Name_____________________________pd_____
AP Biology
Advanced Heredity Problems
1. In snapdragons, flower color is incompletely dominant, when the red (R) and white (R’)
alleles combine they form an intermediate pink. Petal number is controlled by a gene on
a separate chromosome and is completely dominant. Allele N is the dominant number, 6;
allele n is the recessive number, 5. As a parent cross, a pure red, 6-petal flower is
crossed with a white 5-petal flower.
a. What is the genotype and phenotype of the F1 generation?
b. An F1 flower is self-pollinated. If 64 seeds from the F2 generation are collected
and planted, how many of each phenotype would you expect to see?
2. In corn plants, a dominant allele I inhibits kernel color, while the recessive allele i
permits the color when homozygous. At a different locus, the dominant gene P causes
purple kernel color, while the homozygous recessive genotype pp causes red kernels.
a. If plants heterozygous at both loci (IiPp) are crossed, what will be the
phenotypic ratio of the F1 generation?
b. The fact that there is “switch” gene regulating the color genes is referred to
as what?
3. In gremlins the following traits were observed: Long tail = T (dominant), short tail = t
(recessive); Big eyes = B (dominant), Beady eyes = b (recessive). A gremlinologist
crossed a pure long tail, big eye gremlin with a pure short tail, beady eye gremlin. An F1
x F1 cross was performed. Compare the F2 phenotype ratios considering the following
circumstances:
a. T/t and B/b are found on different chromosomes
b. T & B; t & b are linked (however, assume no cross-over)
c. Assuming T & B and t & b are linked, a test cross with a heterozygous long tail,
big eye gremlin and a short tail, beady-eye gremlin took place. According to the
data below, what would the recombination frequency for these two genes? (2 pts)
Long tail/Big eye – 75
Short tail/Beady eye – 70
Long tail/Beady eye – 30
Short tail/Big eye – 25
4. Determine the sequence of genes along a chromosome based on the following
recombination frequencies (3 pts)
A – B – 0.08
A – C – 0.28
A – D – 0.25
B – C – 0.20
B – D – 0.33
5. In fruit flies, the phenotype for eye color is determined at a certain locus. E (wild-type)
is the dominant and e (white eye) is recessive. A cross between a male wild-type and a
female white-eyed* produced the following results
F1 Male Wild type
0
Female Wild type
55
Male White Eye
45
Female White Eye
0
The wild type and white eyed individuals from the F1 generation were crossed to produce the following results:
F2 Male Wild type
23
Female Wild type
31
Male White Eye
22
Female White Eye
24
a. Determine the genotypes of the original parent cross (P-generation*) and explain
your reasoning.
b. Use a Chi-square test on the F2 generation data to analyze your prediction of the
parental genotypes. Show all your work and explain the importance of your final
answer.
Formula for Chi-square 2 =  ((o-e)2/e); o = observed; e = expected
Critical Values for Chi-square distribution
Probability (p)
Degrees of Freedom (df)
1
2
3
0.05
3.84
5.99
7.82
4
9.49
6. In a court case involving a paternity dispute, a man claims that a male child (blood type
B) born to a woman is his son and wants custody. The woman claims he is not the father.
The man has type O blood and the woman has type A blood. Can the child be his son?
Explain your reasoning.
7. Hemophilia is a recessive X-linked trait. Jill Smith does not have hemophilia. Her
father was a hemophiliac and her mother is not a hemophiliac. Jill marries John Jones
who is also not a hemophiliac.
a.
Label the pedigree below with the names of the people bolded above. Shade in
the squares/circles of those individuals who have hemophilia:
b.
If they have a boy, what is the likelihood that he will have hemophilia?