Genetics Practice Problems
Answers

1. An Aa individual will produce A and a gametes in equal proportions.

2.

Cross	Genotypic ratio	Phenotypic ratio
a) Aa x aa	1 Aa : 1 aa	1 dominant : 1 recessive
b) Aa x Aa	1 AA : 2 Aa : 1 aa	3 dominant : 1 recessive
c) AA x aa	all Aa	all dominant

3. 

Progeny	Parents
a) half dominant and half recessive	Aa x aa
b) all heterozygous	AA x aa
c) 3/4 dominant and 1/4 recessive	Aa x Aa

4. When looking at small numbers, it is best not to trust ratios although this ratio is indicative of the cross. It is better to realize that a recessive trait can appear from two dominant parents, but a dominant trait can never appear from two recessive parents. Therefore, tongue rolling must be a dominant trait.

5. A heterozygous red-flowered pea plant (Rr) crossed with a white-flowered plant (rr) should produce half white-flowered progeny. Half of 500 is 250.

6. If one parent develops Huntington's disease, then that parent is assumed to be heterozygous, since no cases of homozygous Huntington's disease have been reported. Therefore, the parents are Hh x hh, and the probability that a child of this cross will develop Huntington's disease is 1/2.

7. Although there is no mention of Huntington's disease being present in the parents, the fact that one of the offspring has the disease means that it must have been present in one of the parents. Since the mother is apparently normal, it must have been the father that was killed that carried the Huntington's allele. Therefore, the probability that the younger sister will also develop Huntington's disease is 1/2.

8. Because neither parent has cystic fibrosis and yet they give birth to a son with cystic fibrosis, the disease must be recessive, and both parents must be heterozygous (Cc x Cc).

9. Let R = rowdy and r = quiet. Let S = serious and s = playful. Now the homozygous, seriously rowdy dad will be RRSS and his quietly playful wife will be rrss. Their daughter will be RrSs. Her husband has the same genetic history, so he will also be RrSs. When they have kids, the cross will be RrSs x RrSs (a cross of two dihybrids).

a) The daughter and her husband will be seriously rowdy.

b) Their kids will be:

9/16 seriously rowdy

3/16 seriously quiet

3/16 playfully rowdy

1/16 playfully quiet

c) Adding rowdy groups, you get 9/16 + 3/16 = 3/4 rowdy

d) Adding playful groups, you get 3/16 + 1/16 =1/4 playful

Note that for c and d, these are the same answers you would have gotten if you had been following only one gene at a time.

10. Because the parents were homozygous, the F1 progeny are all dihybrids and are short-haired and solid color. Therefore, these two traits must be dominant. So let L = short-haired and l = long-haired; let C = solid color and c = spotted. The genotypes of the parents were llCC and LLcc and the F1 progeny were LlCc. When these dihybrids were crossed among themselves, the rarest class of progeny would be the class with both traits recessive, i.e., the long-haired, spotted class would be 1/16 of the progeny.

11. A short-haired, solid color guinea pig could be any of 4 genotypes (LLCC, LLCc, LlCC, or LlCc). The testcross (llcc) indicates 1/4 of each phenotype. The only one of the four genotypes capable of producing 4 different kinds of progeny in a testcross is LlCc.

12. Let A1 = blue, A2 = red, and A3 = white.

a) When light blue is crossed with pink (A1A3 x A2A3), their progeny should be 1/4 purple, 1/4 light blue, 1/4 pink and 1/4 white.

b) When purple and red were crossed (A1A2 x A2A2), their progeny should be 1/2 purple and 1/2 red.

c) When purple and pink were crossed (A1A2 x A2A3), their progeny should be 1/4 purple, 1/4 light blue, 1/4 red and 1/4 pink.

d) When red and light blue were crossed (A2A2 x A1A3), their progeny should be 1/2 purple and 1/2 pink.

13. Since the baby is type O, both the mother and the father of the baby must have given it a "i" allele. The lover cannot have a "i" allele and the ex-husband could. Assuming that these are the only two possible fathers for the baby, the baby must be the ex-husband's.

14. a) 1/2 A and 1/2 AB c) 1/4 A : 1/2 AB : 1/4 B e) all AB

b) 1/2 A and 1/2 B d) all O

15. Each child has a 1/2 chance of being a boy or a girl. This doesn't depend on the number of previous children of either sex.

16. Hemophiliac male's genotype: X^hY Normal female's genotype: X^HX^H

The probability that their son will have hemophilia is 0. The father cannot pass his X chromosome to his son. If he did, the child would be a daughter. Their daughter will also not be a hemophiliac, since she will have only one copy of the hemophilia allele. But she will be a carrier.

17. a) Because there is a difference in the phenotypic ratios between sons and daughters and females can show the trait, the colorblindness gene must be X-linked.

b) Because the daughters are all heterozygous and none is colorblind, the allele must be recessive.

18. Hairy ears must be Y-linked. The trait shows up differently in males and in females and the father passed the trait to his sons. Therefore, it must be on the Y chromosome.

19. The woman must be heterozygous, since she inherited the hemophilia allele from her father. When she marries a man with hemophilia, the cross would look like: X^hX^H x X^hY. Half the daughters and half the sons would be expected to have hemophilia.

20. ggRR x GGrr results in F1's that are all normal for both traits (GgRr). Because of the linkage, each F1 can produce only 2 kinds of gametes: gR and Gr. Putting these into a Punnet square results in 1 glassy-eyed normal-nosed : 2 normal : 1 normal-eyed red nosed progeny.