E4 - E1. Mexican hairless dogs are heterozygous for a...

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E1. Mexican hairless dogs are heterozygous for a dominant allele that is lethal when homozygous. In a cross between two Mexican hairless dogs, we expect 1/4 to be normal, 1/2 to be hairless, and 1/4 to die. E2. Chinchilla 1 is heterozygous c ch c . Chinchilla 2 is heterozygous c ch c h . Chinchilla 3 is heterozygous c ch c . Chinchilla 4 is probably c ch c ch because it always produces chinchilla offspring when mated to chinchilla 1 or 2, which are heterozygous. However, it is possible that chinchilla 4 is also heterozygous c ch c h or c ch c . Chinchilla 5 is c ch c h or c ch c . As noted, we are not sure about the genotypes of chinchillas 4 and 5. E3. There may be two redundant genes that are involved in feathering. The unfeathered Buff Rocks are homozygous recessive for the two genes. The Black Langhans are homozygous dominant for both genes. In the F 2 generation (which is a double heterozygote crossed to another double heterozygote), 1 out of 16 offspring will be doubly homozygous for both recessive genes. All the others will have at least one dominant allele for one of the two (redundant) genes. E4. The first offspring must be homozygous for the horned allele. The father’s genotype is still ambiguous; he could be heterozygous or homozygous for the horned allele. The mother’s genotype must be heterozygous because her phenotype is polled (she cannot be homozygous for the horned allele) but she produced a horned daughter (who must have inherited a horned allele from its mother). E5. The reason why all the puppies have black hair is because albino alleles are found in two different genes. If we let the letters A and B represent the two different pigmentation genes, then one of the dogs is AAbb and the other is aaBB. Their offspring are AaBb and therefore are not albinos because they have one dominant copy of each gene. E6. It is a sex-limited trait where W (white) is dominant but expressed only in females. In this cross of two yellow butterflies, the male is Ww but is still yellow because the white phenotype is limited to females. The female is ww and yellow. The offspring would be 50% Ww and 50% ww. However, all the males would be yellow. Half of the females would be white ( Ww ) and half would be yellow ( ww ). Overall, this would yield 50% yellow males, 25% yellow females, and 25% white females. E7. The sandy variation may be due to a homozygous recessive allele at one of two different genes in these two varieties of sandy pigs. Let’s call them genes A and B. One variety of sandy pig could be aaBB and the other AAbb. The F 1 generation in this cross would be heterozygotes for both genes and are all red. This tells us that the A and B alleles are dominant. In the F 2 generation, 6 out of 16 will be homozygous for either the aa or bb alleles and become sandy. One out of 16 will be doubly homozygous and be white. The remaining 9 will contain at least
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This note was uploaded on 04/07/2009 for the course BIO 325 taught by Professor Saxena during the Spring '08 term at University of Texas at Austin.

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E4 - E1. Mexican hairless dogs are heterozygous for a...

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