Question

Помогите срочно!!!!!Задача 1. У томата гладкая кожица плодов доминирует над опушенной. Гомозиготная форма с гладкими плодами скрещена с растением, имеющим опушенные плоды. В F1 получили 54 растения, в F2 – 736. 1. Сколько типов гамет может образовывать растение с опушенными плодами? 2. Сколько растений F1 могут быть гомозиготными? 3. Сколько растений F2 могут иметь гладкие плоды? 4. Сколько растений F2 могут иметь опушенные плоды? 5. Сколько разных генотипов может образовываться в F2? Задача 2. Черный цвет щетины у свиней доминирует над рыжим. Какое потомство следует ожидать от скрещивания черной свиньи с генотипом FF и черного хряка с генотипом Ff? Задача 3. Нормальный слух у человека обусловлен доминантным геном S, а наследственная глухонемота определяется рецессивным геном s. От брака глухонемой женщины с нормальным мужчиной родился глухонемой ребенок. Определите генотипы родителей. Задача 4. У кроликов шерсть нормальной длины доминантна, короткая – рецессивна. У крольчихи с короткой шерстью родились 7 крольчат – 4 короткошерстных и 3 с нормальной шерстью. Определите генотип и фенотип отца. Задача 5. От скрещивания комолого быка айширской породы с рогатым коровами в F1 получили 18 телят (все комолые), в F2 – 95. Каково количество комолых телят в F2?

Answer 1

Task 1

1. To determine the number of types of gametes that can be formed by a plant with hairy fruits, we need to consider the genotype of the plant. Since the plant with smooth fruits is homozygous (having two identical alleles) for the smooth trait, it can only produce gametes with the smooth allele. On the other hand, the plant with hairy fruits is heterozygous (having two different alleles) for the trait, so it can produce two types of gametes: one with the smooth allele and one with the hairy allele. Therefore, the plant with hairy fruits can form two types of gametes.

2. In the F1 generation, 54 plants were obtained. To determine the number of plants in the F1 generation that can be homozygous, we need to consider the genotype of the plants in the F1 generation. Since the F1 generation is the result of crossing a homozygous plant with smooth fruits (genotype: SS) and a plant with hairy fruits (genotype: Hh), all the plants in the F1 generation will be heterozygous (genotype: Ss) for the trait. Therefore, none of the plants in the F1 generation can be homozygous.

3. In the F2 generation, 736 plants were obtained. To determine the number of plants in the F2 generation that can have smooth fruits, we need to consider the possible genotypes of the plants in the F2 generation. The possible genotypes in the F2 generation are SS, Ss, and ss. Out of these, only the plants with the genotypes SS and Ss will have smooth fruits. Therefore, the number of plants in the F2 generation that can have smooth fruits is unknown without additional information.

4. In the F2 generation, 736 plants were obtained. To determine the number of plants in the F2 generation that can have hairy fruits, we need to consider the possible genotypes of the plants in the F2 generation. The possible genotypes in the F2 generation are SS, Ss, and ss. Out of these, only the plants with the genotype ss will have hairy fruits. Therefore, the number of plants in the F2 generation that can have hairy fruits is unknown without additional information.

5. To determine the number of different genotypes that can be formed in the F2 generation, we need to consider the possible combinations of alleles. In this case, the possible genotypes in the F2 generation are SS, Ss, and ss. Therefore, there are three different genotypes that can be formed in the F2 generation.

Task 2

In this task, we are given that the black color of bristles in pigs is dominant over the red color. The genotype of the black pig is not provided, but we are given the genotype of the black pig as FF. The genotype of the black pig is FF, which means it is homozygous dominant for the black color trait. The genotype of the black pig is Ff, which means it is heterozygous for the black color trait.

When a black pig with the genotype FF is crossed with a black pig with the genotype Ff, the possible genotypes of the offspring can be determined using a Punnett square. The Punnett square for this cross would look like this:

   F     F
F  FF    FF
f  Ff    Ff

From the Punnett square, we can see that all the offspring will have the genotype FF, which means they will all have the black color trait. Therefore, the expected offspring from crossing a black pig with the genotype FF and a black pig with the genotype Ff will all have the black color trait.

Task 3

In this task, we are given that normal hearing in humans is determined by the dominant gene S, while inherited deafness is determined by the recessive gene s. From the given information, we know that the deaf-mute woman is homozygous recessive for the deafness trait (genotype: ss) and the normal man is homozygous dominant for the normal hearing trait (genotype: SS).

When a deaf-mute woman with the genotype ss is married to a normal man with the genotype SS, the possible genotypes of the offspring can be determined using a Punnett square. The Punnett square for this cross would look like this:

   S     S
s  Ss    Ss
s  Ss    Ss

From the Punnett square, we can see that all the offspring will have the genotype Ss, which means they will all have the normal hearing trait. Therefore, the genotypes of the parents are ss for the deaf-mute woman and SS for the normal man.

Task 4

In this task, we are given that the length of fur in rabbits is determined by the dominant gene for normal length (N) and the recessive gene for short length (n). We are also given that a female rabbit with short fur (genotype: nn) gave birth to 7 offspring, out of which 4 had short fur and 3 had normal fur.

From the given information, we can determine the genotype and phenotype of the father rabbit. Since the female rabbit has short fur (genotype: nn), she can only pass on the recessive allele for short fur (n) to her offspring. Therefore, the father rabbit must have the genotype Nn, which means he is heterozygous for the fur length trait.

The phenotype of the father rabbit can be determined by looking at the offspring. Out of the 7 offspring, 4 had short fur and 3 had normal fur. Since the father rabbit has the genotype Nn, which means he has one dominant allele for normal fur (N) and one recessive allele for short fur (n), his phenotype will be normal fur.

Therefore, the genotype of the father rabbit is Nn and his phenotype is normal fur.

Task 5

In this task, we are given that the polled (hornless) trait in Ayrshire cows is dominant over the horned trait. When a polled bull (genotype: PP) is crossed with horned cows, all the offspring in the F1 generation are polled.

From the given information, we can determine the genotype of the polled bull. Since all the offspring in the F1 generation are polled, it means that the polled bull can only pass on the dominant allele for the polled trait (P) to its offspring. Therefore, the genotype of the polled bull is PP.

In the F2 generation, 95 offspring were obtained. To determine the number of polled offspring in the F2 generation, we need to consider the possible genotypes of the offspring. The possible genotypes in the F2 generation are PP, Pp, and pp. Out of these, only the offspring with the genotypes PP and Pp will be polled. Therefore, the number of polled offspring in the F2 generation is unknown without additional information.