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Anonymous users2024-02-11Is the case A. Actually, this thing is very simple.
Cut point length of hind enzyme:
Cleavage point length of the BAMAH enzyme:
Description It is a combination of two enzymes to cut out.
The exact length is a combination of Hind digestion and BAMAH digestion.
The length is a combination of Hind digestion and BAMH digestion.
So the new length is a combination of the old length.
There are no new tangents.
So. There is 1 cutting point of the hind enzyme, which are:
There are 2 clectic points of the BAMAH enzyme, which are:
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Anonymous users2024-02-10aAs can be seen from the figure on the left, the hind acts alone to produce and the two fragments indicate that the hind has only one cleavage site; As can be seen from the figure on the right, BAMAH acts alone to produce three fragments, indicating that BAMAH has two digestion sites, and from the middle figure, it can be seen that the two digestion sites of BAMAH are located in the two digestion fragments of Hind.
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Anonymous users2024-02-093）aabb 、aabb、 aa bb
5) Follow, the heredity of red-green color blindness and albinism is an example.
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Anonymous users2024-02-08(1) 1 4 1 4 4 (2) Dominant: Recessive 1:3 Dominant:
Recessive 1:2:1 (3) aabb aabb aabjb 17 50 (5) followed.
Proof: You can read the book that if someone has albinism (autosomal recessive) and hemophilia (X recessive inheritance) at the same time, the genotype of the offspring born to him and a woman follows the laws of heredity.
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Anonymous users2024-02-07The last empty space of the first question should be 4, not 9, I'm sure!
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Anonymous users2024-02-06I want to know what your question means? Didn't finish it?
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Anonymous users2024-02-05a. According to the analysis of the figure, the chromosome of Figure B has been translocated, which belongs to the variation of chromosome structure, and a is correct;
b. Screening is the screening of chromosomal mutation of individuals, which can be used with a light microscope, while screening is the offspring produced by genetic recombination, which cannot be observed under the light microscope, b is correct;
c. According to the chromosome composition of Figure B, it can be seen that the proportion of gamete abnormalities produced by it is 3 4, so the cells of 3 4 Drosophila in F1 contain abnormal chromosomes, and c is wrong;
D. It is known that individuals without this allele cannot develop, so the ratio of gray and black fruit flies in F2 is 2:1, and D is correct
Therefore, c
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Anonymous users2024-02-04Choosing B, the amount of A was 55 and the amount of A was 40, so it was.
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Anonymous users2024-02-03b, times.
Compare the amount of A required in the two cases by using C backwards.
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First, the dominant recessive nature of albinism and sickle anemia was determined, and according to the left half of Figure 1, it can be concluded that anemia is a recessive genetic disease, which is bb; Then look at Figure 2, because if it is normal, it can be cut, and if it is not normal, it can't, so B is normal, C is albino, and both are homozygous, only A is heterozygous (because A can cut out three pieces: DNA is a double helix structure, only one of A can be cut, B can be cut all of them, cut into two short and two long, C is not cut) Because albinism is a recessive genetic disease, it is AA, so it can be known that B is AA and C is AA >>>More
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