Biology Compulsory 2 Urgent, about Biology Compulsory 2

Because black hair is dominant to brown hair, brown hair female B genotype is DD.

Because the black yak has 3 4 is heterozygous and 1 4 is homozygous. Therefore, the parent genotype of the black bull A has the following possible combinations:

dd + dd: The probability is 1 4 x 3 4 + 3 4 x 1 4 = 6 16;

dd + dd：1/4 x 1/4 =1/16；

dd + dd：3/4 x 3/4 = 9/16。

The probabilities of the genotype of the offspring of the various parent combinations are as follows:

dd + dd ——1/2 dd + 1/2 dd

dd + dd—— 1 dd

dd + dd ——1/3 dd + 2/3 dd

So the probability that the black haired male bull A genotype is DD is: 6 16 x 1 2 + 1 16 x 1 + 9 16 x 1 3 = 7 16;The probability of genotype DD is: 6 16 x 1 2 + 1 16 x 0 + 9 16 x 2 3 = 9 16.

If the black male A is DD, the probability of him giving birth to a brown cow is 0 with the brown female B.

If the black-haired male A is DD, the probability of giving birth to a brown-haired calf with the brown-haired female B is 1 2, and the probability of giving birth to a brown-haired cow is 1 4.

Therefore, the total probability of a black male A and a brown female B giving birth to a brown cow is 7 16 x 0 + 9 16 x 1 4 = 9 64.

I think you can take a look at this.

It was very detailed.

Parental: AABB and AABB crossing, the sub-** is theoretically 1AABB, 2AABB, 2AABB, 4AABB, 1AABB, 2AABB, 1AABB, 2AABB, 1AAB, due to the death of the AA or BB individual, so:

The only ones left: 2aabb, 4aabb, 1aabb, 2aabb, i.e.: yellow short tail: gray short tail = 2:1.

aabb aabb aabb(dead, 1 16) + aabb(dead, 2 16) + aabb(dead, 1 16) + aabb(live, 2 16) + aabb(live, 4 16) + aabb(dead, 2 16) + aabb(live, 1 16) + aabb(live, 2 16) + aabb(dead, 1 16) then the ratio of genotypes is: aabb (yellow bobtail): aabb (yellow bobtail):

AABB (Grey Hair Shorttail): AABB (Gray Hair Shorttail) = 2:4:

1:2, the expressive shape ratio is yellow-haired short-tailed: gray-haired short-tailed = 6:

If there is any miscalculation, please ignore it and correct it yourself.

The aabb was crossed with the aabb, which should have been 9:3:3:

1, but because AA is dead, so 1 AAB in 9, 2 AAB does not exist, so 9-3 = 6, and then AAB in 3, theoretically all can survive, because they do not contain AA and BB, and then look at BB, the next 3, all will die, because they are all A BB, containing BB, and the last 1 is AAB, which will definitely die, so the rest is 6:3=2:1

1.Start by determining the type of genetic disorder:

From II, there is no mesogenesis, and the nail disease is dominant, and the disease is inconsistent from II7 and III11, and the nail disease is not accompanied by X dominant, but autosomal dominant;

From II, out of nothing, it is known that B disease is recessive, and it is known that II3 is not a carrier of B disease genes, and B disease is known to be recessive with X.

2.Next, the III9 and III12 genotypes are determined

Since iii knows that ii genotypes are aaxby and aaxbxb, so iii9 genotype is 1 2aaxbxb or 1 2aaxbxb

From III11 to know that II genotypes are AAXBY and AAXB respectively, so III12 genotype is 1 3AAXBY or 2 3AAXBY.

3.Finally, calculate the probability:

The probability of disease A is 1 3 + 2 3 * 1 2 = 2 3, the probability of disease B is 1 8, and the probability of both diseases is 2 3 * 1 8 = 1 12, so the probability of disease is 2 3 + 1 8-1 12 = 17 24

Think about it carefully, 1:6:9, which adds up to exactly 16, is very similar to pea genetics, and take a closer look at the pea genetics on compulsory two**, you can infer that the purple gene is AABB (double hidden), the red is AABB, AABB and AABB (one visible and one hidden), and the blue one is AABB, AABB, AABB, AABB (double display).

Mix red pollen and there are 10 ways to gamete.

It's hard to fight, but I'll write it down to you: aabb aabb=aabbaabb aabb=aabb or aabb

aabb×aabb=aabb

aabb aabb = aabb or aabb

aabb×aabb=aabb

aabb aabb = aabb or aabb

aabb aabb=aabb or aabb or aabb or aabb or aabbaabb aabb=aabb

aabb aabb = aabb or aabb

aabb aabb=aabb or aabb or aabb you didn't say ask for the proportion, ask for it yourself, I'm too lazy to forget it, there are three types of phenotypes, and the genotypes are counted by yourself.

PS: I've played so much, I have to do it!

This has the answer you want, and the next time you have a question, you can look for it on the know first, and you really don't ask any more questions.

Two pairs of genes are controlled, set by two pairs of genes AA, BB, according to blue-violet = F1 is blue, the parental blue is AABB, violet is AAB, F1 is AABB, and appears blue. F1 is inbred, and the F2 ratio should be 9 (double display A-B-blue) 3 (single display A-BB red) 3 (AAB-red) 1 (AAB purple) so the genotype of red plants in F2 is AAB, AABB, AABB, AABB, and the pollen species produced are AB, AB, AB three kinds, and the genotypes of the fusion plants cultivated are AAB (blue), AAB (red), and AAB (red).

There are 3 genes in the offspring: dd, dd, dd. The proportions are: 1 4, 1 4, 1 2

The first two (dd, dd) are homozygous.

