Monohybrid cross
№1. One child in the family was born healthy, and the second had a severe hereditary disease and died immediately after birth.
What is the probability that the next child in this family will be healthy? One pair of autosomal genes is considered.
Solution. We analyze the genotypes of the parents: both parents are healthy, they cannot have this hereditary disease, because it leads to the death of the body immediately after birth.
If we assume that this disease manifests itself in a dominant manner and the healthy trait is recessive, then both parents are recessive. Then they cannot have a sick child, which contradicts the conditions of the problem.
If this disease is recessive, and the gene for a healthy trait is inherited in a dominant manner, then both parents must be heterozygous and they can have both healthy and sick children. We draw up a crossing scheme:
Answer: The ratio in the offspring is 3:1, the probability of having a healthy child in this family is 75%.
№2. A tall plant was pollinated with a homozygous organism having normal stem growth. The offspring produced 20 plants of normal growth and 10 plants of high growth.
What split does this cross correspond to - 3:1 or 1:1?
Solution: A homozygous organism can be of two types: dominant (AA) or recessive (aa). If we assume that normal stem growth is determined by a dominant gene, then all the offspring will be “uniform,” and this contradicts the conditions of the problem.
For “splitting” to occur, a normal-sized plant must have a recessive genotype, and a tall plant must be heterozygous.
Answer: The ratio of phenotype and genotype in the offspring is 1:1.
№3. When black rabbits were crossed with each other, the offspring were black and white rabbits.
Draw up a crossing scheme if it is known that one pair of autosomal genes is responsible for coat color.
Solution: The parent organisms have the same phenotypes – black color, but “splitting” has occurred in the offspring. According to G. Mendel's second law, the gene responsible for the development of black color dominates and heterozygous organisms are crossed.
№4. Sasha and Pasha have gray eyes, and their sister Masha has green eyes. The mother of these children is grey-eyed, although both her parents had green eyes. The gene responsible for eye color is located on a non-sex chromosome (autosome).
Determine the genotypes of parents and children. Draw up a crossing diagram.
Solution: Based on the mother’s body and her parents, we determine that gray eye color is a recessive trait (G. Mendel’s second law).
Because “cleavage” is observed in the offspring, then the paternal organism must have green eyes and a heterozygous genotype.
№5. Mother is brunette; the father is blond, there were no brunettes in his pedigree. Three children were born: two blonde daughters and a brunette son.
The gene for this trait is located on the autosome.
Analyze the genotypes of offspring and parents.
Solution: The genotype of the paternal organism must be homozygous, because In his pedigree there is a pure line in hair color. The homozygous genotype is dominant (AA) or recessive (aa).
If the father’s genotype is homozygous dominant, then the offspring will not have children with dark hair - “uniformity” will appear, which contradicts the conditions of the problem. Therefore, the father's genotype is recessive. The maternal organism must be heterozygous.
Answer: The ratio of phenotype and genotype in the offspring is 1:1 or 50% 50%.
№6. A person develops a disease called sickle cell anemia. This disease is expressed in the fact that red blood cells are not round, but sickle-shaped, as a result of which less oxygen is transported.
Sickle cell anemia is inherited as an incompletely dominant trait, and the homozygous state of the gene leads to the death of the organism in childhood.
In the family, both spouses have signs of anemia.
What is their percentage chance of having a healthy child?
Solution:
Answer: 25% of healthy children in this family.
Dihybrid crossing independent inheritance of genes
№1. Gene mutations that cause shortened limbs (A) and longhair (V) in sheep, are transmitted to the next generation in a recessive manner. Their dominant alleles form normal limbs (A) and short hair (IN). The genes are not linked.
The farm bred rams and sheep with dominant traits and produced 2,336 lambs. Of these, 425 are long-haired with normal limbs and 143 are long-haired with short limbs.
Determine the number of short-haired lambs and how many of them have normal limbs?
Solution. We determine the genotypes of parents based on recessive offspring. According to the rule of “purity of gametes,” in the offspring for each trait one gene is from the paternal organism, the other gene is from the maternal organism, therefore, the genotypes of the parents are diheterozygous.
1). Find the number of long-haired lambs: 425 + 143 = 568.
2). We find the number of shorthairs: 2336 – 568 = 1768.
3). We determine the number of shorthairs with normal limbs:
1768 ---------- 12 o'clock
x ----------- 9 hours x = 1326.
№2. A person has a gene for black skin color (IN) completely dominates the European skin gene (V), and the disease sickle cell anemia is manifested by an incompletely dominant gene (A), and allelic genes are in a homozygous state (AA) lead to the destruction of red blood cells, and the organism becomes unviable.
The genes for both traits are located on different chromosomes.
A pure Negroid woman gave birth to two mulattoes from a white man. One child had no signs of anemia, and the second died of anemia.
What is the probability of having the next child without signs of anemia?
Solution. We draw up a crossing scheme:
Answer: The probability of having a healthy child in this family is 1/4 = 25%
№3. Recessive genes (A) And (With) determine the manifestation of human diseases such as deafness and albinism. Their dominant alleles control the inheritance of normal hearing (A) and synthesis of melanin pigment (WITH).
The genes are not linked.
Parents have normal hearing; brunette mother, albino father. Three identical twins were born with two characteristics.
