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Tasks in genetics from ct. Dominant genes in chickens In chickens, variegated plumage coloration dominates

Task 5
When crossing a rooster and a chicken with variegated feathers, offspring were obtained: 3 black chicks, 7 variegated and 2 white. What are the genotypes of the parents?
Solution:
If, when crossing phenotypically identical (one pair of traits) individuals in the first generation of hybrids, the traits are split into three phenotypic groups in a 1:2:1 ratio, then this indicates incomplete dominance and that the parental individuals are heterozygous.
With this in mind, we write the crossover scheme:

It can be seen from the record that the splitting of traits by genotype is 1:2:1. Assuming that the variegated chicks have the Aa genotype, then half of the first-generation hybrids should be variegated. In the conditions of the problem, it is said that in the offspring of 12 chickens, 7 were variegated, and this really makes up a little more than half. What are the genotypes of black and white chickens? Apparently, black chickens had the AA genotype, and white ones - aa, since black plumage, or, more precisely, the presence of pigment, as a rule, is a dominant trait, the absence of pigment (white color) is a recessive trait. Thus, we can conclude that in this case, the black plumage in chickens incompletely dominates over white; heterozygous individuals have variegated plumage.

Mendel's laws

We all studied at school and half-heartedly listened to the experiments on peas by the fantastically meticulous priest Gregor Mendel in biology lessons. Probably few of the future divorcees guessed that this information would ever be needed and useful.

Let's remember Mendel's laws together, which are valid not only for peas, but for all living organisms, including cats.

Mendel's first law is the law of uniformity of hybrids of the first generation: with monohybrid crossing, all offspring in the first generation are characterized by uniformity in phenotype and genotype.

Let us consider, as an illustration of Mendel's first law, the crossing of a black cat homozygous for the black color gene, that is, "BB" and a chocolate cat, also homozygous for chocolate color, which means "cc".

Parents (P): BB x cc kittens (F1): BB BB BB

With the fusion of germ cells and the formation of a zygote, each kitten received from the father and from the mother a half set of chromosomes, which, when combined, gave the usual double (diploid) set of chromosomes. That is, from the mother, each kitten received the dominant allele of the black color "B", and from the father - the recessive allele of the chocolate color "c". Simply put, each allele from the maternal pair is multiplied by each allele of the paternal pair - so we get all the possible combinations of alleles of the parental genes in this case.

Thus, all born kittens of the first generation turned out to be phenotypically black, since the black color gene dominates chocolate. However, all of them are carriers of the chocolate color, which does not appear phenotypically in them.

Mendel's second law is formulated as follows: when crossing hybrids of the first generation, their offspring give splitting in a ratio of 3:1 with complete dominance and in a ratio of 1:2:1 with intermediate inheritance (incomplete dominance).

Let's consider this law on the example of black kittens we have already received. When crossing our littermate kittens, we will see the following picture:

F1: Vin x Vv F2: Vv Vv Vv Vv

As a result of such crossing, we got three phenotypically black kittens and one chocolate one. Of the three black kittens, one is homozygous for black and the other two are chocolate carriers. In fact, we got a split of 3 to 1 (three black and one chocolate kitten). In cases with incomplete dominance (when the heterozygote shows a weaker dominant trait than the homozygote), the splitting will look like 1-2-1. In our case, the picture looks the same, taking into account chocolate carriers.

Analyzing cross used to determine the heterozygosity of a hybrid for one or another pair of traits. In this case, the hybrid of the first generation is crossed with a parent homozygous for the recessive gene (cc). Such crossing is necessary because in most cases homozygous individuals (BB) do not differ phenotypically from heterozygous ones (BB)
1) heterozygous hybrid individual (BB), phenotypically indistinguishable from homozygous, in our case, black, crosses with a homozygous recessive individual (cv), i.e. chocolate cat:
parent couple: Вв x вв
distribution in F1: Вв Вв вв вв
i.e., a 2:2 or 1:1 split is observed in the offspring, confirming the heterozygosity of the test individual;
2) the hybrid individual is homozygous for dominant traits (BB):
R: BB x cc
F1: Vv Vv Vv Vv - i.e. splitting does not occur, which means that the test individual is homozygous.

Purpose of a dihybrid cross- trace the inheritance of two pairs of traits simultaneously. With this crossing, Mendel established another important pattern - independent inheritance of traits or independent divergence of alleles and their independent combination, later called Mendel's third law.

