Problem

Solutions For An Introduction to Genetic Analysis Chapter 4 Problem 26P

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Solution 1

In the fruit fly Drosophila, the allele dp⁺determines long wings and the allele dp determines short wings. A separate locus e⁺determines the gray color of the body and e determines the ebony color of the body. Both the dp and e loci are autosomal.

Parents: long ebony ? x short, gray ?

F ₁ progeny: long, gray ?

The F₁ is crossed with a homozygous recessive short, ebony ?

The F₂ progeny is:

Long, ebony – 54

Long, gray - 47

Short, gray – 52

Short, ebony – 47

Total – 200

a) To use the ◊^◊◊test, we need to have a hypothesis. The hypothesis is that the genes are not linked. When a homozygous recessive organism is crossed with the F₁ progeny, a 1:1:1:1 ratio is expected.

Step #2 of 6

b) To calculate ◊^◊◊◊we need the expected value along with the observed values. The expected value is 50 from the F₂ progeny data.

Step #3 of 6

c) Since, there are four phenotypes, the number of degrees of freedom would be 3 [4 – 1]. The p value with the ◊^◊◊value of 0.76 and df of 3 would be between 0.50 and 0.90.

Step #4 of 6

d) A p value between 50% and 90% indicates that the hypothesis might be compatible with the results seen.

Step #5 of 6

e) My conclusion would accept the initial hypothesis.

Step #6 of 6

f) Since the ◊^◊◊value was insignificant, we can infer that the genes assort independently. The genotypes of the parents and the resulting crosses would be:

Parents: dp⁺/dp⁺; e/e x dp/dp; e⁺/e⁺

F ₁ progeny: dp⁺/dp; e⁺/e

Tester: dp/dp; e/e

Progeny:

Long, ebony – dp⁺/dp; e/e

Long, gray – dp⁺/dp; e⁺/e

Short, gray – dp/dp; e⁺/e

Short, ebony – dp/dp; e/e

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