High School Biological Genetic Computing, Solving a High School Biological Genetic Computing Problem

Question 1: The genotype of Aa in F1 is 2 3, the genotype of Aa is 1 3, the offspring that can self-cross is the recessive trait is AA, and the offspring is 2 3*2 3*1 4=1 6 of Aa, so it is 5 1

Question 2: It is a descendant of F2, which is different from the first question.

Ratio in 1:2:1 f1 Homozygous in tall stems accounts for 1 3 Heterozygous accounts for 2 3 Homozygous inbred or tall stems 3 4 in heterozygotes are tall stems, so tall stems occupy.

1 3+2 3*3 4=5 6 The ratio of short stems to 2 3*1 4=1 6f1 is still 1:2:1 Homozygous accounts for 1 3 Heterozygous accounts for 2 3 There are many cases of free mating Pure Pure Pure Miscellaneous

Pure Miscellaneous High Accounts for 1 3 * 2 3 + 2 3 * 1 3 = 2 9 Pure Pure High Accounts for 1 3 * 1 3 = 1 9

Miscellaneous high accounts for 2 3 * 2 3 * 3 4 = 1 3

Add high 8 9

1. F1 medium and tall stem peas, 1 3 is DD, 2 3 is DD, and their inbred height ratio is 1:5

Calculation process: short stem: 2 3*1 4=1 6 tall stem: 1-1 6=5 6 The height ratio is 5:1

2. Aa and AA are crossed to obtain F1, and F1 is self-crossed to obtain F2, and the phenotype is the dominant individual to mate freely, and the ratio of the dominant traits of the offspring is 1:8

Calculation process: short stem: 2 3 * 2 3 * 1 4 = 1 9 tall stem: 1-1 9 = 8 9 The height ratio is 8:1

The former has all self-inbred tall stems, while the latter is a dominant individual that mates freely.

Assuming that A is the dominant gene and A is the recessive genetic disease gene, and the normal performing couples carry the genetic disease gene, that is, the probability of having a normal child with the AA gene is 3 4 (AA, AA, AA, AA). Since another normal woman's father is a patient with the disease (i.e., AA), it means that the woman's genes can only be AA. The crux of the matter comes out, the probability that a boy carries the disease-causing gene is 2 3.

Breaking down the genetic probabilities of boys and girls, there are (a, a, a), (a, a), respectively. The results are out, I think it's 3 5.

Phenotypic couples give birth to sick girls, born out of nothing, autosomal recessive diseases, independent of sex, and both parents of the male are carriers;

Men with a normal phenotype have a probability of being carriers of 2 3, and women are carriers.

Children born to couples: sick (2 6) carrying (3 6) not carrying (1 6) children with normal phenotype.

The probability of being a carrier is 3 4.

Am I doing something wrong?

It was indeed miscalculated.

Male genotype AA (1 3) AA (2 3).

Females are carriers and the genotype is AA

1 3 * 1 2 (probability of being a carrier when the male genotype is AA) + 2 3*2 4 (being a carrier when the genotype is AA) = 3 6

The probability of being behaving normal is 5 to 6

So in the end we get (3 6) (5 6) = 3 5

Select option d.

Choosing d can be judged from the first sentence, and this disease is an autosomal recessive disease. (Something out of nothing is hidden, if it is accompanied by x hidden, the daughter is sick, the father must have it, exclude.) Then, husband aa, wife aa.

This boy aa1 3, aa2 3. Normal women have AA (paternal disease. ）

then aa1 3, aa2 3xaa.

Normal child AA: 1 3x1 2+2 3x1 4=2 6AA: 1 3x1 2+2 3x1 2=3 6 Normal child carries the disease-causing gene = carries the disease-causing gene and is normal Normal child = 3 6 (3 6+2 6) = 3 5

Note: The key to this question is the denominator, not 1, but the probability of a normal child.

Start by identifying the explicit and recessive.

1) The pathogenic gene is recessive and creates something out of nothing.

2) The pathogenic gene is dominant, and there is no mesogenesis.

2.The location of the causative gene is then determined.

1) Chang Yin: Born out of nothing, there is a girl who is sick.

2) Chang Xian: It is normal to have middle school students and no daughters.

3) Accompanied by x hidden: the mother and the child will suffer, and the woman will suffer from the father.

4) Accompanied by X: Fathers and daughters will suffer, and children and mothers will suffer.

5) Inheritance with y: father to son, child to grandchild, children and grandchildren are endless.

3.The type of judgment that cannot be determined.

There are no such characteristics in the pedigree, and can only be speculated from the magnitude of the probability.

1) If the generations are continuous, it is likely to be dominantly inherited.

2) If there is no sex difference in the patient, 1 2 for male and female patients, it is likely that the autosomal gene is controlled.

3) If there is a significant sex difference, it is likely that the gene control of the sex chromosomes aPatients are significantly more male than female, and are likely to have a presence with x-recess.

b.Patients are significantly more female than male, and are likely to have X-ray appearance.

4.(1) If it is determined that it is recessive, and the mother's illness is normal for the son or the daughter is normal for the father to be sick, it must not be x hidden but often hidden.

2) If it is determined that it is dominant, and the father is sick and the daughter is normal or the son is sick and the mother is normal, it must not be x obvious, but often obvious.

5.Probability problems are generally used 9:3:3:1, 3:1, and the test is generally 1:1, and it should be noted that the probability of having a daughter (son) with disease is different from the probability of having a daughter (son) with disease, and the latter needs to be multiplied by 1 2

As for the inheritance of x, it should be noted that there is no need to multiply 1 2 when giving birth to a daughter or son must be sick, and in other cases, it is necessary to analyze it specifically, and the general probability problem will not be too difficult.