To illustrate this law, let's introduce the clarification gene "d" into our formula for black and chocolate colors. In the dominant state "D" the lightening gene does not work and the color remains intense, in the recessive homozygous state "dd" the color is lightened. Then the color genotype of a black cat will look like “BBDD” (assume that it is homozygous for the traits of interest to us). We will cross it not with a chocolate, but with a lilac cat, which genetically looks like a clarified chocolate color, that is, “vvdd”. When these two animals are crossed in the first generation, all kittens will be black and their color genotype can be written as BvDd., i.e. all of them will be carriers of the chocolate gene "b" and the clarification gene "d". Crossing such heterozygous kittens will perfectly demonstrate the classic 9-3-3-1 splitting, corresponding to the third law of Mendel.

For the convenience of evaluating the results of dihybrid crossing, a Punnett grid is used, where all possible combinations of parental alleles are recorded (the topmost line of the table - let the combinations of maternal alleles be written in it, and the leftmost column - we will write paternal combinations of alleles in it). As well as all possible combinations of allelic pairs that can be obtained by descendants (they are located in the body of the table and are obtained by simply combining parental alleles at their intersection in the table).

So we cross a pair of black cats with genotypes:

VvDd x VvDd

Let's write in the table all possible combinations of parental alleles and possible genotypes of kittens obtained from them:

So, we got the following results:
9 phenotypically black kittens - their genotypes are BBDD (1), BBDd (2), BbDD (2), BbDd (3)
3 blue kittens - their genotypes are BBdd (1), Bbdd (2) (the combination of the clarification gene with black color gives a blue color)
3 chocolate kittens - their genotypes are bbDD (1), bbDd (2) (the recessive form of black color - "c" in combination with the dominant form of the clarification gene allele gives us a chocolate color)
1 lilac kitten - its genotype is bbdd (the combination of chocolate color with a recessive homozygous clarification gene gives a lilac color)

Thus, we obtained the splitting of traits by phenotype in the ratio 9:3:3:1.

It is important to emphasize that this revealed not only signs of parental forms, but also new combinations that gave us chocolate, blue and lilac colors as a result. This cross showed the independent inheritance of the gene responsible for the lightened color from the coat color itself.

Independent combination of genes and cleavage based on it in F2 in a ratio of 9:3:3:1 is only possible under the following conditions:
1) dominance must be complete (with incomplete dominance and other forms of gene interaction, the numerical ratios have a different expression);
2) independent splitting is valid for genes located on different chromosomes.

Mendel's third law can be formulated as follows: the alleles of each allelic pair are separated in meiosis independently of the alleles of other pairs, combining in gametes randomly in all possible combinations (with monohybrid crossing there were 4 such combinations, with dihybrid crossing - 16, with trihybrid crossing, heterozygotes form 8 types each gametes, for which 64 combinations are possible, etc.).

Cytological foundations of Mendel's laws
(T.A. Kozlova, V.S. Kuchmenko. Biology in tables. M., 2000)

The cytological foundations are based on:

  • pairing of chromosomes (pairing of genes that determine the possibility of developing any trait)
  • features of meiosis (processes occurring in meiosis that provide independent divergence of chromosomes with genes located on them to different cell pluses, and then to different gametes)
  • features of the process of fertilization (random combination of chromosomes carrying one gene from each allelic pair) Additions to Mendel's laws.

    Far from all the results of crossings found during the research fit into the laws of Mendel, hence the additions to the laws arose.

    The dominant feature in some cases may not be fully manifested or even absent. In this case, there is that called intermediate inheritance, when none of the two interacting genes dominates the other, and their action is manifested in the animal's genotype to an equal extent, one trait, as it were, dilutes the other.

    An example is the Tonkinese cat. When Siamese cats are crossed with Burmese kittens are born darker than Siamese, but lighter than Burmese - such an intermediate color is called Tonkinese.

