Question 86 - (a) Male with genotype GGmm and phenotype gray wingless mates with

Question 86 - (a) Male with genotype GGmm and phenotype gray wingless mates with female with genotype ggMM and phenotype black winged in fruit flies. G is dominant to g in color. M is dominant to m in wing shape. If the actual distribution of the second generation of the fruit flies was as follow : 890 gray wingless, 900 black winged, 115 gray winged, 95 black wingless, calculate the recombination frequency betwen the two genes and distance in recombination units. Let 1 map unit = 1 % recombination. (b) A DNA molecule has 180 base pairs and 20 % adenine. How many cytosine nucleotides are present in this molecule of DNA?

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Question 86 - (a) Male with genotype GGmm and phenotype gray wingless mates with female with genotyp..

Answer / kangchuentat

Answer 86 - (a) Phenotypes of parents = gray wingless, black winged. Phenotypes of recombinant children = gray winged, black wingless. Number of recombinant children = gray winged + black wingless = 115 + 95 = 210. Recombination frequency between the two genes = Number of recombinant children / Total number of childen = 210 / (210 + 890 + 900) = 0.105 or 10.5 % = distance = 10.5 map units. (b) Base pairs number of adenine = 180 x 0.2 = 36 = base pairs number thymine. Base pairs number of guanine = base pairs number cytosine = (Total number of base pairs - 2 x Base pairs number of adenine) / 2 = (180 - 2 x 36) / 2 = 54. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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