Rupert has discovered a new eye color in Drosophila and decided to call this ext
ID: 49046 • Letter: R
Question
Rupert has discovered a new eye color in Drosophila and decided to call this extremely bright eye “hot tamale”. To find out if this gene is on chromosome 2 he did a testcross of a heterozygous female to a male that is apterus (ap, chromosome 2), black (b, chromosome 2), and hot tamale (ht). The following progeny resulted.
+ + + 41 ap ht + 301
+ ht + 34 ap + + 307
+ + b 305 ap + b 46
+ ht b 300 ap ht b 36
a. (2 pts) Diagram this cross (using the correct gene order and coupling):
b. (2 pts) Identify the progeny as Parentals (P), recombinant (R), region 1 crossover (R1), region 2 crossover (R2), or double crossover (dco) as appropriate (you may not use all of these designations!)
c. (2 pts) Determine the map distance between all linked genes.
d. (4 pts) What gamete types would Rupert have seen if he crossed a heterozygous male to a homozygous recessive female? Use the inverse of the cross in part a as an example and be specific.
e. (6 pts) Would Rupert have been able to determine if the genes are linked or unlinked if he had used the cross suggested in part d? Set up the experiment by comparing the cross in part d with a cross in which all three genes are linked and provide specific evidence.
Explanation / Answer
Answer
a.& b.
ap + +
307 (parents)
+ + b
305 (parents)
ap ht +
301 (SCO)
+ ht b
300 (SCO)
ap + b
46 (SCO)
+ + +
41(SCO)
ap ht b
36 (DCO)
+ ht +
34 (DCO)
The genotypes with highest number will be no crossovers (NCO) or parental types. The two genotypes with lowest number will be double crossovers (DCO) and the rest re single crossovers (SCO).
The NCOs are:
ap + +
+ + b
The DCO are:
ap ht b
+ ht +
In order for this to be a double crossover, the “ht” allele must be in the center since it is differing and switching between the parental and double crossovers. Therefore, the gene order would be: ap ht b or b ht ap
c.
The map distance between ap and ht:
Recombination frequency = (number of recombinants/ total progeny)* 100%
= (46+41+36+34/1370) *100%
= 0.114*100
= 11.4 cm
The map distance between ap and b:
= (301+300+36+34/1370)*100%
=0.489*100
= 48.9 cm
ap + +
307 (parents)
+ + b
305 (parents)
ap ht +
301 (SCO)
+ ht b
300 (SCO)
ap + b
46 (SCO)
+ + +
41(SCO)
ap ht b
36 (DCO)
+ ht +
34 (DCO)
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