    Along with the intermediate inheritance of traits, there is a different interaction of genes, that is, genes responsible for some traits can affect the manifestation of other traits:
    -mutual influence- for example, the weakening of the black color under the influence of the Siamese color gene in cats that are its carriers.
    -complementarity- the manifestation of a trait is possible only under the influence of two or more genes. For example, all tabby colors appear only in the presence of the dominant agouti gene.
    -epistasis- the action of one gene completely hides the action of another. For example, the dominant white gene (W) hides any color and pattern, it is also called epistatic white.
    -polymerism- a whole series of genes affects the manifestation of one trait. For example - the density of wool.
    -pleiotropy- one gene affects the manifestation of a series of traits. For example, the same gene for white color (W) linked to blue eyes provokes the development of deafness.

    Linked genes are also a common deviation, which, however, does not contradict the laws of Mendel. That is, a number of traits are inherited in a certain combination. An example is sex-linked genes - cryptorchidism (females are its carriers), red color (it is transmitted only on the X chromosome).

    In garden peas, the genes responsible for tendril development and seed surface shape are located on the same pair of homologous chromosomes. The distance between them is 16 M. Rothel homozygous pea plants were crossed, having smooth seeds and antennae (dominant traits) and wrinkled seeds without antennae. Hybrids F 1 subjected to analyzing crossing. What is the % chance that F2 hybrids will have plants with wrinkled seeds and tendrils?

    A gray hen was crossed with a black rooster. The gray allele is dominant. Color genes are located on the X chromosome. Calculate the % of gray roosters in F 2 hybrids.

    When crossing piebald rabbits with uniformly colored ones, only piebald rabbits appeared in the offspring. In F 2 - 24 piebald rabbits and 8 uniformly colored. How many piebald rabbits are likely to be homozygous?

    In humans, myopia dominates over normal vision, and brown eye color dominates over blue. A brown-eyed, short-sighted man whose mother had blue eyes and normal vision married a blue-eyed woman with normal vision. What is the probability of a birth in % of a child with signs of the mother?

    A healthy man married a healthy woman whose father had no sweat glands (a sex-linked recessive trait) and whose mother and her ancestors were healthy. What percentage of the sons of these parents may not have sweat glands?

    A woman with hypoplasia (thinning) of tooth enamel married a man with the same defect. From this marriage a boy was born who did not suffer from this disease. It is known that the hypoplasia gene is dominant and localized on the X chromosome. Determine the probability in % of the appearance of a girl with an enamel defect in this family.

    Albinism in humans is inherited as an autosomal recessive trait. In a family where one of the spouses is an albino and the other is normal, brothers were born, one of whom is normal and the other is an albino. What is the % chance that the next child will be an albino?

    In corn, smooth seeds dominate over wrinkled ones, and colored seeds dominate over colorless ones. These genes are located on the same chromosome at a distance of 3.6 morganids. A plant with a diheterozygous genotype was crossed, at which it inherited the genes for smoothness and coloration from one parent, with a homozygous plant with wrinkled colorless seeds. Calculate the % of plants with smooth colored seeds in F 1.

    In canaries, the green color of plumage dominates over brown and is determined by the gene localized in the X chromosome, and the short beak dominates over the long one and is determined by the gene localized in the autosome. When crossing a green male with a short beak and a short-billed brown female, offspring were obtained with a different combination of all phenotypic traits. What % of offspring will have green plumage with a long beak?

    In chickens, the variegated color of the plumage dominates over white and is determined by a gene located on the X chromosome, and feathered legs dominate over naked ones and are determined by a gene localized on the autosome. When crossing a motley-colored rooster with feathered legs and a white hen with feathered legs, offspring were obtained with a different combination of all phenotypic traits. What percentage of the resulting offspring will have variegated plumage and bare legs?

    In humans, cataract and polydactyly are determined by dominant autosomal genes located at a distance of 32 M from each other. One of the spouses is heterozygous for both traits. At the same time, he inherited cataracts from one parent, and polydactyly from the other. The second spouse has a normal transparent lens and a normal five-fingered hand. What is the probability (in %) of the birth in the family of a child with a normal transparent lens and a five-fingered hand?

    From a cross between a homozygous gray long-winged male Drosophila and a homozygous black female with rudimentary wings in F 1, offspring with a gray body and long wings are obtained. The genes for body color and wing length are inherited linked, and the distance between them is 19 M. What is the probability (in%) of the appearance of gray flies with rudimentary wings when a Drosophila female from the F 1 generation is crossed with a black male with rudimentary wings.

