If a plant that is heterozygous for all four characters were allowed to self-fer
ID: 312194 • Letter: I
Question
If a plant that is heterozygous for all four characters were allowed to self-fertilize, what proportion of the offspring would be expected to be as follows? What proportion of homozygous for the five dominant traits is expected? What proportion of homozygous for the five recessive traits is expected? What proportion of heterozygous for all five traits is expected? What is the number of different F_2 phenotypes, given complete dominance? What is the number of different F_2 genotypes, given complete dominance?Explanation / Answer
Answer:
AaBbCcDd x AaBbCcDd
Aa x Aa = AA (1/4), Aa (1/2) & aa (1/4)
Bb x Bb = BB (1/4), Bb (1/2) & bb (1/4)
Cc x Cc = CC (1/4), Cc (1/2) & cc (1/4)
Dd x Dd= DD (1/4), Dd (1/2) & dd (1/4)
1). The proportion of homozygous for 4 dominant trait is expected= AABBCCDD
= 1/4 * 1/4 * 1/4 * 1/4 = 1/256
2).
The proportion of homozygous for 4 recessive trait is expected= aabbccdd
= 1/4 * 1/4 * 1/4 * 1/4 = 1/256
3).
The proportion of heterozygous for 4 dominant trait is expected= A_B_C_D_
= 1/2 * 1/2 * 1/2 * 1/2 = 1/16
4). 16 different phenotypes are produced
Each heterozygous tetraploid produces 16 gametes (24).
Number of gametes is equal to number of phenotypes. So 16 different phenotypes are produced
5). Tetraploid heterozygous plant produces 81 different genotypes.
Each single heterozygous(concerned for single gene with two types of allele) plant produce three genotypes.(3)
Double heterozygous plant produces 9 different genotypes. (32)
Triple heterozygous plant produces 27 different genotypes. (33)
Tetraploid heterozygous plant produces 81 different genotypes.(34)
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