In humans, the sickle-cell trait is caused by a single defective allele, but sickle-cell disease only occurs in individuals that are homozygous for the sickle-cell allele. A man and woman each carry the trait, but do not have sickle-cell disease. What is the probability that their first two children will both have sickle-cell disease?

Question

A.  1/16
B.  1/8
C.  1/4
D.  3/8
E.  1/2

Clarify Question
· What is the key concept addressed by the question?
· What type of thinking is required?
· What key words does the question contain?

Gather Content
· What do you already know about monohybrid crosses? About determining the probability of two independent events?

Consider Possibilities
· What other information is related to the question? Which information is most useful?

Choose Answer
  · Given what you now know, what information and/or problem solving approach is most likely to produce the correct answer?

Reflect on Process
· Did your problem-solving process lead you to the correct answer? If not, where did the process break down or lead you astray? How can you revise your approach to produce a more desirable result?
 

Answer 1

A.  1/16

Clarify Question
· What is the key concept addressed by the question?
        o This question asks about a monohybrid cross and combines that with a probability problem.
· What type of thinking is required?
        o You need to Analyze the information to determine the probability of both the two children having sickle cell disease.
· What key words does the question contain?
        o Sickle-cell trait is the phenotype of heterozygous carriers of the allele (HbA / HbS). They have slightly altered blood cells but usually no apparent health effects. Sickle-cell disease is the phenotype of people homozygous for the mutant allele (HbS / HbS) – they have anemia due to poor circulation and oxygen transport.

Gather Content
· What do you already know about monohybrid crosses? About determining the probability of two independent events?
        o Since the parents carry the trait but do not have the disease, they must be heterozygous (HbA / HbS). The question asks for the probability that the first two children will both have the disease (both be HbS / HbS).

Consider Possibilities
· What other information is related to the question? Which information is most useful?
        o Did you recognize that this cross is a monohybrid cross, much like the example with Mendel’s purple pea flowers? Do you remember the classic phenotype ratio?
        o Punnett squares are also a useful way to determine both genotype and phenotype ratios.

Choose Answer
  · Given what you now know, what information and/or problem solving approach is most likely to produce the correct answer?
        o Try drawing a Punnett square. Write the possible gametes on the top and the side.
        o The gametes for each parent are HbA and HbS. There are four categories of offspring: HbA / HbA, HbA / HbS, HbS / HbA, and HbS / HbS.
        o Only the last category will have sickle cell disease: HbS / HbS. So there is a ¼ probability for each child.
        o Since each child is an independent event, we must multiply the two probabilities to get the overal probability. ¼ x ¼ = 1/16

Reflect on Process
· Did your problem-solving process lead you to the correct answer? If not, where did the process break down or lead you astray? How can you revise your approach to produce a more desirable result?
        o This question required you to Analyze the information given, using logic, to dissect the problem and determine the answer.
        o Were you thrown by the unfamiliar terminology?
        o Did you try writing out the genotypes and the cross? Did you draw a Punnett square?
        o Did you remember to treat each child as an independent event? The outcome of the first child does not affect subsequent children.
        o Did you read carefully? Would the answer have been different if the question said “What is the probability that either of their first two children will have sickle-cell disease?”