genprobans - GENETICS PROBLEMS BIOLOGY 16 - ANSWERS 1. a....

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GENETICS PROBLEMS BIOLOGY 16 - ANSWERS 1. a. genotype = all Ww; phenotype = all white b. genotype = 1 Ww : 1 ww; phenotype = 1 white:1 yellow c. genotype = 1 WW :2 Ww : 1 ww; phenotype = 3 white : 1 yellow 2. Let W be the allele for white color and w be the allele for yellow color. Anther is the male gamete former so this genotype is Ww and the female gametes are formed by ww cells. Therefore, the cross is Ww X ww. genotype = 1 Ww: 1ww; phenotype = 1 white : 1 yellow. 3. Let B be for brown eyes and b for blue eyes. The man must be bb but the woman is B-. Since her father is bb, she must have a b allele. So she is Bb. Now, the cross is bb X Bb and 50% blue eyes are predicted. 4. This time the man is B- and the woman is bb. The cross is either Bb X bb, or BB X bb. The first cross would yield 50% blue-eyed children. The second cross would yield all brown-eyed children; every child would receive a dominant B allele from the father. The father is probably homozygous, but one cannot be sure that this is true. Furthermore, an eleventh blue-eyed child is not predictive of the father’s genotype. It may be that there is some other factor (i.e., gene) prohibiting a B allele from being expressed. Remember, too, the father may be heterozygous and each of the first ten children got the B allele. 5. The brown-eyed man is Bb because his mother was bb. His father is either BB or Bb. His wife is bb, so her brown-eyed parents must both be heterozygous, Bb. The blue-eyed son is bb. 6. Probably a number of explanations are possible. But, it is best to assume only one genetic locus at first. This is probably true since only tail length is involved. The ratio is 3:6:2, which is similar to 1:2:1 (like a monohybrid cross). Assume that T 1 T 1 yields a long-tailed cat and T 2 T 2 yields no tail. This would mean that the parental short-tailed cats are heterozygous T 1 T 2. 1
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Assume that there is no dominance, with short tails, a hybrid. So, the progeny would be 1 T 1 T 1 : 2T 1 T 2 : 1 T 2 T 2 . 7. Assume only a single locus. The ratio of pups (haired : hairless : deformed) is 1:2:? - much like a monohybrid cross ratio. This implies that the hairless dogs are monohybrid, H 1 H 2 . Haired pups would be H 1 H 1 . The H 2 allele would be lethal in the homozygous condition. A normal X hairless cross (H 1 H 1 x H 1 H 2 ) would yield 1H 1 H 1 :1H 1 H 2 , which agrees with the first statement. 8.
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genprobans - GENETICS PROBLEMS BIOLOGY 16 - ANSWERS 1. a....

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