Senior 2 Biology Basic Laws of Heredity Please answer in detail, thank you 19 9 23 41

Method: The test and homozygous were mated.

Objective: To test whether they are homozygous or heterozygous.

Offspring are homozygous if only dominant traits are expressed, heterozygous if there is a 1:1 trait ratio, and recessive homozygous if only recessive homozygous.

The purpose of the assay is to determine the type and proportion of gametes produced by F1 or an individual, and to describe the behavior of genes in F1 or an individual when gametes are formed.

Method: The individual is crossed with another recessive homozygous.

Results**:1If there are 2 n-second phenotypes in the offspring, and the ratio is 1:1:1......, indicating that the individual is heterozygous and follows Mendelian's laws of inheritance when forming gametes.

2.If there is only one dominant trait in the offspring, it means that the individual is dominant homozygous.

Objective: To test genotype.

Method: Hybridization with recessive homozygous.

Principle: It is simple and straightforward to test genotype with recessive homozygous crossing.

For the recessive homozygous of the parent, the gamete provided does not contain the dominant gene, so the gene expression of the gamete provided by the other parent in the genotype of the inbred progeny is not masked. In this way, the phenotype of the inlaid offspring can directly reflect the gamete genotype of another parent, and the type and proportion of the phenotype of the inbred offspring can be directly deduced by the type and proportion of the phenotype of the inbred offspring. (And then deduce its genotype.)

Results**:1The offspring have two phenotypes, dominant and recessive, and the parental genotype is AA2If the offspring only have a dominant phenotype, the parental genotype is AA

3.If the offspring have only a recessive phenotype, the parental genotype is AA

Crossing is the use of recessive genotypes and heterozygous mating, resulting in shape separation.

Dominant recessive genes can be distinguished by crossing.

For example, when AA and AA mate, the offspring are all dominant traits.

AA and AA mate, and the offspring will have shape separation.

Select D.

The effect of controlling genes here is an additive effect. It's like the genes that control the shade of the human skin.

AABCC weighs 210, AABCC weighs 120If you get a difference of 90, you will be evenly divided into six dominant genes, that is, if there is one dominant gene in the genotype, it will be 15 heavier than AABCC

The fruit of F1 is 135-165. Make a difference with 120 and divide by 15Get 1-3That is, there are 1 to 3 dominant genes in the f1 genotype.

In the parent, the genotype of A is AABCC, and the four answers in the option are crossed with A.

a。The offspring can have a maximum of two dominant genes i.e. AABCC does not satisfy 1-3 dominant genes.

b。There are up to four offspring i.e. AABCC does not meet 1-3.

c。There are a maximum of three offspring, i.e., AABCC, but at least two, i.e., AABCC does not meet 1-3.

d。There are a maximum of three offspring, i.e., AABCC, and a minimum of one, i.e., AABCC

According to AABCC210 and AABCC120, a dominant gene can increase the weight of 15 grams, F1 fruit is 135-165, it can be seen that it contains 1-3 dominant genes, and A provides a dominant gene for F1, and B can provide 1-2 dominant genes for F1, so D is selected.

The first is that the husband is a because his brother is a patient (AA) and the rest of the family is not sick. The parents of the launch husband are both AASo the chance that the husband is AA is 2 3 (the odds of not being sick don't need to be counted as AA).

Then the wife is also A, and her maternal grandmother is a patient (AA), and the wife's mother is not sick, so the wife's mother must be AA. The wife's father is considered homozygous and the non-diseased genotype is (AA). The probability of finding a wife AA is 1 2

So the probability of their son getting sick is: 2 3*1 2*1 4=1 12

The answer is: a

I think it's better for you to ask online than to buy a reference book.

To really figure out the type of questions, you can use reference books and teachers!

Many dactyly, so there must be a d, and the child born does not have many dd, so it must not be all d, that is, there is a d.

dd solution. Albinism is recessive, and the child manifests it, indicating that everyone has one C, and neither parent shows this symptom, indicating that the other is C.

cc solution. In this way, only polydactyly is DDC The DDD probability is 1 2, and the C probability is 3 4, so it is 3 8.

Only whitening ddcc, dd probability 1 2, cc probability 1 4, so it is 1 8.

Both suffer from DDCC, DD1 2, CC1 8.

Neither ddc, dd1-2, c3-4.

From the meaning of the title, the father is d c and the mother is ddc, because there is a child who is ddcc, then d *dd gives birth to dd, then the father is dd, and in the same way, c *c gives birth to cc, then it must be cc*cc.

