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The height of a type of bean plants is determined by five unlinked genes called A,...

The height of a type of bean plants is determined by five unlinked genes called A, B, C, D, and E. Each gene has two alleles: additive (uppercase letter) and nonadditive (lowercase letter).

How many phenotypic classes would you expect?

The shortest plants are 130 cm. The tallest plants are 220 cm. Estimate how many centimeters each allele contributes to the height difference.

The genotypes are known for two bean plants. Plant 1 is genotype AABbccDdEE. Plant 2 is genotype aaBBCcDdEE. What are their heights?

If the two plants from above are crossed, what is the probability they will produce a plant that is taller than either parent?

Solutions

Expert Solution

phenotypic classes would be expected = 11 (all allele A,B,C,D,E have same effect on plant height and a, b, c, d, e have same effect we can make total 11 cases so total 11 classes would be expected.

shortest plant = 130 cm

Genotype of the shortest plant = aabbccddee

height contributed by each allele = 130/10 = 13 cm (total number of allele present =10)

Tallest plant =220 cm

Genotype of the tallest plant = AABBCCDDEE

height contributed by each allele = 220/10 =22 cm (total number of allele present =10)

height of AABbccDdEE genotype = (22+22)+(22+13)+(13+13)+(22+13)+(22+22) = 184

height of aaBBCcDdEE genotype = (13+13)+(22+22)+(22+13)+(22+13)+(22+22) = 184

probability they will produce a plant that is taller than either parent

the plant are taller than their parent only if they have following genotype

1. AABBCcDDEE

2.AABbCcDDEE

3. AABBCcDdEE=

(AA and EE has no effect on probability as they are same in all progeny plant)

BB x Bb = 2 BB and 2Bb = probability of each genotype = 2/4

cc x Cc = 2 cc and 2Cc= probability of each genotype = 2/4

Dd x Dd = 1 DD: 2 Dd : 1 dd

probabibilty of DD = 1/4, probabibilty of dd = 1/4, probabibilty of Dd = 1/4,

probabibilty of AABBCcDDEE genotype = probabibilty of BB x probabibilty of Cc x probabibilty of DD =2/4 x 2/4 x 1/4 = 4/64

probabibilty of AABbCcDDEE genotype = probabibilty of Bb x probabibilty of Cc x probabibilty of DD =2/4 x 2/4 x 1/4 = 4/64

probabibilty of AABBCcDdEE genotype = probabibilty of BB x probabibilty of Cc x probabibilty of Dd =2/4 x 2/4 x 2/4 = 8/64

probability they will produce a plant that is taller than either parent = probabibilty of AABBCcDDEE genotype + probabibilty of AABbCcDDEE genotype + probabibilty of AABBCcDdEE genotype

4/64 + 4/64 + 8/64 =16/64 = 1/4 = 0.25

gladiator answered 10 months ago
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