If it can't be stably inherited, it's YY, so it's calculated by probability. Let's assume one, then calculate. Count another one.

YY was crossed with YY of 2 3 to produce YY of 1 6, YY of 2 6, YY of 1 6, and YY of 1 6

YY is crossed with 1 3 YY to produce 1 6 YY and 1 6 YY

Therefore, there are 6 offspring, of which 5 are Y-, and 3 of the 5 are not stably inherited (i.e., heterozygous), so the answer is 3 5.

Both parents are sick, all daughters are sick, and the ratio of sons to sick is 1:2The disease is inherited dominantly on the X chromosome.

The father's genotype is XAY, the mother's genotype is XAXA or XAXA, and the son's normal XAY: diseased XAY = 1 2, indicating that the mother produces gametes XA: XA = 1:2, that is, the probability of XA gametes is 1 3, the probability of producing XA gametes when the mother's genotype is XAXA is 1 2, the probability of producing XA gametes when the mother's genotype is XAXA is 0, and the probability that the mother is heterozygous 1 2=1 3, then the probability that the mother is heterozygous 2 3

== The title tells us that all parents in the family are sick, so we can rule out the possibility that the mother is xbxb. Then let's first analyze the genotype of the mother in the problem, if the heterozygous is XBXB, then the genotype of the son who is not sick is XBY, and the sick is XBY, and the sons born from the heterozygous mother are half sick and not sick, so it is said that the non-sick son given in the condition is born by the heterozygous mother, and one of the 2 parts of the sick son is also born by the heterozygous mother, and then the remaining one is born by the homozygous mother (XBXB), so the children born to the heterozygous mother account for 2 3, and the proportion of homozygous mothers is 1 3, so the proportion of heterozygous mothers is 2 3

If the son is not sick, the mother must be heterozygous, and the son is sick, the mother must not be heterozygous, and the heterozygous ratio should be 2 3

The son's lack of illness is equal to one to two, which means that all the mothers provide xa:xa=2:1=4:

2. When xaxa:xaxa=1:2, the mother provides xa:

xa = 4:2, so the heterozygous ratio is 2 3.

The first question and the second question look similar, but in fact there are still differences, the biggest difference is that the first question is random mating, while the second question is self-breeding, if this question is understood, then the answer is clear.

1.Since it is random mating, knowing that the aa genotype is the loss of mate and reproductive ability, so the genotype probability of aa and aa is 3 respectively, then the gene frequency of a is 1 3 + 2 3 * 1 2 = 2 3, and the gene frequency of a is 2 3 * 1 2 = 1 3, so the probability of random mating to produce a a can only be a case, that is, 1 3 * 1 3 = 1 9;

2.The analysis method is the same as the first question, the genotype probability of AA and AA is 3 respectively, but if it is inbred, there is no doubt that only AA inbreeding can produce AA offspring, and the probability of producing AA at this time is 2 3 * 1 4 = 1 6 (where 1 4 is the probability of AA inbreeding to produce AA).

3.The solution method is the same as the first question, because they are all free mating, the answer is omitted, and the answer is 1 9 is correct.

The answer is one in two.

Suppose a is used for gray body and a for black body.

Then the genotype of the offspring should be AA

The gray body aa in the second generation of the child: aa = 1:2

The meaning of free mating is that all male and female fruit flies can mate.

So there are four ways to mate.

1. Male and female AA mate with each other.

2. Males and females aa mate with each other.

3. Female AA and male AA mate.

4. Female AA and male AA mate.

If you count them one by one, it will be very troublesome.

Introduce a simple point algorithm.

It's okay to put it on any topic.

It's the gamete method.

You see, one-third of the second-generation gray flies are homozygous, and two-thirds are heterozygous.

Then one-third of both male and female fruit flies are homozygous, and two-thirds of heterozygous homozygosity can only produce type A gametes, because homozygotes account for one-third of the parents, so it produces one-third of the gametes of type A gametes.

Heterozygous can produce two kinds of gametes, each accounting for half, because heterozygous accounts for two-thirds of the parents, so it produces type A gametes accounting for one-third of the total gametes, type A gametes accounting for one-third of the total gametes In summary, type A gametes account for two-thirds, type A gametes account for one-third of both male and female gametes, then you use the checkerboard method to calculate, then the dominant homozygous in the offspring: heterozygous: recessive homozygous = 4:4:1 so the offspring of the black body Drosophila homozygous accounts for one-half.

Hope it helps.

This method is most suitable for parents with two or more genotypes, a variety of mating combinations, and a large amount of computation.

Directly using gametes to calculate, it saves a lot of computational effort.

"Purebred gray fruit flies are crossed with black fruit flies, and the offspring are all gray bodies. Judging from this sentence, the gray body is explicit, and if it is set to A, then the black body is A. The genotype of the offspring is AA, and the offspring are allowed to mate freely to produce the second generation of offspring, and the genotypes of Drosophila gray in the second generation are 1 3AA, and 2 3AA.

In the second generation of children, the probability of a is 2 3, and the probability of a is 1 3, so the probability of a a free mating of the second generation of children is 2 3 * 2 3 = 4 9.

This question does not need to be calculated, 100%. Because the black body of Drosophila is recessive gene control, the black body is homozygous.

If you want to ask how much homozygous is in the gray body fruit fly, you need to calculate. The result is 1 2

If the genes are represented by a and a, it is obvious that the second generation of gray fruit flies is 1 3aa + 2 3aa, and it mates freely to obtain 4 9aa + 4 9aa + 1 9aa, so the gray body flies account for half of a and a aa.

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