Question

P1: A female heterozygous for white (w), yellow (y) and miniature (m) – (all 3 recessive mutations) is crossed to a male who is mutant for all three traits.

The F1 male progeny phenotypes are:
+ + m	2278
w y +	2157
w y m	1203
+ + +	1092
+ y m	49
w + +	41
+ y +	2
w + m	1
Total =	6823 What are the map distances between all three genes? Round all numbers to two decimal places. m to y is 1.36%, y to w is 33.68% and m to w is 35.04% w to y is 18.39%, y to m is 16.64% and w to m is 35.03% y to w is 1.36%, w to m is 33.68% and y to m is 35.04% w to y is 1.36%, y to m is 33.68% and w to m is 35.04% y to w is 1.32%, w to m is 33.63% and w to m is 34.95%

Answer 1

Genotype with Minimum progeny represents double cross-over, whereas, Genotype with the higher number represents the parental type.

Genotype + + m and w y + are parental type

Genotype + y + and w + m is the double crossover

To find the gene order make a double cross in parental genotype if it gives the genotype same as the double crossover then your gene order is correct.

Double cross of Gene order y w + and + + m gives same as given double crossover

No to find the linkage distance

Make the first crossover between two parental genes in its corrected gene order you will get

+ w + and m + y genotype

Now look at the table which matches genotype and add those number then divide it by total progeny

So Frequency of first crossover between y and w = ( 41 + 49)/6823 = 0.01319

Therefore distance between y and w= 0.01319 x 100 = 1.319% = 1.32%

  Similarly for second cross

It will give genotype + + + and m w y

Do the same as we did for first cross over

Frequency of second cross over between w and m = ( 1203 + 1092)/6823 = 0.3363

Therefore distance between w and m = 0.3363 x 100 = 33.63%

And distance between y and m will be = 1.32% + 33.63% = 34.95%

therefore, last option is the correct option.