Question

question 17
hemophilia, a blood clotting disorder in humans, is due to an x-chromosome mutation. what will be the results of mating between a normal (non-carrier) female and a hemophiliac male?
(hint: this is a sex - linked trait)
○ a all sons are normal, and all daughters are carriers (heterozygous).
○ b half of daughters are normal and half of sons are hemophiliac.
○ c half of sons are normal and half are hemophiliac; and all daughters are (heterozygous).
○ d all daughters are normal, and all sons are carriers (heterozygous).

question 18
in rabbits, white fur (w) is dominant to black (w), and long ears (e) are dominant to short ears (e). a breeder mates two rabbits that are heterozygous (wwee) for both traits.
what is the chance that the offspring will be white with long ears?
(you may use the easy quick method to solve. hint = set up 2 monohybrid crosses and multiply)
image of a white rabbit punnett square with wwee and columns/rows we, we, we, we

Explanation:

Question 17

Step 1: Determine Genotypes

• Normal (non - carrier) female: \(X^H X^H\) (where \(H\) is the normal allele, \(h\) is the hemophilia allele)
• Hemophiliac male: \(X^h Y\)

Step 2: Analyze Gametes

• Female gametes: All \(X^H\)
• Male gametes: \(X^h\) and \(Y\)

Step 3: Predict Offspring

• Sons: Get \(X^H\) from mother and \(Y\) from father, genotype \(X^H Y\) (normal)
• Daughters: Get \(X^H\) from mother and \(X^h\) from father, genotype \(X^H X^h\) (carrier, heterozygous)

Step 1: Analyze Fur Trait (Ww x Ww)

• Using Punnett square for \(Ww\times Ww\):
• Genotypes: \(WW: Ww: ww = 1:2:1\)
• Phenotypes: White (\(WW, Ww\)) probability \(=\frac{3}{4}\), Black (\(ww\)) probability \(=\frac{1}{4}\)

Step 2: Analyze Ear Trait (Ee x Ee)

• Using Punnett square for \(Ee\times Ee\):
• Genotypes: \(EE: Ee: ee = 1:2:1\)
• Phenotypes: Long ears (\(EE, Ee\)) probability \(=\frac{3}{4}\), Short ears (\(ee\)) probability \(=\frac{1}{4}\)

Step 3: Combine Probabilities

• Probability of white (\(\frac{3}{4}\)) and long ears (\(\frac{3}{4}\)): Multiply the two probabilities \(\frac{3}{4}\times\frac{3}{4}=\frac{9}{16}\)

Answer:

a. All sons are normal, and all daughters are carriers (heterozygous).

Question 18