    Rh-positiveness and elliptocytosis are determined by dominant autosomal genes located at a distance of 3 M. One of the spouses is heterozygous for both traits. At the same time, he inherited Rh-positiveness from one parent, and elliptocytosis from the other. The second spouse is Rh-negative and has normal red blood cells. What is the probability (in%) of the birth in the family of a child with a positive Rh factor and normal erythrocytes?

    Rh-positiveness and elliptocytosis are determined by dominant autosomal genes located at a distance of 3 M. One of the spouses is heterozygous for both traits. At the same time, he inherited Rh-positiveness from one parent, and elliptocytosis from the other. The second spouse is Rh-negative and has normal red blood cells. What is the probability (in %) of the birth in the family of a child with a negative Rh factor and elliptocytosis?

    In minks, the length of the coat and its color are inherited independently. From the crossing of dihomozygous short-haired dark minks with dihomozygous long-haired whites, short-haired minks with light fur color and a black cross on the back (kohinoor minks) are born. When crossing F 1 hybrids, 64 minks were obtained. Determine how many of them were long-haired Koh-i-Noor minks, if the splitting corresponded to the theoretically expected.

    In minks, the length of the coat and its color are inherited independently. From the crossing of dihomozygous short-haired dark minks with dihomozygous long-haired whites, short-haired minks with light fur color and a black cross on the back (kohinoor minks) are born. When crossing F 1 hybrids, 64 minks were obtained. Determine how many of them were short-haired dark minks, if the splitting corresponded to the theoretically expected.

    In horses, height and coat color are inherited independently. From crossing dihomozygous high bay (red) horses with dihomozygous low albinos, tall foals are born with a golden-yellow body color with an almost white mane and tail (palamino color). When crossing F 1 hybrids, 32 foals were obtained. Determine how many of them were low bay foals, if the splitting corresponded to the theoretically expected.

    In horses, height and coat color are inherited independently. From crossing dihomozygous high bay (red) horses with dihomozygous low albinos, tall foals are born with a golden-yellow body color with an almost white mane and tail (palamino color). When crossing F 1 hybrids, 32 foals were obtained. Determine how many of them were short palamino foals, if the splitting corresponded to the theoretically expected.

    In sheep, coat color and ear length are inherited independently. From the crossing of dihomozygous long-eared dark sheep with dihomozygous earless light sheep, short-eared dark lambs are born. When crossing F 1 hybrids, 32 individuals were obtained. Determine how many of them were short-eared dark lambs, if the splitting corresponds to the theoretically expected.

    In Drosophila, the yellow body color gene and the white-eyed gene are linked and are located on the X chromosome, while the number of normal and crossover gametes is formed in equal parts. The corresponding wild-type dominant alleles determine gray body color and red eyes. In the experiment, females of wild-type pure lines and males recessive in both genes (heterogametic sex) were crossed. Then the hybrids of the first generation were crossed with each other, and 40 eggs were obtained. Calculate how many eggs will produce yellow-bodied males with red eyes.

    In Drosophila, the gene for "ragged" wings and the gene for "pomegranate" eyes are linked and are located on the X chromosome, while the number of normal and crossover gametes is formed in equal parts. The corresponding wild-type dominant alleles determine normal wing length and red eyes. In the experiment, females of wild-type pure lines and males recessive in both genes (heterogametic sex) were crossed. Then the hybrids of the first generation were crossed with each other, and 56 eggs were obtained. Calculate how many eggs will produce males with "cut" wings and "pomegranate" eyes.

    In Drosophila, the yellow body color gene and the white-eyed gene are linked and are located on the X chromosome, while the number of normal and crossover gametes is formed in equal parts. The corresponding wild-type dominant alleles determine gray body color and red eyes. In the experiment, females of wild-type pure lines and males recessive in both genes (heterogametic sex) were crossed. Then the hybrids of the first generation were crossed with each other, and 64 eggs were obtained. Calculate how many eggs will produce males with a gray body and white eyes.

    In Drosophila, the yellow body color gene and the white-eyed gene are linked and are located on the X chromosome, while the number of normal and crossover gametes is formed in equal parts. The corresponding wild-type dominant alleles determine gray body color and red eyes. In the experiment, females of wild-type pure lines and males recessive in both genes (heterogametic sex) were crossed. Then the hybrids of the first generation were crossed with each other, and 48 eggs were obtained. Calculate how many eggs will produce females with a gray body and red eyes.