The figure shows the P F1, testcross, and testcross progeny from crosses involvi

Question

The figure shows the P F1, testcross, and testcross progeny from crosses involving three linked genes located on the X chromosome in Drosophila. Using this figure for reference, complete the sentences using the words and phrases (not all words and phrases are used). F1 (all identical) vg b pr /vg b pr Testcross SP vg b pr/vg' b pr x vg b pr/vg b pr 1779 vg b pr Parental combinations for Testcross 1654 vg b pr all three genes progeny 252 vg b pr Recombinants for vg relative to 241 vg b pr parental combinations for b and pr 131 vg b pr L Recombinants for b relative to 118 Vg b pr parental combinations for vg and pr 13 vg b pr Recombinants for pr relative to 9 vg b pr J parental combinations for vg and b 4197 vg pr and vgr pr vg b and vgr b 17.7 band pr bt pr and b pr vg and pr vg brand vg+b 8.8 recombinant vg and b the total of the the total number of all parental classes progeny

Explanation / Answer

Answer:

The order of the three genes can be deduced by analysis of the frequencies of recombinant clsses

Analysis begins with the selection of any pair of genes. This example will begin by using the vg and b genes but any two pairs of genes would do. The recombinants are vg b+ and vg+ b; the third pair of alleles can be ignored for the moment.

The sum of all progeny vg b+ and vg+ b, is divided by 41970 to obtain a value of 0.01768. To obtain the result in map units this is then multiplied by 100 to obtain the result of 17.7 m.u.

Analysis is then performed on another pair of genes. In this example the pair vg and pr, ignoring b. The recombinant classes are vg pr+ & vg+ pr. The same calculation is performed with the values of these classes. This result is 12.3 m.u.

The same process is used for the remaining pair, b and pr. The recombinant classes are b pr+ and b+ pr; the result is 6.4 m.u.

Finally, compring the three addivtive RF values, it is apparent that vg & b are the most widely separated genes. Therefore pr must be in the middle.

Explanation:

Imagine the order is vg b pr / vg+ b+ pr+

1. If single cross over (SCO) occurs between vg & b

Normal order= vg --------- b & vg+ ----- b+

After cross over= vg-----b+ & vg+------b

vg-----b+ recombinants are 252 + 131= 383

vg+------b recombinants are 241+118 = 359

Total recombinants = 742

RF = (742/4197)*100 =17.68%

2. If single cross over (SCO) occurs between b &pr

Normal order= b---------pr & b+----pr+

After cross over= b-----pr+ & b+------pr

b-----pr+ recombinants are 131+13= 144

b+------pr recombinants are 118+9= 127

Total recombinants = 271

RF = (271/4197)*100 = 6.46%

3. If single cross over (SCO) occurs between vg & pr

Normal order= vg--------pr & vg+------pr+

After cross over= vg-----pr+ & vg+------pr

vg-----pr+ recombinants are 241+13= 254

vg+------pr recombinants are 252+9=261

Total recombinants = 515

RF = (515/4197)*100 = 12.3%

% RF = Map unit distance

The order of genes is -----

vg--------12.3 m.u.-----pr--6.4 m.u.---b

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