In a two-point cross to map genes A and B, you obtained 98 recombinant types and 902 parental types among the offspring. How far apart are these genes?
A. 9.8 cM
B. 0.98 cM
C. 90.2 cM
D. 9.02 cM
E. .098 cM
Clarify Question
· What is the key concept addressed by the question?
· What type of thinking is required?
· What key words does the question contain and what do they mean?
Gather Content
· What do you already know about two-point mapping?
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?
A. 9.8 cM
Clarify Question
· What is the key concept addressed by the question?
o This question addresses a two-point mapping cross.
· What type of thinking is required?
o This question is asking you to take what you already know and apply it to this unfamiliar situation.
· What key words does the question contain and what do they mean?
o A two-point cross is a cross involving the alleles of two genes.
o Recombinant types are chromosomes with new allele combinations, different than the parents carried.
o Parental types are chromosomes with the same allele combinations as the parents carried.
Gather Content
· What do you already know about two-point mapping?
o Genetic distances are described in centiMorgans (cM).
o One cM represents 1% recombination between the genes.
o Recombination events are recognized by counting the number of recombinant chromosomes.
Consider Possibilities
· What other information is related to the question? Which information is most useful?
o To calculate the distance in cM, just determine what % of progeny have recombinant chromosomes.
Choose Answer
· Given what you now know, what information and/or problem solving approach is most likely to produce the correct answer?
o From this cross, the progeny were 98 recombinant types and 902 parental types.
o 98 + 902 = 1000 total progeny
o Recombinant progeny = 98/1000 = 9.8/100 = 9.8%
o Since 1 cM represents 1% recombination, we can say the distance is 9.8 cM.
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 asked how many cM apart are genes A and B, based on the recombination between them.
o The question required you to take what you already know and apply it to this unfamiliar situation.
o Did you recognize that the distance in centiMorgans is just the percent recombination?
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