Question

1.)

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).  

A.) How many phenotypic classes would you expect?

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

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

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

2.)

A plant breeder has determined the following variances for yield of corn in his fields:

Total Phenotypic Variance	100
Additive Genetic Variance	40
Dominant Genetic Variance	20
Epistatic Genetic Variance	20
Environmental Variance	20

A.) Calculate the total genetic variance.  

B.) Calculate both the broad-sense heritability and narrow-sense heritability indicated by this data.

C.) The breeder wishes to improve yield. If the average yield in the starting population is 400 and he selects for breeding plants with an average yield of 500, what will be the expected average yield among the offspring of the selected plants?

D.) If the breeder continues to select breeding plants with the highest yield, will he continue to improve yield in the offspring?

3.)

When the phenotype of an organism is the result of the interaction between two different genes each exhibiting alleles with Simple Dominance, we can account for this genetic interaction in the F2 phenotypic categories seen in a 9:3:3:1 ratio.

A.) Explain how the phenotypic categories would change for a standard dihybrid cross when the trait examined is a Quantitative Trait. (Here the dominant alleles of each gene would act equally and additively to contribute to the phenotype.)

B.) Show the identifying genotype for each phenotypic class.

C.) Show how the phenotypic ratio for a standard dihybrid cross involving a Quantitative Trait would be derived.

Answer 1

Please post the questions separately.

Question-1

A. Phenotypic classes of bean plants can be calculated by using the formula (2n+1) where n represents the total number of genes.

             2n+1 = 2(5) +1

=11

B. The tallest plants have genotype AABBCCDDEE. They are all additive alleles.

Size of the short plant (µ) is 130 cm.

µ+nX =220cm where, n is the total number of genes.

µ+10X =220cm

130+10X =220cm

X=220-130/10

X=9

Therefore, the average height difference contribution of each allele is 9 cm.

C. To find the height of Plant-1, multiply the number of capital letters by the value that each allele contributes to the height difference, the resulted value is then added with the size of the short plant.

µ +6X = 130cm + (6 × 9cm)

=184 cm

  The size of plant-2 will be as same as the plant-1, due to the presence of 6 additive alleles (capital letters)

D. Answer is 5/16 because, a plant that is taller than either parent would have 7 contributing alleles or 8 contributing alleles.They are: AaBBCcDdEE (1/8) or AaBbCcDDEE (1/16) or AaBBccDDEE (1/16) and AaBBCcDDEE (1/16)

Calculate the probability of combinations for all 5 alleles in this way. A cross between parent 1 (Aa) x parent 2 (aa), resulting in 1/2 of offspring being Aa and 1/2 being aa. This means that the probability of offspring receiving the allele combination Aa or aa is 1/2. By this way we can calculate the probability of combinations for all 5 alleles AA: 0, Aa: 1/2, aa: 1/2, BB: 1/4, Bb: 1/2, bb: 1/4, CC: 1/2, Cc: 1/2, cc: 0, DD: 0, Dd: 0, dd: 1, EE:1/4, Ee: 1/2, ee: 1/4