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Anonymous users2024-02-08When YYSS (yellow round grain) is crossed with YYSS (green round grain) to obtain F1 generation as YYSS (yellow round grain) and then self-crossed, the obtained yellow round grain (genotype YYSS or YYSS or YYSS or YYSS) accounts for 9 16, yellow round grain (genotype YYSS or YYSS) accounts for 3 16, green round grain (genotype YYSS or YYSS) accounts for 3 16, and green wrinkle grain accounts for 1 16The recombinant type is yellow wrinkled grains and green round grains, so it accounts for 3 16 + 3 16 = 3 8
However, when the parents were yellow wrinkles (yyss) and green round grains (yyss), the hybrid generation F1 was yellow round grains (YYSS), and the F1 generation was self-crossed, and the obtained yellow round grains (genotype YYSS or YYSS or YYSS or YYSS) accounted for 9 16, yellow round grains (genotype YYSS or YYSS) accounted for 3 16, green round grains (genotype YYSS or YYSS) accounted for 3 16, and green wrinkles accounted for 1 16At this time, the recombinant type is yellow round grains and green wrinkled grains, so they account for 9 16 + 1 16 = 5 8
Similar question: Because YYSS accounts for 1 16 in F2, a total of 80 (1 16) = 1280 plants are produced, and the heterozygosity that is different from the parent (i.e., different from the yellow round grain and the green wrinkled grain) in the F2 phenotype is YYSS and YYSS is 1
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Anonymous users2024-02-07Recombinant individuals in the F2 generation are those with a different phenotype from the female parent in F2.
When p is aabb aabb or aabb aabb, these two phenotypes are removed from f2 as 3 8
p is aabb aabb.
F2 with these two phenotypes removed is 5 8
From Mendel's genetic theorem 9:3:3:1 yyss is 80 so y-s- is 80*9 y-ss 80*3 yys- 80*3
The phenotype is different from that of the parent, and the sum is 80*3+80*3=480
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Anonymous users2024-02-06When p is aabb aabb or aabb aabb, these two phenotypes are removed from f2 as 3 8
p is aabb aabb.
F2 with these two phenotypes removed is 5 8
That's it
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Anonymous users2024-02-05It is possible to know that the causative gene is on x and is recessive.
Then the genotype of the normal couple is x big a y x big ax small a, so the genotype of the offspring is x small ay, and the probability is 1 in 4.
Because the incidence rate in women is 4 in 10,000, the probability of X small A in women is 2% (2% squared is 4 in 10,000), so the probability of offspring suffering from the disease is 2% times 1 in 4 equals 1 200 choose C
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