tons 14 and 15 reter to the folloing: nDrosophila, fomales heterozygous for the
ID: 271940 • Letter: T
Question
tons 14 and 15 reter to the folloing: nDrosophila, fomales heterozygous for the recessive alleles a and were crossed to males homozygous recessive for these tnre mutant genes. The phenot ypes of the resulting progeny were: 324 b e 26 26 ab+ 146 146 dobe arc Total progeny 1000 14. How were the three pairs of alleles arranged in the heterozygotes females? a) +b+b) abe (c) d) ab+ atc 15. What is the order of the genes? In this question, the order used in the above data may be incorrect. a )sacb b) abc c) bacExplanation / Answer
Answer:
14) c). +bc/a++
15). a). acb
Explanation:
Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.
Hence, the parental (non-recombinant) genotypes is +bc/a++
1).
If single crossover occurs between a&b…
Normal combination: +b/a+
After crossover: ++/ab
++ progeny= 26+146=172
ab progeny = 26+146=172
Total this progeny = 344
The recombination frequency between a&b = (number of recombinants/Total progeny) 100
RF = (344/1000)100 = 34.4%
2).
If single crossover occurs between b & c..
Normal combination: bc/++
After crossover: b+/+c
b+ progeny= 146+4=150
+c progeny = 146+4=150
Total this progeny = 300
The recombination frequency between b&c = (number of recombinants/Total progeny) 100
RF = (300/1000)100 = 30%
3).
If single crossover occurs between a & c..
Normal combination: +c/a+
After crossover: ++/ac
++ progeny= 26+4=30
ac progeny = 4+26=30
Total this progeny = 60
The recombination frequency between a&c = (number of recombinants/Total progeny) 100
RF = (60/1000)100 = 6%
Recombination frequency (%) = Distance between the genes (cM)
a----------6cM--------c-----------30cM--------------b
16). a). 30
17). d). 10
Explanation:
Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.
Hence, the parental (non-recombinant) genotypes is mqz/+++
1).
If single crossover occurs between m&q…
Normal combination: mq/++
After crossover: m+/+q
m+ progeny= 39+9=48
+q progeny = 41+11=52
Total this progeny = 100
The recombination frequency between m&q = (number of recombinants/Total progeny) 100
RF = (100/1000)100 = 10%
2).
If single crossover occurs between q & z..
Normal combination: qz/++
After crossover: q+/+z
q+ progeny= 142+11=153
+z progeny = 138+9=147
Total this progeny = 300
The recombination frequency between q&z = (number of recombinants/Total progeny) 100
RF = (300/1000)100 = 30%
3).
If single crossover occurs between m &z..
Normal combination: mz/++
After crossover: m+/+z
m+ progeny= 142+39=181
+z progeny = 138+41=179
Total this progeny = 360
The recombination frequency between m&z = (number of recombinants/Total progeny) 100
RF = (360/1000)100 = 36%
Recombination frequency (%) = Distance between the genes (cM)
m----------10cM--------q-----------30cM--------------z
18). e). Proline
19). a). diploid
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