High school biological genetics questions, senior one biological genetics questions

Updated on educate 2024-08-06
6 answers
  1. Anonymous users2024-02-15

    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

  2. Anonymous users2024-02-14

    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).

  3. Anonymous users2024-02-13

    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

  4. Anonymous users2024-02-12

    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.

  5. Anonymous users2024-02-11

    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.

  6. Anonymous users2024-02-10

    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.

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