Question 5 points 10 out of 10 if william and linda

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Question 5 Points: 10 out of 10 If William and Linda have a daughter, what is the probability that she will have this condition? 50% 75% 0%
100% Feedback Correct. We know William’s genotype is XhY, and Linda’s is XHXh. Crossing XhY and XHXh results in 50% of daughters that are XHXh and do not have the condition (but are carriers), and 50% that are XhXh and do have the condition. Question 6 Points: 10 out of 10 Traits that are sex-linked and recessive are ________________________.
Question 7 Points: 0 out of 10 What is the genotype of the unshaded individuals in the pedigree below?
Question 8 Points: 10 out of 10 Sickle cell anemia is an autosomal recessive disorder. A person with the disorder and a person that is a carrier mate. ( A carrier has the allele for the trait but does not express it .)
What is the chance that their offspring will develop sickle cell anemia?
Question 9 Points: 10 out of 10 Nondisjunction in meiosis II results in: ______________________ Two cells missing a chromosome and two cells with an additional chromosome. One cell missing a chromosome, one cell with missing a chromosome and two healthy cells.
Two cells missing a chromosome and two healthy cells. Two cells with an additional chromosome and two healthy cells. Feedback Correct. Meiosis I occurred correctly, so 2 of the cells will have the correct number of chromosomes.

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