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Anonymous users2024-02-07There are two obvious errors in the answer Impossible, unless you are approximately equal, which is indeed more suitable for the answer Phenotypic plants so p(aa)=aa (aa+aa)=aa (1-aa)=(2 0 02 0 98) ( )= Disclaimer I also have that multiple-choice question at hand, and the answer should be wrong.
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Anonymous users2024-02-06The answer to the original question is correct. From the aa 4*10 -4 in the question, we can see that p(a)=, p(a)=, and then obtained.
p(aa)=（，p(aa)=2*
The proportion of AA in phenotypically normal individuals was then found.
aa (aa+aa)=aa (1-aa) [2*The probability of albino seedlings in phenotypic normal offspring is:
1/4*(aa)^2＝1/4*[2*
So the answer is normal.
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Anonymous users2024-02-05The answer is wrong. That's not a small calculation...
Yours is right. I see you know how to be too lazy to say it.
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Anonymous users2024-02-04a-a 1-a is the frequency of albino genes in phenotypically normal plants, divided by 1- a, which is equivalent to a (1+ a).
Squared, i.e. the answer.
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Anonymous users2024-02-03We might as well take the square of t as a, which is the albino gene frequency.
Since it is inbred in normal seedlings, it should be noted that all normal seedlings are frequent, so the gene frequency of Baihua is (t-t 2) (1-t 2) = t (1+t), and it is sought by one square.
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Anonymous users2024-02-02For reference only, because the hybridization of aa and aa should be considered, aa self-inbreeding, the calculation method is p(aa) squared 1 4
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Anonymous users2024-02-01This question seems to be a genetic problem, and I'll do the math for you later.
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Anonymous users2024-01-31Analysis: Examines the calculation of genotype frequencies. The albino gene is recessive, set to a, its gene frequency is set to q, and the gene frequency of osmosis a is set to p.
The plant population is in equilibrium, so the Harwin equilibrium can be used to solve the problem. The frequency of aa is Q2 = 4 10 4, Q = , P = 1 - Q = . The phenotypic normal plant genotype is AA or, and only the inbred offspring of AA will show albino seedlings.
The genotype frequency of AA in normal plants is 2PQ (P2+2PQ)=2, and the frequency of AA in inbred offspring is 1 4 [2]. The High School Attached to the National People's Congress? )
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Anonymous users2024-01-30Albino recessive. aa 1/100 a 1/10
So aa probability.
So aa probability.
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