Biology questions about family pedigree: Albinism a Color blindness b .

1) I1 and I2 genotypes: I1 YXBAA, I2 XBXBAA

2) What kinds of egg cells can I2 produce XBA, XBA, XBA, XBA The proportion of egg cells containing AB is 1 4

3) I1 and I2 may have several genotypes in a girl born XBXBAA (1 8), XBXBAA (1 8), XBXBAA (1 8), XBXBAA (1 8), XBXBAA (1 4), XBXBAA (1 8), several expressive (normal girls, albino girls).

4) 3 and 4 Probability of having an albinistic child in marriage (detailed process).

ii3 xbxbaa, xbxbaa, ii4 yxbaa, yxbaa, albino 2 3*1 2=1 3

5) If 3 and 4 give birth to a boy with two diseases at the same time, they are in the probability of having a normal boy (detailed process) question add:

yxb1/2*1/2*1/2=1/8

aa 2/3*1/2=1/3

Normal boy 7 8 * 2 3 = 7 12

1)aaxbxb aaxby

2) axb axb axb axb 1 44) Because one of 4's sisters has albinism, his parents have a genotype AA aa, 4 has a probability of being aa and a probability of 1 4 and a probability of 1 2And because the genotype of 3 is AA, the probability of having a child with albinism is 1 2 1 2 = 1 4

5) The genotype of 3 is aaxbxb 4 is aaxby The probability of having a normal boy is 1 2 1 2 1 2 = 1 8

Albinism is an autosomal recessive disorder, and color blindness is a recessive disorder with an X chromosome.

7 is color blindness (the color blindness gene in men can only come from the mother), and the causative gene comes from the mother, that is, the second generation of the No. 4 individual, and 8 is an albinism patient but does not suffer from color blindness, so its genotype is aaxbxb or aaxbxb, each accounting for 1 2.

If the second generation has genotype 5 aaxby, genotype 6 is aaxby, and 11 is normal, then its genotype is (1 3) aaxby or (2 3) aaxby. The probability of having a child with color blindness alone = 1 2*1 3*1 4+1 2*2 3*1 2*1 4=1 12 (color blindness alone equals the probability of not having albinism multiplied by the probability of color blindness, there are only two cases: (1 2)aaxbxb and (1 3)aaxby or (1 2)aaxbxb and (2 3)aaxby).

The genotype according to Question 7 is AAXBY; The 10 genotypes are (1 3) aaxbxb or (2 3) aaxbxb. The probability of having a child with albinism = 2 3*1 4=1 6, and the probability of having a completely normal child = 1 2*1 2*1 4+2 3*1 4*1 4=5 48.

Albinism is an autosomal recessive disorder, and I use a a to denote normal and white genes. Therefore, the genotype of the disease must be AA, and the normal individual may be AA or AA

Such topics are generally divided into two steps.

One pushes the various genotypes and probabilities of the parents, and the other calculates the probability of the offspring.

One: First look at —2

The first step of genotype inference method: write out the known gene according to the phenotype: from the graph, we know that -2 is normal, so there must be a normal A gene.

Step 2: Push Unknown Genes: Method:

Recessive homozygous breakthrough method. Both from her parents or offspring to find AA, from this topic, the base parents do not have AA diseased individuals. So at this time, it is not certain what its genotype must be.

However, probable genotypes and probabilities can be derived based on the parent's genotype.

The parents must be AA and AA when the parents are normal but give birth to a diseased daughter, so the parents give birth to a child with three genotypes: AA1 4, AA1 2, and AA1 4. However, because the phenotype of -2 is normal, it cannot be AA that is to eliminate AA, so the genotype probability of -2 should be 1 3AA, 2 3AA

In the same way —3 is also 1 3aa, 2 3aa

Therefore, the two of them are married, and there should be four situations according to different genotypes.

Both: 1 3aa*1 3aa

1/3aa*2/3aa

2/3aa*1/3aa

2/3aa*2/3aa

Note: The former one is —2 and the latter one is —3

The probability of giving birth to a sick child in the first three of the above four situations is 0, and the probability of giving birth to a sick child in the fourth case is: 2 3*2 3*1 4 1 9

It's over, I'm writing according to the general method of doing genetic questions, which may be a bit redundant for this question, but I hope it will be useful to you. Able to deal with all of these questions in this way. Be able to draw inferences.

Word by word, tired!

"Something out of nothing"is recessive. Therefore, all generations are heterozygous, so the probability that -2, -3 is heterozygous is 2 3 (because there is no disease, the double hidden will be excluded). The probability of heterozygosity to give birth to AA double hidden is 1 4.

Therefore, if you want to be able to give birth to a diseased child, then = —2 and —3 are both heterozygous probability * chance of giving birth to double hidden = 2 3 * 2 3 * 1 4

1) Recessive inheritance.

