He decides to map one of these mutants- white 1 (wl-). He finds that the mutatio

Question

He decides to map one of these mutants- white 1 (wl-). He finds that the mutation maps to chromosome2. To map this mutation precisely he uses two RFLP loci located on chromosome2 and performs a three point cross. RFLP locusA gives either a 2kb (A2) or a 3kb band (A3) on a Southern blot with probeA. RFLP Locus B gives either a 5kb (B5) or a 7kb (B7) band on a Southern with probeB. A heterozygous slug for all three loci (w1-/W1+, A2/A3, B5/B7) is test crossed to a w1/w1, A2/A2, B5/B5 homozygous recessive slug. The following progeny are obtained: (The RFLP marker is only scored for band size in the heterozygous slug) What is the correct order of these three loci? Also what are the map distances between the W1 gene and the two RFLPs

Explanation / Answer

Answer:

Correct order if three loci

A3 ---------12 m.u.----------w1- ---- 10 m.u.----B7

A2 ---------12 m.u.----------W1+ ---- 10 m.u.----B5

Distance between W1+ & B5 = 10 mu

Distance between W1+ & A1 = 12 mu

Explanation:

Parental combinations are more so, the parental combinations are w1- A3 B7 & W1+ A2 B5

1. If single cross over (SCO) occurs between w1- & A3 and W1+ A2

Normal order= w1- --------- A3 & W1+ ----- A2

After cross over= w1- ----- A2 & W1+ ------ A3

w1- ----- A2 recombinants are 50+9=59

W1+ ------ A3 recombinants are 50+11= 61

Total recombinants = 120

RF = (120/1000)*100 =12%

2. If single cross over (SCO) occurs between A3 & B7 and A2 & B5

Normal order= A3 --------- B7 & A2 --------- B5

After cross over= A3 --------- B5 & A2------ B7

A3 --------- B5 recombinants are 41+50= 91

A2------ B7 recombinants are 39+50 = 89

Total recombinants = 180

RF = (180/1000)*100 = 18%

3. If single cross over (SCO) occurs between w1- & B7 and W1+ & B5

Normal order= w1- --------- B7 & W1+ ------ B5

After cross over= w1- --------- B5 & W1+ --------- B7

w1- --------- B5 recombinants are 41+9= 50

W1+ --------- B7 recombinants are 39+11= 50

Total recombinants = 100

RF = (100/10000)*100 = 10%

% RF = Map unit distance

The order of gene is -----

A3 ---------12 m.u.----------w1- ---- 10 m.u.----B7

A2 ---------12 m.u.----------W1+ ---- 10 m.u.----B5

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