (vg). She crosses a fly homozygous for
spastic and vestigial traits with a fly homozygous for the wild-type traits and then uses
the resulting F1 females in a testcross. She obtains the following flies from this testcross.
vg+ sps+
230
vg
sps
224
vg
sps+
97
vg+ sps
99
a. Are the genes that cause vestigial wings and the spastic mutation linked?
Genes are linked and not assorted independently
7. Recombination rates between three loci in corn are shown here.
Loci
Recombination Rate
R and W
17%
R and L
35%
W and L
18%
What is the order of the genes on the chromosomes?
R – W – L or L – W – R
8. Waxy endosperm (wx), shrunken endosperm (sh), and yellow seedling (v) are encoded b
three recessive genes in corn that are linked on chromosome 5. A corn plant homozygous
for all three recessive alleles is crossed with a plant homozygous for all the dominant
alleles. The resulting F1 are then crossed with a plant homozygous for the recessive
alleles in a three-point testcross. The progeny of the testcross are:
a. wx sh V
87
b. Wx Sh v
94
c. Wx Sh V
3479
d. wx sh v
3478
e. Wx sh V
1515
f. wx Sh v
1531
g. wx Sh V
292
h. Wx sh v
280
a. Determine the order of these genes on the chromosome. V is in the middle
b. Calculate the map distances between the genes.
Wx  V = 7 m.u.
Sh  V = 30 m.u.
Wx  Sh = 37 m.u.
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9. Assume that long ear lobes in humans are an autosomal dominant trait that exhibits 30%
penetrance. A person who is heterozygous for long ear lobes mates with a person who is
homozygous for normal ear lobes. What is the probability that their first child will have
long ear lobes?
.15 or 15%
10. A variety of poppy having lacerate leaves was crossed with a variety that has normal
leaves. All the F1 had lacerate leaves. Two F1 plants were interbred to produce the F2. Of
the F2, 249 had lacerate leaves and 16 had normal leaves. Give genotypes for all the
plants in the P, F1 and F2 generations. Explain how lacerate leaves are determined in the
poppy.
The F1 progeny tell us that lacerate is dominant over normal leaves. In the F2, 249:16 does not
come close to a 3:1 ratio. Let’s see if these numbers fit a dihybrid ratio. Dividing 265 total
progeny by 16 (because dihybrid ratios are based on 16ths), we see that 1/16 of 265 is 16.56.
Therefore, the F2 progeny are very close to 15/16 lacerate, 1/16 normal, a modified dihybrid
ratio. If we symbolize the two genes as A and B, then:
F1
Aa Bb × Aa Bb all lacerate
F2
9/16 A- B- lacerate (like F1)
3/16 A- bb lacerate
3/16 aa B- lacerate
1/16 aa bb normal
A dominant allele at either gene A or gene B, or both, results in lacerate leaves. Finally, the
parents must have been AA BB lacerate × aa bb normal. Note that only AA BB for the lacerate
parent would result in F1 that are Aa Bb.
1060 Hixson-Lied Student Success Center  515-294-6624  sistaff@iastate.edu  http://www.si.iastate.edu