Biology genetics, Biology genetics

There are 6 types. yyrr,yyrr,yyrr,yyrr,yyrr,yyrr.

We only study one pair of alleles at a time.

i.e. study yy x yy. first

They can produce 2 genotypes.

Re-study RR x RR

They can produce 3 genotypes.

So a total of 2x3=6 types are generated.

The genes that control different relative traits are separated, calculated separately, and then multiplied by theorems.

There are 2 genotypes of yy yy progeny and 3 genotypes of rr rr progeny and they are freely combined, and the genotype of the offspring is 2 3 = 6

Another example is Yyrr inbred, the genotype of the offspring is 3 3 = 9, and the phenotype is 2 2 = 4

yyrr,yyrr,yyrr,yyrr,yyrr,yyrr.Study yy x yy firstThey can produce 2 genotypes.

Re-study RR x RRThey can produce 3 genotypes.

So a total of 2x3=6 types are generated.

yyrr yyrr

yyrr yyrr

yyrr yyrr

The six examples are actually relatively simple, and if you use the calculation formula, I think it's a bit more troublesome.

It depends

Use the checkerboard method. It's not very troublesome, but brother, don't worry, the exam won't test this kind of dead knowledge. I've just entered my second year of high school, and my biology is okay... Khan, untruthful and modest... Really) If there's a problem, I'd be happy to discuss it with you

Trait segregation ratio of hybrid inbred offspring: Taking AABB as an example, the trait segregation ratio of inbred progeny should be 4:2:

2:1, the rest is to bury dates 1:1, and it is impossible to bend and tear down 1:

1：1：1；The proportion of gamete types produced by hybrids is still taken as an example of ABB, and its gamete types are ab, ab, ab, and ab, and their proportions are all 1 4, so their proportion is 1:

1：1：1；The phenotypic proportion of hybrid offspring, taking AABB AABB as an example, the genotype of its offspring is AABB, AABB, AABB, AABB, and their phenotypes are different, and the proportion of their offspring is 1 4, so the proportion is 1:

1:1:1, which can be interpreted and ; The proportion of genotypes in the genotype of hybrid inbred offspring is still taken as an example of AABB AABB, whose offspring have nine genotypes, and it is impossible to appear 1:

1:1:1, if it is AAB AABB, then there are only three genotypes, and the same is true for AABB AABB.

That's all I have to say, if there is anything inappropriate, I hope you will forgive me!

The ratio of trait segregation of self-inbred offspring of Shenling cavity hybrids: 9:3:3:1 The ratio of gamete categories produced by hybrids: 1:1:1:1

Phenotypic ratio of hybrid offspring: 1:1:

Genotype ratio of 1:1 hybrid inbred offspring: 4:

1:1 (you can refer to the checkerboard).

Genotype ratio of hybrid offspring: 1:1:1:1

So choose B

The first is AABB AABB, with a trait segregation ratio of 9:3:3:1, and the specific examples are in the book.

The second He Meng is the gamete of Zen Huqiao, ab, ab, ab, ab, the third is aabb aabb, and the offspring have four genotypes as aabb, aabb, aabb, aabb, aabb, and there are four kinds of expressions, each with a ratio of 2 1 2 1

There are 9 types of the fourth genotype.

Assuming the parent is (AABB).

1. The proportion of hybrid inbred offspring should be.

2. The gametes produced by hybrids are 1:1:1:

1（ab:ab:ba:

ab) 3, you can see that his cousin is noisy (aabb:aabb:aabb:

AAB)4, the genotype ratio of hybrid self-staring offspring is: 9:3:3:15, lateral crossing is that heterozygous parents and recessive parents have children, that is, hybrid parents can produce four kinds of gametes that are produced by 2, while recessive can only produce one gamete.

If you are a Shanghai textbook, page 11 of the third volume should be helpful to you.

b The ratio of trait separation to width of hybrid self-bred progeny should be 9; 3;3;1. Exclude AC

The phenotypic ratio of the heteronormal species side is 1; 1；1；1

Set the hybrid genotype AABB, then the inbred offspring 9:3:3:1, which breed so 1 wrong. The proportion of offspring phenomorphism is not 1 1 1 1. If the genomorph of Brother Li is ABB, the proportion of offspring is 1 1 1 1.

