In the fruit fly Drosophila, there is a dominant gene for normal wings and its recessive allele for vestigial wings. At another gene locus on the same chromosome, there is a dominant gene for red eyes and its recessive allele for purple eyes. A male that was heterozygous at both gene loci was mated with a female that was homozygous for both recessive alleles and the following results were observed among the offspring:Normal wings and red eyes - 420Vestigial wings and red eyes - 80Normal wings and purple eyes - 70Vestigial wings and purple eyes - 430According to these data, what is the distance, in centimorgans, between these 2 gene loci? (Enter the number only without the units. For example, 100 cM would be entered as 100)

Question

Fill in the blank(s) with the appropriate word(s).

Answer 1

15

Clarify Question
• What is the key concept addressed by the question?
        o This question addresses linkage mapping by recombination.
• What type of thinking is required?
        o Although you may not have seen this example before, you know enough about the concept to Apply your knowledge and understanding to this unfamiliar situation.
• What key words does the question contain?
        o Dominant alleles are expressed in the heterozygote; recessive alleles are only expressed in the homozygote. Centimorgans are a unit of map distance. 1 centimorgan is equivalent to a 1% recombination frequency.

Gather Content
• What do you already know about linkage mapping by recombination?
        o Linkage mapping uses recombination frequencies as a proxy for distance along the chromosome.

Consider Possibilities
• What other information is related to the question? Which information is most useful?
        o Centimorgans are the unit of map distance. 1 centimorgan is equivalent to a 1% recombination frequency.

Choose Answer
• Given what you now know, what information and/or problem solving approach is most likely to produce the correct answer?
        o To analyze the data, we should first determine which were the likely parental chromosomes in the father. Since vg+ pr+ / vg pr (normal wings and red eyes) and vg pr / vg pr (vestigial wings and purple eyes) are the most common classes, those are the original paternal chromosomes.
        o So the recombinant chromosomes are vg pr+ / vg pr (vestigial wings and red eyes) and vg+ pr / vg pr (normal wings and purple eyes).
        o The recombinant chromosomes constitute the fraction of progeny: 80 + 70 / 420 + 80 + 70 + 430, or 150 / 1000.
        o To determine the percentage recombination frequency, 150 / 1000 = 15 / 100 = 15%.
        o Since 1% recombination = 1 cM, then 15% recombination is 15 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 required you to Apply your knowledge and understanding to this unfamiliar situation.
        o Did you remember that 1% recombination = 1 centimorgan?
        o Did you recognize that you needed to determine which two classes were the recombinant classes?
        o Did you determine the percent recombination using the equation : 80 + 70 / 420 + 80 + 70 + 430, or 150 / 1000?