So, then draw one to get homozygous (dd or dd) with a chance of 1 2 If both are homozygous, then 1 2 * 1 2 = 1 4 The same is true of mathematics, which is just a calculation tool.

Biological calculations are just and also based on mathematical formulas.

I don't know what the landlord's "two stoppers" means, but I understand that there are 2 genes aadd There are 9 genes in the hybrid offspring: aadd, aadd, aadd, aadd, aadd, aadd. The proportions are:

The homozygous is: aadd, aadd, aadd, aadd, so, then draw one to get homozygous with a chance of 1 4

If both are homozygous, then 1 4 * 1 4 = 1 16 I hope it can help you.

Extract two possibilities: (list all possibilities, including duplicates, so that it is easy to see the result) dd dd, dd dd, dd dddd, dd dd, dd dd, dd dd, dd dd

4 16 = 1 4 is the answer.

dd dd, dd dd, dd dd, dd dddd, dd dd, dd dddd, dd dd, dd dd, dd dd, dd dd The results that meet the requirements are:

dd dd, dd dd, dd dd, dd dd biology at the beginning of the Mendelian hybridization experiment, didn't you learn to draw**, right** draw it, list all the possibilities, don't be afraid of trouble, it will help you solve the problem!

I also didn't learn this knowledge very well, but then the teacher taught me, and if I do more questions, it will be I hope it will help you

Offspring seeds dd:dd:dd 1:2:1

Probability that both grains are homozygous 1 2 1 2 1 4 Extraction situation:

dd dd；dd dd；dd dd；dd dd；dd dd；dd dd

It's a good match.

Seed combinations: dd dd dd dd There are four types of seeds.

Among them, there are two types of homozygous: dd and dd.

So it's 50%.

Only the hybrid combination of DD is given in the question, and the 16 types must be a hybrid combination of DDDD.

The answer should be B

Since the RR all die before flowering and does not affect the next generation, the parent generation is RR and RR account for 1 2 each. Then Rr produces R-containing gametes that account for 1 and produce an equal number of R and R gametes that account for 1 2, that is, each accounts for 1 4 of the total number of gametes. Then the gametes contain 3 4 and 1 4 in the gametes.

The probability of another zygote is equal to the product of the probability of the gamete. Therefore, susceptible (RR) plants account for 1 16 in the offspring.

RR Total Death is not considered so only RR RR can mate in a ratio of 1 to 1

Using the gamete method, if the rr ratio is 1, then 3 r's and 1 small r

then rr=1 4 1 4=b

d,1/8

All susceptible plants die before flowering, so the resistant plants RR and RR in the parent generation can produce offspring, each accounting for half, and the ratio of RR of resistant plants to RR of susceptible plants is 1:2:1

The landlord's account of this topic is incomplete.

According to grandparents and fathers, it can be known that this is a recessive disease.

However, it is not possible to tell whether it is autosomal recessive or concomitant X recessive.

The inference is as follows. If it is accompanied by X chromosome, then the paternal genotype is XAY, the grandmother is XAXA, the father is XAY, and the mother is XAXA

Both sons have a genotype of XAY

Full Compound Question Requirements.

If it is with autosomal recessiveness, then the paternal genotype is AA, grandmother, aa, father, mother, aa

Both sons had a normal genotype AA.

It is also fully compounded with the requirements of the topic.

In the book of high school compulsory two.

Chapter 2, Section 3 deals with companion inheritance, which is the inheritance of genes on the X and Y chromosomes.

Autosomal inheritance is discussed in Section 2.

If you haven't learned the third section of companion inheritance, it's better to write it as an autosomal recessive disease.

I saw you say that it was the teacher's question.

Then it means that only autosomal recessive properties are considered.

That's 1 2 correct.

As mentioned earlier, both sons have an AA genotype

Married to a sick woman, AA.

The offspring is aa:aa=1:1

AA is the sick individual.

So it's 1 2

You probably didn't learn Compulsory 2, Chapter 2, Verse 3, which talks about companionship.

If you don't consider companion inheritance without learning, then the genetic map is.

Father: BB Mother BB

Normal son bb sick daughter-in-law bb

The probability of the offspring suffering from BB is 1 2

If it is inherited with sex.

Normal son xby sick daughter-in-law xbxb

The probability of the offspring suffering from XBY is 1 2

If you haven't learned about companion inheritance, don't consider it).

First, let's see if the disease is dominant or recessive. From the grandparents are normal, to be able to have a sick father, saying that it is a recessive genetic disease (out of nothing). So, is it companion or autosomal?

This question cannot be judged, and the score is discussed in two situations. If it is inherited with sex, then the "one of the sons" in this question, marrying the sick person, the offspring will be the son who is sick to be able to get sick, that is, 1 2.

If it is autosome, then this "one of the sons" must be heterozygous, and the heterozygous marries the sick person, and the probability of changing the disease is 1 2. So the answer to this question is 1 2.

First of all, "the grandparents are normal, but the father is sick" indicates that the disease is recessive. Because the mother is normally heterozygous, it is not possible to determine whether the disorder is autosomal or inherited with sex.

If autosomal:

Grandfather: AA – Grandmother: AA

Father: AA – Mother: AA

Son 1: aa Son 2: aa - Sick woman: aa Therefore, the probability of aa of a sick child is: 1 2

In the case of sex chromosomes:

Grandfather: Xay – Grandmother: Xaxa

Father: Xay – Mother: Xaxa

Son 1: xay Son 2: xay - Sick woman: xaxaxaxa xay xay

Therefore, the probability of XAXA or XAY in the sick child is: 1 2

Therefore, regardless of whether the disease is autosomal or co-inherited, the probability of the offspring being infected is 1 2