P1: A female heterozygous for white ( w ), yellow ( y ) and miniature ( m ) – (a
ID: 187241 • Letter: P
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%
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%
Explanation / Answer
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.
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