Genetics Problems (Review)
1. In a particular dog, short hair is dominant over long, and dark is dominant over light. One dog is
heterozygous for hair length and shows the recessive trait for hair color. Another dog is homozygous
dominant for hair length and heterozygous for hair color. Identify the possible phenotypes of the
offspring and give the number of each expected.
A: All will have short hair, ½ of those will be dark, ½ light
2. In fruit flies, straight wings is dominant over curly. Show how you would determine if a straight
winged fly were heterozygous or homozygous?
A: Use a test-cross. Mate the unknown fly to a known homozygous recessive. Think about the results
if the fly was heterozygous vs. if it were homozygous dominant.
3. In guinea pigs rough coat "R" is dominant over smooth coat "r". Black coat "B" is dominant over
white coat "b". R and B are independent genes.
a) Can two black coat parents produce white coat offspring? Can two white coat parents produce black
coat offspring?
A: Yes, only if both are heterozygous
b)Cross a homozygous rough black animal with a smooth white one. What will be the appearance of
the F1; of the F2; of the offspring of a cross of F1 back with the rough black parent; with the smooth
white parent?
A:
F1 = all rough, black guinea pigs
F2 (RrBb x RrBb) = 9 rough, black; 3 rough, white; 3 smooth, black; and 1 smooth, white
F1 x rough black parent = all rough, black guinea pigs
F1 x smooth white parent = 1 rough, black; 1 rough, white; 1 smooth, black; 1 smooth, white
4. In humans, the ability to taste phenylthiourea (PTU) is dominant. "Tasters" (TT) or (Tt) perceive an
extremely bitter taste of PTU, while "non tasters" (tt) experience no sensation, or taste.
a) What are the genotypes of Mr. and Mrs. Meadowmuffin, who can taste PTU, and who have 3
children, one of whom is a non-taster?
A: Both must be Tt
b) What offspring phenotypes would be expected from the following crosses and in what ratios?
i) heterozygous x heterozygous A: 3 tasters to 1 non-taster
ii) homozygous taster x heterozygous A: all tasters
iii) heterozygous x non-taster A: 1 taster to every 1 non-taster
5. A plant of genotype TTYy was crossed with one of genotype ttYy. If the cross yielded 800 seeds,
what number of what phenotypes probably resulted? What were their genotypes?
A: Aprox. 600 (3/4) will be dominant for T and Y traits, and 200 (1/4) will be dominant for T, but
recessive for the y trait.
6. A dominant gene (S) in snapdragons produces bilaterally symmetrical ("irregular") flowers; the
recessive allele (s) produces radially symmetrical flowers. Another pair of alleles controls color,
yeilding red (A1A1), pink (A1A2), or white (A2A2). If a cross is made between two snapdragons of
genotype SsA1A2, what will be the phenotypes of the F1? Give the proportions of each.
A: bilateral/red (3/16), bilateral/pink (3/8), bilateral/white (3/16), radial/red (1/16), radial/pink (1/8),
radial/white (1/16)
7. Two fruit flies with normal wings are crossed. Among 123 offspring, 88 have normal wings and 35
have curly wings. What inheritance pattern is shown by the normal and curly alleles? What were the
genotypes of the two parents?
A: Normal wings are dominant to curly. Parents were heterozygous.
8. Flower color in certain plants is governed by epistasis. P= purple p= white, C= curved petals c=
straight petals. A homozygous recessive petal shape prevents purple color. If parent plants are Ppcc x
PPCc, what is the genotypic and phenotypic ratios of the offspring?
A: ¼ PPCc, ¼ PpCc, ¼ PPcc, ¼ Ppcc; ½ will be purple/curved, ½ will be white/straight
9. Assume that straight hair allele is recessive. The dominant allele gives curly hair. Two curly haired
parents have a straight haired child.
(a) What are the genotypes of the parents? Heterozygous
(b) What is the genotype of the child? Homozygous recessive
(c) Give two genotypes for curly haired people. Homozygous dominant; Heterozygous
(d) Using letter symbols show two different crosses for a curly haired individual crossed with a straight
haired individual.
A
a
a
Aa
aa
a
Aa
aa
A
A
a
Aa
Aa
a
Aa
Aa
10. In pigeons, the checkered pattern results from the dominant gene C and plain coloration on the
recessive gene c. Red color is controlled by a dominant gene R and by the brown r recessive gene.
(a) A homozygous checkered, red bird and a plain, brown bird are crossed. Summarize the expected F1
and F2 results, giving both genotype and phenotype.
