Question

3. If the couple above (question 2) have more children, what is the probability that their subsequent sons will be blood type A? A.096 B. 100% C. 75% D. 50% E.25% 4. HH ferbert is hetetozygous for "male pattem baldness. His wife, Sheri, is also bald, due to the male sex-influenced trait. The baldness allele (m) is dominant in males and recessive in females. (Recall that gene. Their young son, Frank, still has all of his hair. Recall that pattem baldness is a genotypes are m/m, m/+, and +/+. See the Chapter 6 PowerPoint slide about male pattern baldness for more details.) What is the probability that Frank will lose his hair when he is older? a. A. 100% B. 75% C.50% D. 25% 5. If Herbert and Sherri have a daughter, what is the probability she will be bald? A. 100% B. 75% C. 50% D. 25% FS F7 F8 F5 F6 F3

i did number 2 but i need help with 3 the same scenerio applies for question 2&3

Answer 1

Mother's Blood type is A. The allele may be AA or AO. The chance of child gets type A from mother is 75% and Type O is 25%

Father's Blood type is AB. The allele is AB. The chance of child gets type A from father is 50% and type B is 50%.

so

• The chance that their child's genotype is AA is 75% * 50% = 37.5%.
• The chance that their child's genotype is AB is 75% * 50% = 37.5%.
• The chance that their child's genotype is A0 is 25% * 50% = 12.5%.
• The chance that their child's genotype is B0 is 25% * 50% = 12.5%

Add the results for AA and A0 together.

• The chance for A blood group is 37.5% + 12.5%' = 50%.

Note: The 50% of children may males or females