1J (1 pt) The F2 progeny from a particular cross exhibit two phenotypic types in
ID: 247231 • Letter: 1
Question
1J (1 pt) The F2 progeny from a particular cross exhibit two phenotypic types in a ratio of 9:7. What phenotypic ratio would be expected from a testcross of the F1 progeny? 2) (4 pts) Two homozygous strains of corn are hybridized. They are distinguished by six different pairs of genes (ie, carry different alleles for all six genes), all of which assort independently and all of which control different phenotypes. The F1 is selfed to produce an F2 population. a) What is the total number of possible genotypes in the F2? b) How many of these genotypes will be homozygous at all six gene loci? c) If all alleles act in a dominant/recessive manner, what proportion of the F2 wil be homozygous dominant for all six genes? d) What proportion of the F2 will show all dominant phenotypes?Explanation / Answer
1. ANS: The 9:3:3:1 ratio is that of the genotypes A— B—, A—bb, aa B—, and aa bb. The modified 9:7 implies that all of the last three genotypes have the same phenotype. The F1 testcross is between genotypes Aa Bb and aa bb, and the progeny are expected in the proportions 1/4 Aa Bb, 1/4 Aa bb, 1/4 aa Bb, and 1/4 aa bb. The last three genotypes would again have the same phenotype, and so the ratio of phenotypes among the progeny of the testcross is 1:3.
2. ANS:
2a. ANS: For a given number of loci n, there would be possible genotypes and possible phenotypes. As there are 6 loci, then the number of possible genotypes would be determined by the following equation:
Possible genotypes = 36 = 729.
Therefore, there are 729 possible genotypes.
2b. ANS: For one locus, only two of he possible three genotypes are necessary for them to be homozygous therefore, for 6 loci, the number of possible genotypes is given by the following equation:
Possible genotypes = 26 = 64.
Therefore, there are 64 possible genotypes.
2c. ANS: In the F2 generation, because it is a mononohybrid cross, 1/4 of the progeny will show all homozygous dominant phenotypes. Therefore, because there are 6 genes then, the proportion is given as
P (all dominant)= 1/ 46
= 1/4096
Therefore, the proportion of F2 plants will have the homozygous dominant phenotype is 1/4096.
2d. ANS: In the F2 generation, because it is a monohybricl cross, of the progeny will show all dominant phenotypes. Therefore, because there are 6 genes then, the proportion is given as:
(p dominant)= 3/4 X 3/4 X 3/4 X 3/4 3/4 X 3/4
= 729/ 4096 = 17%.
Therefore, the proportion of F2 progeny to show all dominant phenotype is 17%.
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