The manifestation may be 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Genotype ratio of hybrid inbred offspring: 1:1.

Question 1: First of all, the color of the seed coat is related to the gene expression of the parent.

f1：gg x gg

f2： gg

f3:1/4gg 2/4gg 1/4gg

The color of the seed coat of the seeds produced by the F3 plant is the F3 genotype 3 4 gray and 1 4 white.

Therefore, the separation ratio of the seed coat color of the seeds produced by F3 plants is 3:1

Question 2: Since red is not fully visible to white, so AA will not show red, let's think pink!

g：aa x aa

f1： aa

f2: 1/4aa 2/4aa 1/4aa

So safflower only aa accounts for 1 4 in f2

Question 3: G: AA

f1：1/4aa ； 2/4aa； 1/4aa

f2：1/4aa ； 1/8aa 2/8aa 1/8aa； 1/4aa

f3: 1/4aa ； 1/8aa 1/16aa 2/16aa 1/16aa 1/8aa; 1/4aa

So the proportion of heterozygous in the F3 generation is 1 8

The answer is that B is inherited dominantly with X.

The females in the parents are dominant pure and xaxa, while the males are recessive xay, and the offspring are xaxa and the males are xay

Female and male offspring are crossbred with offspring.

xaxa (female), xay (male), xaxa (female), xay (male), (ratio 1:1:1:1).

The separation ratio of male individuals is 1:1

The answer is a, of course, it can be accompanied by x dominant, the female is dominant in the parent, and the male is recessive.

1. Mating is the mating of heterozygous F1 with recessively homozygous individuals.

From 1:1, it can be seen that the allele of heterozygous F1 is separated by minus one, and finally the gamete has one of the two genes with equal probability.

So the measurement of the intersection is to prove the law of separation.

1:2:1 How do you intuitively know that?! All you can get straight to is 3:1

2. Note that "born a girl bald" is not "born bald girl"; The former girl is a prerequisite and does not require an additional x1 2, while the latter requires x1 2

6. It is normal for the C phenotype to be affected by the environment, and it is originally a product of the interaction between genotype and the environment. For example, if people are exposed to the sun for a long time, it will be dark, and this is not because there are black people in your family.

a, the offspring have both recessive and dominant traits, which is a combination of different traits that occur (not necessarily free).

7,b,AAXAA offspring must be AA, so 100% of them are red fruits, which is completely inconsistent with the first set of experiments.

11, missing picture, 20, yellow x yellow offspring is yellow: gray = 2:1

This means that yellow is dominant and gray is recessive.

So yyxyy=yy:yy:yy=1:2:1

Now there is only 2:1 left, which shows that yy is homozygous and lethal, and the lethal part accounts for 1 4 = 25% of the total

1. (1) Autosomal recessive inheritance According to the judgment of the I3 and I4 who do not suffer from nail disease and the number II8 girl who has nail disease, it is judged that (2) with X chromosome dominant inheritance According to the figure, it can be seen that the male parents who suffer from B disease and the offspring of female parents who do not suffer from B disease are both male with B disease, and women are not diseased.

2. Autosomal recessive inheritance Autosomal recessive inheritance Autosomal recessive inheritance.

Autosomal dominant inheritance with x recessive inheritance with y inheritance.

Autosomal recessive inheritance Autosomal recessive inheritance.

Autosomal recessive inheritance Autosomal dominant inheritance with x recessive inheritance.

It's good to try this kind of question a few more times.

49 5000, according to Hardy-Weinberg's law, one in every 2500 people suffers from the disease, so the probability of AA is 1 2500, so the frequency of A is 1 50, and A is 49 50, so the probability of carrier AA is 98 2500. Mary is a healthy couple who have a boy with the disease after marriage, and Mary is the carrier AA, so the chance of a remarried couple having a child with the disease is (98 2500) * (1 4) = 49 5000

If the couple has a healthy patient, it means that the disease is autosomal recessive.

Both David and Mary had AA genotypes, and the child had AA genotypes.

Later, Mary remarried to a normal man, who was either AA or AA, with both odds accounting for 1 2 each

The child born at the time of AA is completely normal.

The probability of having a child with AA is 1 4

So the chance of the remarried couple having a child with the disease is 1 4 * 1 2 = 1 8