High School Biology Genetic Problem Solving, Solving High School Biology Genetic Questions

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

Second, the genotype 5:AABB is easy to write based on the genetic map, because 2 is an albinist and not anaemia, and 1 is a non-albinistic patient and not anaemia, but the offspring of both are both.

1 is AAB 2 is AABB So we know that 6: AAB or AABB and the proportions are 1 3 and 2 3, respectively.

And because the electrophoresis results of inscriptions 3 and 4 are similar to A, it can be inferred that they are AA according to the above inference, and because they are not diseased, but their offspring are anemic, so the genotype is AABB. And because 7 does not suffer from any disease, 7:a b.

According to the inference, 7: aabb aabb aabb aabb and the ratio is 1 9 2 9 2 9 4 9 and 6 and 7 are crossed, which is more complex and can be paired one by one, and the final result is 20 27.

lz you see, there is an easy way to calculate the end, but I forgot, if you have any questions, you can ask again If you are satisfied, give me extra points

Option D, the wife's sister is a patient, and her parents are normal, so the disease is an autosomal recessive disease. The husband's genotype may be 1 2aa or 1 2aa; The wife's genotype may be 1 3aa or 2 3aa. They are married to four hybrid combinations and the genotypes and probabilities of their offspring are as follows

1/2aa×1/3aa=1/6aa ② 1/2aa×2/3aa=1/6aa；1/6aa ③ 1/2aa×1/3aa=1/12aa；1/12aa ④ 1/2aa×2/3aa=1/12aa；1/6aa；1 12aa Because both children are normal according to the title, the probability of them having a normal child is 11 12, and the probability of being homozygous is 6 11, and the probability of heterozygous is 5 11. Two children with a normal phenotype and homozygous are independent events, and in order for them to occur at the same time, their respective probabilities should be multiplied, so there is: 6 11*6 11=36 121, so choose d. Oh.

In this case, only the alleles of E and E need to be considered.

Parental, the genotype of the female parent is EE, the genotype of the father is EE, F1 generation, EE: EE = 1:1, that is, EE accounts for 1 2, EE accounts for 1 2, F1 inbred, 1 2ee inbred offspring should be:

ee=1/4*1/2=1/8，ee=1/2*1/2=1/4，ee=1/4*1/2=1/8；

The offspring of 1 2ee inbred should be: ee=1 2;

In this case, F2 accounts for EE=1 8, EE=2 8, EE=5 8, then the problem should be found, if EE gene homozygous is lethal, then EE should account for 2 3 in F2, if EE gene homozygous is deadly, then EE should account for 2 7 in F2.

So the answer is: the E gene is homozygous and lethal.

30 (8 points).

1) 45 3 8 (2) Homozygous lethality in the E gene (3) Aadd Answer 1: Since Leaf phenotype and its proportions If the ratio of sharp to smooth leaf margins of F2 plants is close to 15:1 If the ratio of sharp to smooth leaf margins of F2 plants is close to 3:

1 Answer 2: Measure the phenotype of the blade and its proportion.

If the ratio of sharp to smooth leaf margins of F2 plants is close to 3:1, the ratio of sharp and smooth leaf margins of F2 plants is close to 1:1

Solve the problems that high school biology students are prone to.

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.

The second void is that the red and white eye genes are on the sex chromosomes. Therefore, the probability of a black body in males is well understood, at 1 4. In males, the genotypes of red and white eyes are B and B (this gene is not present on the Y chromosome).

The probability of occurrence of these two genotypes is determined by the P generation, which is 1:1, so the probability of appearing in a white-eyed male is 1 x 1 2 = 1 8

The first void of the fourth question, since F1 is a mutant, this mutation can be deduced as a dominant mutation, so the wild type must be recessive homozygous. Hence the following 3:1. If the mutant type is used to measure crossing, the unpredictability of F2 is greater.

1.Gray and black body: parent: aa and aa, fi: aa and: aa, aa, aa, aa, the probability of black body is 1 4, red and white eyes: parent: xbxb and xby, f1: xby, xbxb, f2 male: xby, xby, the white eye ratio is 1 2

In the original question, it is 1 2 * 1 4 = 1 8

2.After the hybridization of wild-type and mutant homozygous, FI is mutant, indicating that wild-type carries mutant genes, just like the wild-type AA and mutant AA hybridization, dominant mutations, F1 are all AA, and then self-crossed, satisfying the trait segregation ratio of 3:1, which can prove allele control.

If the mutant type is used for hybridization, the shape separation ratio is not easy to control.

PS: Don't blame me if you're wrong.

(3) The second void: the male's black body and white eyes are aaxby

It is not the proportion of this gene to F2, and it should only be analyzed by looking at males according to the topic.

Again, on a pair-by-pair basis, aa accounts for 1 4 of all body color genes (aa 2aa aa) and xby accounts for 1 2 of male eye color genes (xby xby), so the answer is 1 8.

4) I don't understand the title, and it doesn't say the gender of the mutant type

Solve the problems that high school biology students are prone to.