Question

1. You have a plant with three different genes, a, b, and c. These genes are on the same chromosome and c is in the middle. The distance between a and c is 6 map units and the distance between c and b is 14 map units. You have a plant, heterozygous for all three genes as shown below: a C F + D You cross this plant in a test cross and collect 1000 progeny plants. (48 points) How many progeny would you expect for each class, listed below? (Rounded to the nearest whole number. The two genotypes of each class do not have to have the exact same number; they just have to add up to the correct number. i.e. Parentals = 536 = one parental could have 264 and the other could have 272, total of all progeny need to add up to 1000) Show all calculations to receive full credit. Genotype # of progeny + C + a + b a + + + cb a c b + + + + + b a c +

Answer 1

Expected DCO = %of recombination between a&c ×% of recombination between c&b

0.06×.14=0.0084 ( 0.84 in numbers)

Expected SCO between a&c = distance between a&c - expected DCO

6-0.84= 5.16

Expected SCO between c & b = distance between c&b- expected DCO

14-0.84= 13.16

Expected DCO in numbers =( .84×1000)/100

=8.4

Expected SCO between a&c in number= 5.16×1000/100=51.6

Expected SCO between b&c in number= 13.16×100/100=131.6

+C+=24

a+b=28

a++=5

+cb=3

acb=62

+++=70

++b=410

ac+=400