A:
F1 = all CcRr checkered/red birds
F2 = 9:3:3:1 phenotype ratio (9 checkered/red, 3 checkered/brown, 3 plain/red, 1 plain/brown)
Genotypes: CCRR (1/16), CcRR (1/8), CCRr (1/8), CcRr (1/4), CCrr (1/16), ccRR (1/16),
Ccrr (1/8), ccRr (1/8), ccrr (1/16)
(b) Male checkered, red birds with the genotype CcRr were mated with females of the following
genotypes; (i) CCRR (ii) Ccrr (iii) ccrr. Summarize the expected first generation results of each cross,
giving all ratios.
A:
(i) All checkered/red birds
(ii) ½ checkered/red, ¼ checkered/brown, 1/8 plain/red, 1/8 plain/brown
(iii) ¼ checkered/red, ¼ checkered/brown, ¼ plain/red, ¼ plain/brown
11. A rooster with gray feathers is mated with a hen of the same phenotype. Among their offspring, 15
chicks are grey, 6 are black, and 8 are white. What is the simplest explanation for the inheritance of
color in the chicks? What offspring would you expect from the mating of a grey rooster and a black
hen?
A: Gray is the heterozygous condition for rooster/hen color and it’s a case of incomplete dominance
12. A black guinea pig crossed with an albino guinea pig produces 12 black offspring. When the albino
was crossed with a second black one, 7 blacks and 5 albinos resulted. What is the best explanation for
these results. Give genotypes for parents and all offspring as well as the inheritance pattern.
A: First black guinea pig was homozygous dominant, second one was heterozygous
1st pairing: Black guinea pig (BB) x albino guinea pig (bb) produce all black guinea pigs (Bb)
2nd pairing: Black gp (Bb) x albino gp (bb) produce ½ black (Bb) and ½ albino (bb) guinea pigs
13. The allele that causes albinism (p) is recessive to the allele for normal pigmentation (P). A normal
woman whose father is an albino marries an albino man whose parents are both normal. They have
three children, two normal, one albino. Give the genotype for each individual.
A: Father (bb), Mother (Bb), normal children (Bb), albino child (bb)
14. If a color-blind male marries a carrier female, what are the chances that their sons will be colorblind? What are the chances that their daughters will be carriers?
A: ½ of their sons will be colorblind. ½ of daughters will be carriers, but the other ½ will be
colorblind (so they, too, will carry the gene)
15. A color-blind man and a woman with normal vision whose father was color blind have a son. What
is the probability that their first son will be color blind? What is the probability that they will have a
color-blind daughter? (note: this is not the same as asking “What is the probability of their daughter
being color-blind)
A: 50% chance their first son will be colorblind (if it’s a guaranteed boy). 25% chance they will have
a colorblind daughter (50% if it’s a guaranteed girl).
16. Hemophilia is an X-linked recessive trait. A woman who is seemingly normal but whose father was
a hemophiliac marries a normal man. What proportion of their sons will have hemophilia? Their
daughters? What proportion of their daughters will be carriers? If the couple has four sons, what is the
probability that all four will be born with hemophilia.
A: 50% of their sons will have hemophilia. None of their daughters will have hemophilia, but 25%
may be carriers. There is still a 50% chance for each and every son they have, regardless of previous
children.
17. a) If a carrier female marries a normal male, what are the chances of their offspring having the
disease hemophilia? What are the chances of their sons being normal? What are the chances of their
daughters being a carrier?
A: 50% chance if they have a boy, zero chance if they have a girl. 50% chance their daughters will be
carriers.
b) If a woman's father had hemophilia, what are the chances that she is a carrier? If her mother was a
carrier, what are the chances that she is a carrier? A hemophiliac?
A: 1st question: 100% chance that she is a carrier. Her mother being a carrier cannot increase that
chance, but it can make it a 50% chance that she is a hemophiliac.
18. The gene for wild-type eye color is dominant and sex-linked in Drosophila. The mating of a male
with wild-type eye color with a female of the same phenotype produces offspring that are 3/4 wild-type
eye color and 1/4 white-eyed. Indicate the genotypes of the P and F1 generations.
A: Parent genotypes: XTX, and XTY, F1 genotypes: XTX (½), XTY (¼), and XY (¼)
19. John and Carol, planning their first child, each had a brother who died of a recessively inherited,
autosomal disease which is fatal in early childhood. Find the chance of them having a child with the
disease. If their first child is born with the disease, what is the probability that their second will?
A: John and Carol each have a 75% chance of being a carrier, assuming each of them was, they would
have a 25% chance of having a child with the disease (same for second, regardless of first). All told,
about a 14% chance overall (multiplying risks).