Polydactyly only: The probability of d*dd polydactyly is 1 2, and the probability of not suffering from albino cc*cc is 3 4, then 1 2*3 4=3 8

Albinism only: the probability of dd*dd not suffering from polydactyly is 1 2, while the probability of suffering from albino cc*cc is 1 4, then 1 2*1 4=1 8

The probability of suffering from both albinism and polydactyly is 1 2, and the probability of suffering from albino cc*cc is 1 4, then 1 2*1 4=1 8

This kind of question is not difficult, as long as the two probabilities are separated! ~~

It's quite aabb....Self-inbreeding, trait A has a probability of 3 4, trait b is also, and because n pairs do not affect each other, they are independent events, so this is the result of the nth trait.

Well, I don't remember what the law is, but there are 3 concepts, dominant, recessive and completely dominant, for example, rr and mm these two sets of alleles (uppercase for dominant genes, lowercase for recessive genes), combined with 4 equal probability possibilities, namely rm, rm, rm, rm, complete dominance that is, rm (all dominant), rm, rm, rm is dominant (dominance includes complete dominance), rm is recessive, so the dominant individual is 3 4, when there are n pairs of alleles, it is n times of (3 4).

Complete dominance means, for example, in genes a and a, as long as a is contained in a gene, then such a gene is completely dominant.

Of course, there are also incomplete explicitness, co-dominance, etc., which are not considered here!

If a pair of genes aa is freely combined, then aa is 1 4 is recessive, and the others are dominant, so a pair of genes freely combined dominant is 3 4

Therefore, when the n-pair alleles are freely combined and fully dominant, the proportion of dominant individuals in the F2 generation is (3 4) to the nth power.

After answering, if you don't understand, please ask.

In order to be able to answer this question correctly, it is necessary to understand the conceptual connotation of gene frequency in Chapter 7 [Modern Biological Evolutionary Theory] of Biology Compulsory 2 "Heredity and Evolution", and at the same time, you should know the application of the law of genetic equilibrium in order to answer this question more easily.

Solution: From the description of the title, it can be seen that the genetic disease is a recessive genetic disease, and 1 out of 10,000 people have this disease, so it can be known that the genotype of patients with recessive gene inheritance is AA, and their proportion in the whole population is AA 1 10000, that is, it is equivalent to.

aa a a a a 2 1 10000 so the gene frequency of a a 1 100

According to the law of genetic equilibrium, we can see that a a 1 so the gene frequency of a a 99 100

Once the gene frequencies of a and a are known, the solution can be analyzed.

1) The proportions of each genotype in these 10,000 people are:

The genotype is AA 99 100 99 100 9801 10000

The genotype is AA 2 99 100 1 100 198 10000

The genotype is AA 1 100 1 100 1 10000

2) The probability of a carrying male marrying a female and the offspring getting sick is:

From the analysis of the theme, it can be seen that the genotype carrying the male must be AA, while the female genotype is possible for all three genotypes, that is, there are the following three genetic combinations:

aa aa, this genetic combination produces offspring that will not have the disease.

aa aa , the probability of this genetic combination producing offspring with a diseased person is 1 4 1

aa aa, the probability of this genetic combination producing offspring with a diseased person is 1 2 1

Based on the above analysis, it can be concluded that the probability of a carrying male marrying a female and the offspring will be diseased

3) If a carrying male marries a normal female, the probability of the offspring suffering from the disease is:

From the analysis of the title, it can be seen that the genotype of carrying males must be AA, and the genotype of normal females has two genotypes, namely AA and AA, and only after marriage with a woman whose genotype is AA can the disease occur, and the concept of offspring disease is 1 4, and because the phenotypic normal female, the probability that the individual with the genotype AA accounts for all normal females is:

aa ÷ aa＋aa ) 2/101

Based on the above analysis, it can be seen that if a carrying male marries a normal female, the probability of offspring suffering from the disease is:

aa = 1 10000, i.e. the gene frequency of a is 1 100 and the gene frequency of a is 99 100

The male carrier AA probability is 2*1 100*99 100=198 10000

Females normally have AA or AA probability is squared, 198 10,000 offspring are diseased, that is only possible if both men and women are carriers, that is, 1 4 * male and female gametes each account for 1 100 because the gametes produced by males are all male gametes and females are all female gametes The gene frequencies are the same The probability of producing a is The probability of producing a is The probability of producing a is.

Choose C, because the parents are round, F1 is all flat disks, so the F1 genotype is all ABB, and the heterozygous genes are both ABB genes and flat disks, and only contain one A or B are round, so the parents are C

The inbred results of F2 conform to the ratio of 9:6:1, while the parents are round, and F1 is flat disk, so the flat disk shape is bidominant, the round shape is monodominant, and the oblong is double recessive. The parents are monodominant and homozygous. The answer is c