ll Aha are under 3.5 feet tall. They produce a child of normal size. Describe th
ID: 182423 • Letter: L
Question
ll Aha are under 3.5 feet tall. They produce a child of normal size. Describe the alleles for the father, mother, and their child, respectively. Justify your answer. Essay #15: (OutlinedueJ an. 3; Essay due Jan. 8) The following data was collected using Drosophila melanogaster and two traits. In the parental cr that were homozygous dominant for both traits were crossed with flies that were homoz for both traits. The two traits that were analyzed were wild type wings (dominant) versus vestigial (recessive) and red eyes (dominant) versus cinnabar eyes (recessive). After the initial cross, Fi flies were crossed. e wings (dominant) yerous recessive Phenotypes Parent FliesFi Flies 2 Flies Wild type wings, red eyes Wild type wings, cinnabar eyes Vestigial wings, red eyes Vestigial wings, cinnabar eyes Totals 500 1200 985 45 35 310 1400 500 1000 1200 a. Describe the most probable explanation for how these two traits are inherited b. Explain how this data suggests that mode of inheritance. c. Explain the evidence that shows the recombination of these two traits. d. Explain how this mode of inheritance differs from the conclusions of Gregor MendelExplanation / Answer
Wild type wing= W
Vestigial wing= w
Red eyes = R
Cinnabar eye = r
Wild type wing & Red eyes (W---R / W---R) x Vestigil wings & Cinnabar eyes(w—r / w---r)—Parents
Wild type wing & Red eyes (W---R / w---r) -------------------------F1
It is a Dihybrid cross. According to mendel laws, the phenotypic ratio of F2 generation is 9:3:3:1, when the two genes of the traits assort independently (not linked)
W_R_ = 9
W_rr = 3
wwR_ = 3
wwrr = 1
But the given ratio is not like that. Hence, it is concluded that the gene are linked.
The distance between the genes is equal to the recombination frequency (number of recombinants)
Number of Recombinants = 45 + 35 =80
Non-Recombinants = 985+310 = 1295
Total = 1400 (but in the question sum of all F2 files gives 1395)
Recombination frequency (RF)= (number of recombinants / Total number of progeny) 100
RF= (80/1400) 100
RF = 5.71%
RF (%) = Distance between the genes (map units)
Therefore, the two genes are linked and the distance between the genes is 5.71 map units. Hence, the inheritance of the two traits deviates the Mendel laws.
The linkage reduces the tendency of recombination between the genes. It means that linkage is inversly proportional to recombinantion frequency. The distance between the genes is directly proportional to the recombination frequency.
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