2）5-aa;9-aa

3)aa(1/4),aa(1/2)

4) 8 AA probability 2 3, 9 patient AA, both give birth to patient x1 4, so 1 9

5) The chance of having a daughter is 1 2, and the chance of the offspring getting sick is 1 4, so the chance of having a daughter and getting sick is 1 8

Number 10 is the probability of AA 2 3x patient AA = 2 3 x 1 2 = 1 36)a

If it is accompanied by x hidden, 9 will be sick and 5 will be sick, so it is not accompanied by x hidden, it is often hidden.

Heterozygous AA, patient 9 AA

The genotype of 8 may be AA1 3 or AA2 3 AA 2 3, 9 patient AA, both give birth to patient X1 4, so 1 9 AA4 AA has the chance of giving birth to a patient 1 4 and must be a girl x1 2 = 1 The probability of AA 2 3 x Patient AA=2 3X1 2=1 3

6.Injection of normal genes does not change germ cells, so the offspring AAXAA = AA 100% diseased A

1 Recessive Something out of nothing is recessive.

2 aa aaa

3 aa 1/3 aa 2/3

5 1/8 1/36 a

(1) Known human white.

Baihua disease is an autosomal recessive disorder controlled by a pair of alleles DAOA and A, following Mende.

The law of segregation of the Huier gene

2) In the diagram answer, III7 is patient AA, so the genotype of normal individual 3 is AA (3) Since III7 is patient AA, it can be seen that both parents are AA, so the probability that III8 is heterozygous AA is 23

4) It is known that 4 is AA, since it is a patient, and 6 genotype is AA, so the genotype of III10 is AA, which is exactly the same as the genotype of 4

So the answer is: 1) Gene segregation. 2）aa

(1) The pathogenic gene of human albinism is inherited recessively on autosomes.

2) The genotypes of 3 and 6 in the figure are AA and AA, respectively.

2) The probability that 9 is homozygous in the figure is 0.

3) The genotype of 12 in the figure is AA or AA.

4) The 5 genotypes in the figure are most likely AA. (Actually, both AA and AA are possible, but the following three are all A).

5) If 10 marry a normal man (whose brother is a patient), the probability of their children getting sick is 1 6

Anyway, the fifth question is a little tangled, but according to the meaning of the question, the man's parents are normal, and the man is 1 3 AA

The probability of 1 3aa and aa(10) is 0

The probability of 2 3aa and aa(10) is 2 3*1 4=1 6

It should be the answer, if there are any questions to ask.

If the other party is not a carrier of the same type, it is theoretically normal, at most a carrier.

First, albinism is an autosomal recessive disorder. 3 marries a heterozygous woman, and the son born is an albino, it can be seen that 3 is heterozygous, that is, AA and AA mating this model, the probability of giving birth to AA disease is 1 4, the probability of giving birth to a girl is 1 2, and multiplying is 1 8

Albinism is an autosomal recessive disease, if the disease is set to AA, i1, 2 must be AA, II3 one-third is AA, two-thirds are AA, and the son born of a heterozygous female AA is an albino, then II3 is AA, AA and AA mating This model, the probability of giving birth to AA is 1 4, the probability of giving birth to a girl is 1 2, and multiplying is 1 8

(1) From the above analysis, it can be seen that Y disease is color blindness

2) 2 has color blindness, but does not have albinism, and its genotype is AAXBY, 11 has neither albinism nor color blindness, but both parents carry the color blindness gene, so 11 genotype is

3) 7 genotype is aaxby, 10 genotype and probability is 2 3aaxbxb or 1 3aaxbxb, the probability of having an albinism child after marriage is 2 3 1 4 = 1 6, then the probability of having a child with normal skin color is 1-1 6 = 5 6, and the probability of having a child with albinism but normal color vision is 1 6 1 2 = 1 12 There is one person with albinism for every 10,000 people in a certain area, that is, the probability of aa is 1 10,000, then the gene frequency of a is 1 100,a has a gene frequency of 99 100, so the genotype frequency of aa is (99 100)2, and the genotype frequency of aa is 2 99 100 1 100, of which aa accounts for 99 101 and aa accounts for 2 101, so 7 marrying a well-behaved woman in the area has a probability of having a boy with albinism about 2 101 1 4 1 1 2 1 400

4) Human genetic diseases include monogenic genetic diseases, polygenic genetic diseases and chromosomal abnormal genetic diseases, which can be seen that the root causes of genetic diseases are gene mutations and chromosomal variations

So the answer: (1) color blindness (2) aaxby aaxby or aaxby (3) 5 6 1 12 1 400

4) Genetic mutations on chromosomal variations.