14
1.
1. e
2. b
3. f
4. a
5. g
6. c
7. d
2. In this example, we would use R to indicate the dominant red allele and r for the recessive ebony allele. The heterozygous red-eyed fly has a genotype of Rr, and the ebony-eyed fly must be rr.
The percentage of offspring with the recessive ebony eyes (rr) would be 2⁄4, or 50%.
3. If the male cat is heterozygous for both traits, his genotype must be BbSs. The female shows both recessive phenotypes, so her genotype must be bbss. When determining the odds that the kitten will have white, long fur (bbss), create two separate Punnett squares to determine the odds of the kitten having either white fur or long fur.
In regards to fur color, the odds of producing a white kitten are 50%, or ½. Now set up a second Punnett square looking only at the fur length:
In regards to fur length, 25% (or ¼) of the kittens would have long fur. The question is asking you about the odds of producing a white kitten with long fur, so using the rule of multiplication, multiply the odds of producing each of these traits independently:
½ × ¼ = ⅛ . . . there’s a 1 in 8 chance of these two cats producing a white, long-haired kitten!
4. The man is XCY (he has the gene for normal color vision), and you know that the woman must be XCXc because her father is color-blind (and the father provides the Xc chromosome to his daughter).
Of their four offspring, they could produce a colorblind boy (¼ chance).
5. The purple allele is dominant because it masks the white allele in the second generation (they are all purple).
6. False. It will always result in all the offspring having the dominant phenotype (because all the offspring are heterozygous and the dominant allele will be the “winner”).
7. The white rooster (recessive phenotype) is heterozygous for a large comb, so his genotype is bbLl. The heterozygous hen is BbLl. In regards to feather color:
The odds of creating a chicken with white feathers is ½.
In regards to comb size:
There is a ¾ chance that the offspring will have a large comb. Therefore, the odds of creating a chick with both white feathers and a large comb is ½ × ¾ = ⅜.
8. The color-blind man has the genotype XcY. The woman must be XX. Therefore:
In order for a daughter to be a carrier, they must have one allele for color blindness (XcX). There is a 100% chance that their daughter will be a carrier for color blindness.
9. d. If you cross a homozygous dominant (e.g., AA) with a heterozygous individual (Aa), all offspring would have at least one of the dominant alleles; therefore, all offspring would exhibit the dominant phenotype.