Linkage and Recombination 1. A plant of genotype AB/ab is testcpssed to ab/ab. I
ID: 218902 • Letter: L
Question
Linkage and Recombination 1. A plant of genotype AB/ab is testcpssed to ab/ab. If the two loci are 10 map units apart, what proportion of progeny will be AB/ab? 2. The mutations cinnabar (on, bright red eyes) and vestigial (wg, malformed wings) are linked on the second chromosome of Drosophila. Among 1000 progeny of the cross cn vg++ femesXcn g cn g males, the following genotypes of progeny were observed. From these data, estimate the frequency of recombination between the cn and yg genes. 445 45 +cn yg 455 3. In addition to go and yg, the genes curved (c, curved wings) and plexus (px, extra wing veins) are linked in the second chromosome of Drosophila. In a cross of cp c px / + + + females X gn c px / n c px males, the following progeny were counted. cn c px/cn cpx 296 +cn c px cn c+cn cpx 63 + pxI cn c px 329 82 cncn cpx 119 + c px /cn c px 86 Total 1000 a. What is the frequency of recombination between cn and c? b. What is the frequency of recombination between c and px? c. What is the frequency of recombination between cn and px?Explanation / Answer
Answer:
1).
AB/ab x ab/ab ------Test cross
Distance between the genes = Recombination frequency
10 mu = 10% recombinants
AB & ab are non recombinants = 90% (each 45%)
Ab & aB are recombinants = 10% (each 5%)
ab
AB
AB/ab—45%
Ab
Ab/ab—5%
aB
aB/ab—5%
Ab
ab/ab—45%
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2).
Hint: Always recombinant progeny are less in number.
Recombination frequency = (no. of recombinants / Total progeny) 100
= (100/1000)100
= 10%
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3).
Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.
Hence, the parental (non-recombinant) genotypes is cn c px / + + +
1).
If single crossover occurs between cn & c.
Normal combination: cn c / ++
After crossover: cn +/+ c
cn + progeny= 119+10 = 129
+ c progeny= 86+15=101
Total this progeny = 230
The recombination frequency between cn&c = (number of recombinants/Total progeny) 100
RF = (230/1000)100 = 23%
2).
If single crossover occurs between c&px.
Normal combination: c px / ++
After crossover: c+/+px
c+ progeny= 63+15=78
+px progeny = 82+10=92
Total this progeny = 170
The recombination frequency between c&px = (number of recombinants/Total progeny) 100
RF = (170/1000)100 = 17%
3).
If single crossover occurs between cn&px.
Normal combination: cn px/++
After crossover: cn+/+px
cn + progeny= 63+119=182
+ px progeny = 82+86=168
Total this progeny = 350
The recombination frequency between cn&px = (number of recombinants/Total progeny) 100
RF = (350/1000)100 = 35%
Recombination frequency (%) = Distance between the genes (cM)
cn----------27cM------c-----17cM------px
Expected double crossover frequency = (RF between cn & c) * (RF between c & px)
= 0.27 * 0.17 = 0.0459
The observed double crossover frequency = 10+15 / 1000 = 0.025
Coefficient of Coincidence (COC) = Observed double crossover frequency / Expected double crossover frequency
= 0.025 / 0.0459
= 0.55
Interference = 1-COC
=1-0.55= 0.45
ab
AB
AB/ab—45%
Ab
Ab/ab—5%
aB
aB/ab—5%
Ab
ab/ab—45%
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