Quantitative Genetics Worksheet: Part 1 1. In Kumquat trees, plant height is det
ID: 202405 • Letter: Q
Question
Quantitative Genetics Worksheet: Part 1 1. In Kumquat trees, plant height is determined by 3 genes as in our simplified model of polygenic inheritance. Each locus has a contributing allele and a non-contributing allele. A tree with the genotype AABBCC grows to be 100cm tall. The other homorygote grows to be 40cm tall. What is the contribution for each contributing allele? a. 60cm b. 50cm C.30cm d. 20cm e. 10cm 1, 2. If you crossed a 100cm tree with a 40cm tree and then crossed the Fi to get the F generation, what proportion of the Fa offspring will grow to be 70cm tall? b. 20/64 c. 20/128 d. 10/64 e. 1/8 2. Considering again the F2 produced in question 2, what proportion of the offspring will be 70cm tall or taller? 3. b. 15/64 c. 6/64 d. 42/64 e. 1/64 3. 4. Determine the limit of Kumquat tree height in the offspring of the crosses listed below: Minimum Height Maximum Height Cross AABBCc x aaBBcc aaBbcc x AabbCc AaBBcc x AabbCc AaBbCC x AaBbCc 5. On a remote Pacific island there are Kumquat trees with quantitative genes that allow the trees to bear enormous fruit. One strain of tree bears "huge" 20 pound kumquats. Another strain bears "normal" 4 pound fruit (still very largel). Assume that this trait (fruit weight) is controlled by 4 gene loci. From a cross between the "huge" and "normal" strains, what weight fruit will the progeny have? a. 12 b. 16 c. 10 d. 4 e. 18 6. Following our simplified model of polygenic inheritance, what is the contribution per allele? a. c. 4 d. 6 6 7. If the F2 generation were produced, what proportion of the offspring would you expect to produce 14 pound fruits? b. 128/256 c.56/256 d. 47/128e. 1/256 7. When two 12 pound F2 trees were crossed they produced offspring which all produced 12 pound fruit. However, when these offspring were crossed randomly, the fruit produced weighted from 4 to 20 pounds. What can you say about the possible genotypes of the original F2 trees? 8. 65Explanation / Answer
1) 10 cm - Each locus should be contributing 20cm and the dominant allele should be contributing 10cm, because the other homozygote tree is 40 cm tall.
2) 20/64 - Atleast three dominant alleles need to be present in a genotype to produce 70cm tree. When crossing F1 heterozygous, 20 offsprings of the 64 total offsprings will be od 70cm tall.
3) 42/64 - As mentioned above atleast three dominant allleles are needed to be present in 70cm or a taller tree. So, in F2 generation 42 out of 64 offspring will be 70cm or taller.