Professor Farnsworth now wants you to create a genetic map f

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.)

Solution

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