Use this information for questions 18-20: Sickle cell anemia is an autosomal rec

Question

Use this information for questions 18-20: Sickle cell anemia is an autosomal recessive disease. A couple with a family history of sickle cell anemia is contemplating having a baby. The couple had genetic testing performed and discovered that they are both carriers for the disease. 18. If the couple has one child, what is the probability that this child will have sickle cell anemia? b. i c. 19. If the couple has two children, what is the probability that the first and second child will both have sickle cell anemia? a. V c. 1/8 d. 20. If the couple has two children, what is the probability that the first and second child will both have normal blood cells? b. 3/4 Bonus (2 pts): If the couple decides that they plan to have three children, what is the probability that one of the three children will have sickle cell anemia. (This might help you. Think of this situation as the probability of having one sickle cell child and two normal children with either the first child, the second child, or the third child being the one to have the disease. We did an example like this in class.) Show your work. b. 9/64 c. 27/64

Explanation / Answer

Please find the answers below:

According to the information, both the parents are carrier for the disease i.e. they are heterozygous for the trait. Thus, their genotypes would be given by Ss and Ss, each, where S reprsents dominant allele and s represents recessive allele. Thus, the Punnett square for the couple can be designed as below:

Genotypic ratio: SS: Ss: ss :: 1:2:1

Phenotypic ratio: Normal : carrier : affected :: 1:2:1 or normal: affected :: 3:1

Answer 18: Choice a (as shown in the Punnett square above, the probability of having an affected child is 1/4)

Answer 19: Choice b (the probability for having first and second affected childs would be 1/4 *1/4 each i.e. 1/16)

Answer 20: Choice b (similar to the probability of having two affected children, the probability of having two children with normal cells would be 1/4*1/4 each or 1/16)

Answer: Choice b (the probability of having only one children affected is 1/4, the probability of having second child diseased 1/4 and the probability of having second and third children normal is 3/4 * 3/4. Thus the total probability is 1/4*3/4*3/4 or 9/64 )

S s S SS Ss s Ss ss

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