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Section 4: Graded Questions K 111 e Mendelian Pigs a NOTES QUESTION Section 4: G

ID: 166483 • Letter: S

Question

Section 4: Graded Questions K 111 e Mendelian Pigs a NOTES QUESTION Section 4: Graded Questions Graded Questions These questions will be graded by your instructor. Your score will be available on the Home screen when your instructor publishes scores for this module. Click "Submit All' at the bottom of the page to turn in your answers. HINT Please pay particular attention to the words homozygous and heterozygous in the questions below. Students sometimes confuse these terms and get the answer wrong even when they know the comectanswer Q4.1. suppose a gene has two alleles, one of which is dominant over the other. An individual whose genotype is homozygous for the dominant allele has which of the following? Two copies of the allele that determines phenotype whenever the allele is present Two copies of the allele whose effect is hidden unless the other allele is absent e One copy of the allele that determines phenotype when present and one copy of the other allele One copy of each of two alleles that both contribute equally to determining phenotype Use the following information for the next 2 questions. Pigeons have two alleles of a single gene that determines whether they have feathers on their lower legs. The alleles are called no grouse (associated with the featherless phenotype) and grouse (associated with feathery legs) Q4.2. Suppose a pigeon that is homozygous for the no grouse allele mates with a heterozygous pigeon. What is the expected frequency of the homozygous no grouse GENOTYPE in the offspring? 50% e 75% 100% Q4.3. The no grouse allele is dominant over the grouse allele. If a pigeon homozygous for the no grouse allele mates with a heterozygous pigeon, what is the expected frequency of the feathery-legged (grouse PHENOTYPE in the offspring?

Explanation / Answer

Q4.1. Two copies of the allele that determines the phenotype whenever the allele is present

Q4.2.50%

Explanation

N=no grouse

n=grouse

NN x Nn

NN, Nn, NN,Nn

Q4.3. 0%

NN x Nn

NN, Nn, NN,Nn (all are no grouse)

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