Homework set 3: Genetics Homework 2 All work must be shown to get credit.
1. In a cross with homozygous dominant and homozygous recessive parents, what proportion of the offspring (F-2
generation) would exhibit the phenotypic character of the recessive parent?
F1
DD x dd
↓
Dd x dd
↓
1 out of 4 would be homozygous recessive or 25%
2. In a test cross, the individual with the unknown genotype is crossed with an individual with what genotype?
homozygous recessive
3. In human blood types, which type best represents codominance?
AB
4. Which human blood type is recessive to two different alleles?
O
5. In snapdragons the two alleles for flower color belong to one gene and this gene exhibits incomplete dominance.
What will be the phenotypes of a cross of a red snapdragon with a white snapdragon?
Pink
6. Albinism is an autosomal recessive trait. A normally pigmented man marries an albino woman. If their first child
was an albino what would their (the parents) genotype and their child's genotype be? If they have more children what
other genotypes could there be?
Normal man must be Aa x albino woman is aa
Children can be Aa or aa
7. An albino man marries normally pigmented woman. They have 13 children all of whom are normally pigmented. What
are the probable genotypes of the children and parents?
Woman is unknown therefore Aa or AA,
Because all 13 children are normal she is most likely AA
By chi-squared there is a less than 5% chance that these results are due to
random chance
8. A normally pigmented marries an albino woman. The man's father was an albino. The woman's parents were both
normally pigmented. The man and woman have three children, two normally pigmented and one albino. List each of
these persons and give their probable genotypes..
note: children can be any sex
In Drosophila, the fruit fly, wild type long wings "G" is dominant over vestigial wing (g).
9. In a cross between heterozygous long-winged male and a vestigial-winged female, what types and proportions of
genotypes would be expected?
g
G
g
x g
g
g
G Gg Gg
g gg
gg
1:1
10. In a cross between two heterozygous long-winged flies, what types and proportions of genotypes would be expected?
G
g
G GG Gg
g Gg gg 1:2:1
11. If a heterozygous long-winged fly mated with a homozygous long-winged one, what percentage of the offspring
would be expected to be homozygous long-winged? Would there be any vestigial-winged flies?
G
G
G GG GG
g Gg Gg
50% homozygous long-winged
None would be vestigial winged
12. A long-winged fly mated with a vestigial-winged fly producing 35 long-winged and 35 vestigial-winged offspring.
What are the genotypic formulas of the parents? Of the long-winged offspring? Of the vestigial-winged offspring?
parents are:
G/g x g/g
Long winged are all
G/g
vestigial winged are all g/g
13. Mating a long-winged fly with a vestigial-winged fly produced 256 offspring all of which were long-winged. What
were the probable genotypes of the parents? Mating of the F1 long-winged flies together would produce what types of
offspring genotypically and in what proportions?
Parents are most likely: G/G x g/g
Progeny are:
G/G
G/g
g/g
at 1:2:1
otherwise at least some offspring would be different
14. In human beings, a downward-pointed frontal hairline ("widow's peak") is a heritable trait. A person with widow's
peak always has at least one parent who also has this trait, whereas persons with a straight frontal hairline may occur in
families in which one or even both parents have widow's peaks. When both parents have a straight frontal hairline, all
children also have a straight hairline. Using A and a to symbolize alleles for this trait, what is the genotype of an
individual without widow's peak? (Show your reasoning)
A = Widow’s Peak (d)
a = straight hairline ®
aa – straight must be recessive. Whenever a trait shows up when both parents
are non-expressors the trait must be recessive
15. The first of three children of two normally pigmented parents has albinism, a recessive trait that results from lack of
the pigment melanin.
a. Given that the normal allele is A and the albino allele is a, draw this pedigree and label both the phenotypes and
genotypes.
b. What is the probability that the second child, who shows normal pigmentation, is a carrier? 2/3
c. What is the probability that both unaffected children are carriers?
= P 2nd child is Aa x P 3rd child is Aa
= 2/3 x 2/3 = 4/9
d. What is the probability that only one of the two unaffected children is a carrier?
= P 2nd child is AA x P 3rd child is Aa OR P 2nd child is AA x P 3rd child is Aa
= 1/3 x 2/3 + 2/3 x 1/3
= 2/9 + 2/9 = 4/9