In the fruit fly, Drosophila melanogaster, wild type female fly is mated to a ma

Question

In the fruit fly, Drosophila melanogaster, wild type female fly is mated to a male with the three autosomal recessive traits scarlet (red body, ss), long (long body, ll) and hairless (no thoracic bristles, hh). Phenotypically wild-type F1 female progeny were mated to fully homozygous (mutant) males. a. If the genes were completely linked, what phenotypic ratio would you observe? b. If the genes were completely linked, what phenotypic ratio would you observe? c. The following progeny (1000 total) were actually observed. Draw a genetic map showing the gene order and distances between gene pairs. d. Calculated the interference.

Explanation / Answer

1. in case of unlinked gene, phenotypic ratio observed was 27:9:9:9:3:3:3:1. In the case of trihybrid cross the 8 female gamete cross with 8 male gametes will give 64 progeny. Here, all the gene are unlinked and seggregare independently.

2 In the case of linked gene, four pair of phenotypic classes formed

One parental class : most frequent

Two single recombinant (I &11) classes between the middle locus and each outside locus

Double recombinant class between middle locus and both outside loci

SsHhLl X sshhll in this following cross

Phenotype Class

SHL parental

SHl double cross over

ShL single cross over II

Shl single cross over I

sHL single cross over I

sHl single cross over II

shL double cross over

shl parental

3. To construct a genetic map , you have to first calculate cross over frequency or recombination frequency

slH 404 II

SLh 422 II

lSH 18 D

sLh 16 D

lSh 75 I

sLH 59 I

SLH 2 P

slh 4 P

For S &L

SL and sl are parental type

Sl and sL are recombinant type I

Count all class with this type

SHl +shL+sHL+Shl

18+16+59+75/1000

.168 = 16.8% = 17% = 17cM

For H & L

HL and hl are parental type

Hl and hL are recombinant type II

SHl + shL+ShL+sHl

18+16+422+404/1000

.86 = 86% = 86cM

For S& H

ShL+sHl+Shl+sHL

422+404+75+59/1000

.96 = 96% = 96cM

The distance between S & H is greatest so these are outside

S....17cM.........L........................................86cM..................................................H

but .96< (.17+.86)

because in double recombinants resembles parental type but each represents 2 cross over between outside S &H So corrected count of cross over is 422+404+75+59+2(18+16)

= 102.8 = 103

d) interference occur when cross over decreases the probability of others nearby

1- (observed/expected)

1-(18+16)/.17).86)1000

1-.23 = .77

here interference occur due to oserved not equal to expected

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