The data below involve three mutations: brown eyes ( be ), black body ( blk ) an
ID: 321696 • Letter: T
Question
The data below involve three mutations: brown eyes (be), black body (blk) and vestigial wings (vg). The wild-type alleles be+ (red eyes), blk+ (gray body), and vg+ (long wings) show complete dominance. In each cross, repulsion heterozygotes were mated to flies homozygous for the recessive alleles of the genes. The progeny were sorted as having either a parental phenotype or a recombinant phenotype.
Calculate the frequency of recombination for each cross:
Parental
Recombinant
Recombination
Cross
Heterozygote
Progeny
Progeny
Frequency %
1
be blk+/be+ blk
1105
579
2
blk vg+/blk+ vg
1077
400
3
be vg+/be+ vg
1054
145
The alelles for all three mutations are linked. How would you know this?
Parental
Recombinant
Recombination
Cross
Heterozygote
Progeny
Progeny
Frequency %
1
be blk+/be+ blk
1105
579
2
blk vg+/blk+ vg
1077
400
3
be vg+/be+ vg
1054
145
Explanation / Answer
Answer:
The distance between the genes is equal to the frequency.
% of recombination frequency = distance between the genes (map units)
If the distance between the genes is 50 map units or more, then the genes are not linked. If the distance is less than 50 mu, then the genes are linked.
Parental Recombinant Recombination Heterozygote Progeny Progeny Total progeny Frequency % 1 be blk+/be+ blk 1105 579 1684 34.38% 2 blk vg+/blk+ vg 1077 400 1477 27.08% 3 be vg+/be+ vg 1054 145 1199 12.09%Related Questions
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