Professor Farnsworth now wants you to create a genetic map for the space wasps.
ID: 172411 • Letter: P
Question
Professor Farnsworth now wants you to create a genetic map for the space wasps. Consider the following three genes. Each gene has two alleles with different phenotypic effects. The wild-type alleles (indicated by) are dominant. Gene names are the same as the name of the recessive allele. Gene s: allele s - short stinger; allele s^+ - long stinger Gene w. allele w - two wings; allele w^+ - four wings Gene b: allele b - small body; allele b^+ - giant body A female heterozygous at all three genes is testcrossed with a male carrying the recessive allele for all three genes. Their daughters are then classified and counted: Long-stinger, four-wings, giant-body - 170 Short-stinger, two-wings, small-body - 150 Long-stinger, two-wings, giant-body - 5 Short-stinger, four-wings, small-body - 3 Short-stinger, four-wings, giant-body - 710 Long-stinger, two-wings, small-body - 698 Long-stinger, four-wings, small-body - 42 Short-stinger, two-wings, giant-body - 38 Draw a genetic map between these genes, labeled with map distances. (You do not need to correct for multiple crossovers.)Explanation / Answer
Answer:
Explanation
Phenotype
Shorthand
Number of progeny
Long, four, gaint
s+ w+ b+
170
Short, two, small
s w b
150
Long, two, gaint
s+ w b+
5
Short, four, small
s w+ b
3
Short, four, gaint
s w+ b+
710
Long, two, small
s+ w b
698
Long, four, small
s+ w+ b
42
Short, two, gaint
s w b+
38
Total= 1816
Imagine, the parental gene combination = s w+ b+/ s+ w b
1. If single cross over (SCO) occurs between s & w+ and s+ w
Normal order= s---------w+ & s+-----w
After cross over= s-----w & s+------w+
s-----w recombinants are 150+38 = 188
s+------w+ recombinants are 170+42=212
Total recombinants = 400
RF = (400/1816.)*100 =22.03%
2. If single cross over (SCO) occurs between w+ & b+ and w&b
Normal order= w+---------b+ & w----b
After cross over= w+-----b & w------b+
w+-----b recombinants are 3+42=45
w------b+ recombinants are 5+38=43
Total recombinants = 88
RF = (88/1816)*100 = 4.85%
3. If single cross over (SCO) occurs between s & b+ and s+&b
Normal order= s---------b+ & s+------b
After cross over= s-----b & s+------b+
s-----b recombinants are 150+3=153
s+------b+ recombinants are 170+5=175
Total recombinants = 328
RF = (328/1816)*100 = 18.06%
% RF = Map unit distance
The order of genes is -----
s-----------18.06 m.u.------------b---4.85 m.u.----w
Gene configuration = s b+ w+/ s+ b w
Phenotype
Shorthand
Number of progeny
Long, four, gaint
s+ w+ b+
170
Short, two, small
s w b
150
Long, two, gaint
s+ w b+
5
Short, four, small
s w+ b
3
Short, four, gaint
s w+ b+
710
Long, two, small
s+ w b
698
Long, four, small
s+ w+ b
42
Short, two, gaint
s w b+
38
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