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Medgar Evers College Bio 302.001 ESSAY. write your answer in e space provided or

ID: 307352 • Letter: M

Question


Medgar Evers College Bio 302.001 ESSAY. write your answer in e space provided or heet paper. a male that is claret a separate of mated to Dnesawhile melanogaster, a spineless (no female fly is were F1 female progeny dark eyes) and hairless (no thoraeie Phem homozygous mutam) males and the following progeny (tox spineless claret, spineless 18 claret hairless hairless, daret, spineless with respect to the three genes mentioned in the problem what are the genotypes ofthe bo parents used in making the phenotypically wild (c) What are the map distances between the three genes? (d) What is the coefficient of coincidence?

Explanation / Answer

Answer:

(1) (a) Spineless (s + +) and Claret hairless (+ c h) are the parentals. Claret (+ c +) and hairless,spineless (s + h) are the double cross overs.

So, Parentals: s + + + c h

Double crossovers: + c + s + h

We get, s+ s+, +c +c and h+ +h.

Since, h is different, h (hairless) is in the middle.

(b) The genotypes of the homozygous parents used in making the phenotypically wild F1 heterozygote:

s + +/ s + + and + c h/ + c h

(c) The map distances have been calculated as under:

s-h (Spineless – hairless): (32 + 38 + 12 + 18 (Number of recombinants)/1000) * 100

= (100/1000) * 100 = 10% = 10 mu

h-c (hairless-claret): (140 + 130 + 12 + 18)/1000) * 100

= (300/1000) * 100 = 30% = 30 mu

(d) Expected double cross overs = [recombination frequency in region 1 (map units / 100)] X [recombination frequency in region 2 (map units / 100)]

= .(10/100) * (30/100)

= 0.1 * 0.3 = 0.03

Coefficient of coincidence = observed double crossovers/expected double crossovers

= [(18 + 12)/1000] / 0.03

= (30/1000) / 0.03

= 0.03 / 0.03 = 1