Biogenetic Questions 20, Biogenetic Questions.

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!

The answer is: c.

The offspring of F1 are pink tall stalks: pink dwarf stalks: white tall stalks

White dwarf stalk = 1:1:3:

3. It can be known that the genotype of F1 is AABB:AABB 1:1, and then we can know that the parents can produce two gametes:

AB and AB, so the genotype of the parent is AAB.

2) 1 2, does not follow (3) 6, 5:1:5:1 , 1 4 Analysis: (2) Aa inbred offspring are homozygous for 1 2, and a pair of alleles follows the law of segregation.

3) AAAA can produce two kinds of gametes, and BBBB can produce three kinds of gametes, for a total of 6 gametes.

The F1 phenotypic ratio (1AA:1AA) (5b--:1bb)=5:1:5:1a gene frequency is still 1 4

In the case of AAAA inbred, F1 homozygous accounts for 1 9 because the gamete ratio is (1AA:4AA:1AA).

Set albinism as AA

A normal woman may be AA or AA, because it is an albinistic child who is born, so the mother must have a and the mother is AA

The same is true for men, who are also AA

The man is also referred to as BB, indicating that B is the causative gene. The woman is normal, which means that the woman is a BB

So aabb and aabb, gametes: ab, ab 1 2 each, father ab, ab, ab 1 4 each

1 conjunctive finger, indicating that it contains a and b, when looking at a alone, the offspring contains a is 3 4, and if you look at b alone, the offspring contains b as 1 2Multiply it by 3 8

2 are white, indicating that the offspring is AAB, and the probability of looking at a alone is 1 4 and the probability of looking at b alone is 1 2, and the multiplication is 1 8

3 albino and syndacty, indicating that the offspring contains b and aa, the probability of aa is 1 4, the probability of containing b is 1 2, the probability of boys is 1 2, and the multiplication is 1 16

4 is a disease, adding 1 and 2 to 1 2

5. The possibility of disease in the offspring can be ruled out by the method of not having the disease first. The offspring is BB with A

The probability of containing a is 3 4, the probability of bb is 1 2, and the multiplication of not getting sick is 3 8, so the disease is 5 8

It's finally over, I'm tired, and if there's anything you don't understand, please ask me again. Especially this kind of biocomputing problem. Hee-hee.

I'll write it.

From the question, the genotype of the parent can be inverted: (the albinism gene is represented by a) aabb xaabb

Then: The probability that a child has albinism (AA) is 1 4

The probability that a child is syndactyly (bb) is 1 2

The probability of both, 1 4x1 2=1 8

4) i.e. (1) + (2) = 3 8 + 1 8 = 1 2 (5) i.e. albinism + syndactyly - (3) = 5 8

8 Because bb syndactyly, it means that syndactyly, which is a dominant genetic disease, and the father b b and the mother only have b b, so there is a probability of 1 2 syndactyly.

The probability of not having albinism is 3 4 (explained later) So: the probability of only having syndactyly is 1 2 * 3 4 = 3 8

2. Suppose that albinism is c because the parents did not have a child with albinism (recessive), which means that both parents have a child with albinism cc, the probability is 1 4, and the probability of syndactyation and non-syndactyation is 1 2, so only the probability of albinism is 1 2 * 1 4 = 1 8

3, also 1 8

4. The probability of only suffering from albinism is 1 2*1 4=1 8 The probability of suffering from syndactya alone is 1 2*3*4=3 8

So the probability of suffering from only one disease is 1 8 + 3 8 = 1 2

5. The possibility of offspring not getting sick is 1 2*3 4=3 8, so the possibility of offspring getting sick is 1-3 8=5 8

3 8 1 8 1 16 1 2 5 8 It's been a long time I don't know if it's right.

Because there is no picture, this is the answer I brought out from the question bank:

1) Gene mutation manuscript beam (2) 7 16 (nucaroed reed 3) 2n (2 to the n power) ddrr m 4