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According to the normal parents of the couple but the child has a disease, it can be judged as a recessive genetic disease (out of nothing is recessive); The husband's sister is sick and the husband's father is normal, indicating that it is an autosomal genetic disorder (female disease paternity disease). So the probability that the couple's parents are AA, and the probability that both of them are AA is 2 3, and the probability of AA is 1 3. The probability of the sick child is 2 3 * 2 3 * 1 4 * = 1 9.
The probability of a son is 1 9 * 1 2 = 1 18
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1/18。According to the normal parents of the couple but the birth of a sick child, it can be judged as a recessive genetic disease; The husband's sister is sick and the husband's father is normal, indicating that it is an autosomal genetic disorder. So the probability that both husband and wife are AA is 2 3.
The probability of a boy with the disease is 2 3 * 2 3 * 1 4 * 1 2 = 1 18. (Note that it's the son, don't forget 1 2).
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It is easy to infer that the disease is autosomal recessive, (taking AA as an example) that the couple has a probability of 2 3 per 2 3 and that it is 2 3 * 2 3 * 1 4 * 1 2 = 1 18 if they have a diseased son
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Answer: (1) The genotype of parent A is: The phenotype of parent B of AAB is: broad-leaved yellow-green stem.
2) The proportion of individuals with phenotypes different from their parents in F1 is 1 4;In F1, the proportion of individuals with stable genetics in yellow-green stem and broad-leaved plants was 2 3.
3) Conduct a test experiment.
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This kind of problem can be calculated using the four-grid table: in a four-grid table, the probability of occurrence of a situation in each grid is theoretically equal.
Answer 1: The genotype of the test plant is mr 0
i Figure A was crossed with a test plant and the result is:
So the ratio of broad-leaved safflower to broad-leaved white flower is 1:1
ii Figure B was crossed with a test plant and the result is:
So the ratio of broad-leaved safflower to broad-leaved white flower is 2:1
iii Diagram C was crossed with a test plant and the results were:
Therefore, the ratio of broad-leaved safflower to narrow-leaved white flower is 2:1
I won't explain the situation in the second answer in detail, I hope you can understand. If you don't understand, please ask.
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Narrow-leaved white-flowered plants missing chromosome 2: (m,r), forming gametes (m,r) and (0,0), and (m,r) and (m,0) of plant B. The genotype of its progeny is (m,r)*(m,r) broad-leaved safflower:
m,0)*(m,r)Broad-leaved white flowers: (0,0)*(m,0)Dead (r missing):(0,0)*(m,r)Broad-leaved safflowers, so the ratio of broad-leaved safflowers to broad-leaved white-flowered plants is =2:1.
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
Hehe. It's fun, but there's a good way to help you remember. You see, if the daughter is sick, the daughter already knows that it is the daughter, so there is no need to multiply 1 2, if it is a sick daughter, it means that she knows the disease and does not know the gender, so you have to multiply!
There are 6 types. yyrr,yyrr,yyrr,yyrr,yyrr,yyrr.
We only study one pair of alleles at a time. >>>More
Insect resistance genes don't necessarily have antibiotic resistance, right? Moreover, the marker gene is not necessarily introduced into cotton cells along with the insect resistance gene. Here, the antibiotic gene, the insect resistance gene, and the marker gene are three different genes and cannot be confused. >>>More
p17, the enzyme that catalyzes the hydrolysis of lipase.
Lipase"It is an enzyme, and the enzyme is a protein, so it is natural to use protease. >>>More