Question

ДАЮ 100 БАЛЛОВ! ПОМОГИТЕ ПОЖАЛУЙСТА!!! Буду очень благодарна) Практична робота №3 " Складання схем схрещування " Задача № 1 У собак чорний колір шерсті домінує над кавовим. Який відсоток чорних щенят можна чекати від схрещування двох особин, одна з яких гетерозиготна чорна, а друга –гомозиготна кавова? Задача № 2 У великої рогатої худоби ген чорного забарвлення домінує над геном коричневого забарвлення. Якими будуть генотип і фенотип потомства від схрещування чорного гетерозиготного бика з чорною гетерозиготною коровою? Задача №3 У гарбуза біле забарвлення плодів домінує над жовтим, а кругла форма - над видовженою. Яким буде розщеплення за фенотипом при дигібридному схрещуванні батьківських особин з генотипами АаВВ х ааВb? Задача №4 Довгошерстого чорного самця морської свинки схрестили з чорною короткошерстою самкою. Отримано 15 свинок з короткою чорною шерстю, 13 - з довгою чорною, 4 - з короткою білою, 5 - з довгою білою. Визначте генотипи батьків, якщо чорна і довга шерсть є домінуючими проявами ознак. Сформулюйте і запишіть висновок.

Answer 1

Ответ:

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Answer 2

Task 1: Crossbreeding of Dogs with Dominant Black Coat Color and Homozygous Coffee Coat Color

To determine the percentage of black puppies that can be expected from the crossbreeding of two individuals, one of which is heterozygous black and the other is homozygous coffee, we need to consider the inheritance pattern of coat color in dogs.

In dogs, the black coat color is dominant over the coffee coat color. This means that if an individual has at least one copy of the black coat color gene (B), it will have a black coat. On the other hand, an individual needs to have two copies of the coffee coat color gene (bb) to have a coffee coat.

When we crossbreed a heterozygous black individual (Bb) with a homozygous coffee individual (bb), the possible genotypes and phenotypes of the offspring can be determined using a Punnett square.

The Punnett square for this crossbreeding would look like this:

| | b | b | | ---- | ------ | ------ | | B | Bb | Bb |

From the Punnett square, we can see that all the offspring will have the genotype Bb, which corresponds to a black coat. Therefore, 100% of the puppies can be expected to have a black coat [[1]].

Task 2: Crossbreeding of Cattle with Dominant Black Color Gene and Brown Color Gene

To determine the genotypes and phenotypes of the offspring from the crossbreeding of a heterozygous black bull with a heterozygous brown cow, we need to consider the inheritance pattern of coat color in cattle.

In cattle, the black color gene (B) is dominant over the brown color gene (b). This means that if an individual has at least one copy of the black color gene, it will have a black coat. On the other hand, an individual needs to have two copies of the brown color gene to have a brown coat.

When we crossbreed a heterozygous black bull (Bb) with a heterozygous brown cow (Bb), the possible genotypes and phenotypes of the offspring can be determined using a Punnett square.

The Punnett square for this crossbreeding would look like this:

| | B | b | | ---- | ------ | ------ | | B | BB | Bb | | b | Bb | bb |

From the Punnett square, we can see that there are three possible genotypes for the offspring: BB (black coat), Bb (black coat), and bb (brown coat). Therefore, the genotypic ratio of the offspring will be 1 BB : 2 Bb : 1 bb. The phenotypic ratio will be 3 black coat : 1 brown coat.

In conclusion, the genotypes and phenotypes of the offspring from the crossbreeding of a heterozygous black bull with a heterozygous brown cow will be as follows: - 25% of the offspring will have a homozygous black genotype (BB) and a black coat. - 50% of the offspring will have a heterozygous black genotype (Bb) and a black coat. - 25% of the offspring will have a homozygous brown genotype (bb) and a brown coat [[2]].

Task 3: Splitting by Phenotype in Dihybrid Cross of Pumpkin Plants

To determine the splitting by phenotype in the dihybrid cross of pumpkin plants with white dominant fruit color and round dominant fruit shape, we need to consider the inheritance patterns of these traits.

In pumpkin plants, white fruit color (A) is dominant over yellow fruit color (a), and round fruit shape (B) is dominant over elongated fruit shape (b).

When we crossbreed two parental individuals with the genotypes AaBB and aaBb, the possible genotypes and phenotypes of the offspring can be determined using a Punnett square.

The Punnett square for this crossbreeding would look like this:

| | Aa | Aa | | ---- | ------ | ------ | | Bb | AaBB | AaBb | | Bb | AaBB | AaBb |

From the Punnett square, we can see that there are four possible genotypes for the offspring: AaBB (white and round), AaBb (white and round), aaBB (yellow and round), and aaBb (yellow and elongated). Therefore, the phenotypic ratio of the offspring will be 2 white and round : 1 yellow and round : 1 yellow and elongated.

In conclusion, the splitting by phenotype in the dihybrid cross of pumpkin plants with the genotypes AaBB and aaBb will be as follows: - 50% of the offspring will have white fruit color and a round shape. - 25% of the offspring will have yellow fruit color and a round shape. - 25% of the offspring will have yellow fruit color and an elongated shape [[3]].

Task 4: Determining the Genotypes of Guinea Pig Parents

To determine the genotypes of the parents in the crossbreeding of a long-haired black male guinea pig with a short-haired black female guinea pig, we need to consider the inheritance patterns of coat length and color in guinea pigs.

In guinea pigs, the long-haired coat (L) is dominant over the short-haired coat (l), and the black color (B) is dominant over the white color (b).

From the given information, we know that the offspring from the crossbreeding are as follows: - 15 offspring have short black hair. - 13 offspring have long black hair. - 4 offspring have short white hair. - 5 offspring have long white hair.

Based on this information, we can deduce the genotypes of the parents.

Since the long-haired coat is dominant over the short-haired coat, the male guinea pig must be heterozygous for coat length (Ll) because it has both long-haired and short-haired offspring.

Similarly, since the black color is dominant over the white color, both the male and female guinea pigs must be heterozygous for coat color (Bb) because they have both black and white offspring.

Therefore, the genotypes of the parents can be determined as follows: - The male guinea pig: LlBb (heterozygous for coat length and coat color) - The female guinea pig: LlBb (heterozygous for coat length and coat color) [[4]].

In conclusion, the genotypes of the parents in the crossbreeding of a long-haired black male guinea pig with a short-haired black female guinea pig are both LlBb.