Biological genotype calculation, how to calculate genotype

The probability of ff is 1 3, and the probability of ff is 2 3, so we can know that ff:ff=1:2 f has 4, f has 2 and has 6 in total

f is 4 6 = 2 3f f is 2 6 = 1 3f with a chessboard. 2/3f 1/3f

1/3f 4/9ff 2/9ff

2 3f 2 9ff 2 9ff (death).

The genotype of the offspring ff:ff=1:1

ff accounts for 4 9 ff accounts for 4 9

I don't know if it's going to be messy, this is a college entrance examination question this year, but the meaning is the same, the question has changed, and the letters have changed.

The probability of a population having genotype FF is 1 3, the FF probability is 2 3, and there are 4 cases of intra-population hybridization in a population.

1/3ff×1/3ff→1/9ff

2/3ff×2/3ff→4/9(1/3ff,2/3ff)1/3ff×2/3ff→2/9(1/2ff,1/2ff)→2/18ff,2/18ff

2 3ff 1 3ff 2 18ff, 2 18ff "There is a difference between this group and the previous group, the paternal parent and the female parent are different".

So the offspring. ff is 1 9 + 4 27 + 4 18 = 13 27

ff is 14 27

In this question, if ff dies, the gene frequency will change after intra-group hybridization, which is not suitable for calculation.

Free mating within a population is generally calculated using gene frequencies, because there are no special circumstances, and the gene frequencies of the population do not change. Here, there are special cases: FF death, just count it as a general situation first, and then eliminate this situation (you may not understand what I'm talking about, but please see me demonstrate).

The steps are as follows: in the original population.

The gene frequency of f is 1 3 + 2 3 * 1 2 = 2 3 The gene frequency of f is 2 3 * 1 2 =1 2 free mating within the population.

f 2/3 f 1/3

f 2/3 ff 4/9 ff 2/9

f 1/3 ff 2/9 ff 1/9

After free mating, unless otherwise specified, the genotype frequency of the offspring is.

ff = 4 9 (i.e. 4 out of 9).

ff = 2 9 + 2 9 = 4 9 (i.e. four out of nine) ff = 1 9 (i.e. one out of nine).

Then consider the special case: FF death, in the nine offspring except for the death of FF, that is, in the offspring actually survived eight shares, of which FF and FF each accounted for four copies.

So, the gene frequency of the offspring is:

ff = 1/2

ff = 1/2

Other special cases of free mating can also be used in the above general method).

First, the genotype is ff and ff, then the combination of parents is: ff and ff, ff and ff, ff and ff, and the corresponding ratio is: 1 9, 4 9, 4 9

Second, the genotype of its offspring is FF; ff vs. ff; ff, ff and ff and their corresponding ratios are: 1; 1/2,1/2；1/4,1/2,1/4

Third, because the frequency of FF type death is 4 9 * 1 4 = 1 9, so.

The frequency of ff is: 1 9*1+4 9*1 2+4 9*1 4=4 The frequency of 9ff is: 4 9*1 2+4 9*1 2=4 9

Algorithm of genotype.

Genotype type: AABB is crossed with AABB, and its progeny genotype types are: 3 times 2 = 6. Phenotypic type: AABB and AABB hybridization, representing the current type of species: 2 times 1 = 2.

Expression of genotype.

Genotype is a general term for all the genetic combinations of an individual in an organism. It reflects the genetic makeup of an organism, i.e., the sum of all genes obtained from both parents. The genotype used in genetics is often the genotype of a trait.

As long as two organisms have one locus that is different, then their genotypes are not the same, so genotype refers to all combinations of all loci of all alleles of an individual. In the hybridization test, it refers specifically to the gene combinations studied that are related to the segregation phenomenon, such as purebred high-stemmed peas (DD), hybrid high-stemmed peas (DD), etc. The genotype is generally not directly visible, and needs to be inferred from the phenotype through hybridization (crossing) test space-length test.

Categories: Education Science >> Entrance Examination >> College Entrance Examination.

Problem description: Aa and Aa were crossed to obtain F1, F1 was self-crossed to obtain F2, and the phenotype was taken as dominant to mate freely, and the ratio of dominant recessive traits in the offspring was (A).

Analysis: The title "free mating" refers to free interbreeding Li Zhen, right?

The dominant genotypes in F2 are dd (1 copy) and dd (2 copies), and the probability of producing recessive coarse traits from their free crosses is calculated

1.As long as one party is dd, there is no recessive trait, only two dds can be generated, and the probability of producing two dds in this step is 2 3 * 2 3 4 9 (the probability of the first person obtaining dd is 2 3, and the probability of the second place obtaining dd is 2 3, which is multiplied by the principle of step-by-step counting and multiplication).

2.The probability of two DDs producing a recessive trait (equivalent to F1 self-breeding) is 1 4

Therefore, the probability of producing recessive traits in the whole process is 4 9*1 4 1 9 (multiplication principle), so the probability of dominant traits is 8 9, and the ratio is 8:1.

a is the total number of genes.

p(a) Total number of A genes Total number of A genes.

aa is the number of individuals with genotype.

aa genotype frequency The total number of individuals in this diploid population is 2The known gene frequencies were found for the frequencies of the gene acacia type.

Gene Mingpein type: ABB and AABB crossed, and its offspring genotype types were: 3 times liter only 2=6 species.

Phenotypic type: ABB crossed with AABB, and its progeny phenotypic species were: 2 times 1 = 2.

Epigenotype, also known as hereditary type, reflects the genetic makeup of an organism, i.e., the sum of all genes obtained from both parents. It is estimated that there are about 35,000,000 pairs of structural genes in humans.

Therefore, the genotype of the whole organism cannot be represented, and the specific genotype used in genetics often refers to the genotype of a certain trait, such as the genotype of albinism is cc, which only indicates that this pair of alleles cannot produce coolinase. So the genotype is acquired from the parent and may develop into the genetic basis of a certain trait. <>

How genotype frequencies are calculated.

To calculate by proportion, it is necessary to look at the genes of the parents to calculate the approximate number of genotypes of the offspring.

The F2 separation ratio of a pair of relative traits is 1:2: Min Lu rotten 1, town teasing.

The f2 segregation ratio of two pairs of relative traits is the square of attack (1:2:1), which is 1:

2:1, i.e., the ratio of F2 bridging type in Mendel's hybridization experiment of two pairs of relative traits, 1AAB, Dao 2AAB, 2AAB, 4AAB, 1AABB, 2AAB, 1AABB, 2AABB, 1AAB.

Do you know (a+b)x(a+b) and what's after that? b+b=1, if what is after the square of b+b. You might as well do the math yourself. You will understand why B is for and B is for the sake of the question you are asking.

If the frequency of B gene in the population is x and the frequency of B gene is y, then x+y=1

Then the frequency of the BB gene in the population is X, the frequency of the BB gene is 2Xy, and the frequency of the BB gene is Y [this is a general rule].

In this question, the frequency of bb is directly given as 1%, so x=10%, y=90%, and 2xy=18% are calculated