What is the probability that the next child in this family will have both diseases?
Solution.
According to the rule of “purity of gametes,” it was determined that the parents are diheterozygous:
Answer: The probability of having a child with both diseases is 1/8 = 12.5%
№4. Two pairs of autosomal genes exhibiting independent inheritance are being studied.
A rooster with a rose comb and feathered legs is bred with two hens with a rose comb and feathered legs.
From the first hen, chickens with feathered legs were obtained, some of them had a rose-shaped comb, and the other part had a simple comb.
The chickens from the second hen had a pink comb, and some of them had feathered legs and some had unfeathered legs.
Determine the genotypes of a rooster and two hens.
Solution.
According to the conditions of the problem, both parents have the same phenotypes, and in the offspring from two crosses, splitting occurred for each trait. According to G. Mendel's law, only heterozygous organisms can produce “splitting” in their offspring. We draw up two crossing schemes.
Interaction of nonallelic genes
№1. Two pairs of non-allelic unlinked genes that determine fur color in ermine are being studied.
Dominant gene of one pair (A) determines the color black, and its recessive allele (A)- blue color.
Dominant gene of the other pair (IN) contributes to the manifestation of pigmentation in the body, its recessive allele (V) does not synthesize pigment.
When black individuals were crossed with each other, the offspring turned out to be individuals with blue fur, black and albinos.
Analyze the genotypes of the parents and the theoretical ratio in the offspring.
Solution.
Answer: 9 black, 3 albino, 4 blue.
№2. The inheritance of plumage color in chickens is determined by two pairs of non-allelic unlinked genes located on the autosome.
Dominant gene of one pair (A) determines the synthesis of the melanin pigment, which ensures the presence of color. Recessive gene (A) does not lead to the synthesis of pigment and the chickens turn out to be white (feather albinism).
Dominant gene of the other pair (IN) suppresses the action of the genes of the first pair, as a result of which pigment synthesis does not occur, and chickens also become albinos. Its recessive allele (V) does not have a downward effect.
Two organisms heterozygous for two pairs of alleles are crossed.
Determine the ratio of chickens with colored plumage and albinos in the offspring.
Solution.
Answer: 13 white, 3 painted.
№3.
In oats, grain color is determined by two pairs of non-allelic unlinked genes.
One dominant gene (A) determines black color, another dominant gene (IN)- grey colour. The black gene suppresses the gray gene.
Both recessive alleles determine the white color of the grains.
When pollination of diheterozygous organisms occurred, the offspring produced plants with black, gray and white grains.
Determine the genotypes of parental organisms and the phenotypic ratio in the offspring.
Solution.
Answer: 12 black, 3 gray, 1 white.
Inheritance of genes located on sex chromosomes
№1. Normal blood clotting gene (A) in humans it is inherited in a dominant manner and is linked to X-chromosome. Recessive mutation of this gene (A) leads to hemophilia - inability to clot blood.
U-chromosome does not have an allelic gene.
Determine the percentage probability of having healthy children in a young family if the bride has normal blood clotting, although her sister has signs of hemophilia. The groom's mother suffers from this disease, and his father is healthy.
Solution. 1) Determine the genotype of the bride. According to the conditions of the problem, the bride's sister has a recessive genotype X a X a , which means both sisters receive the hemophilia gene (from their father). Therefore, a healthy bride is heterozygous.
2) Determine the genotype of the groom. Mother of the groom with signs of hemophilia X a X a , therefore, according to the chromosomal theory of sex, she passes the recessive gene to her son X a U .
Answer: the phenotypic ratio is 1:1, 50% of children are healthy.
№2. One pair of allelic genes is studied in X-chromosome that regulates color vision in humans.
Normal color vision is a dominant trait, while color blindness is a recessive trait.
Analyze the genotype of the maternal organism.
It is known that the mother has two sons, one of them has a sick wife and a healthy child. The second family has a daughter with signs of color blindness and a son whose color vision is normal.
Solution. 1) Determine the genotype of the first son. According to the conditions of the problem, he has a sick wife and a healthy child - it can only be a daughter X A X a . The daughter received the recessive gene from her mother, and the dominant gene from her father, therefore, the genotype of the male body is dominant (X A U) .
2) Determine the genotype of the second son. His daughter is sick X a X a , which means that she received one of the recessive alleles from her father, therefore the genotype of the male body is recessive (X a U -) .
3) We determine the genotype of the mother’s organism based on her sons:
Answer: mother's genotype is heterozygous X A X a .
№3. Albinism in humans is determined by a recessive gene (A) located on the autosome, and one form of diabetes is determined by a recessive gene (V) , linked to the genitals X -chromosome.
Dominant genes are responsible for pigmentation (A) and normal metabolism (IN) .
U -chromosome does not contain genes.
The couple have dark hair color. Both mothers suffered from diabetes, and fathers were healthy.
One child was born with two symptoms.
Determine the percentage probability of having healthy and sick children in this family.
Solution. Applying the rule of “gamete purity” we determine the genotypes of the parents by hair color - heterozygous genotypes Ahh.
According to the chromosomal theory of sex, it was determined that the father has diabetes X in Y - and the mother is healthy 10th century
We make a Punnett grid - the gametes of the father's body are written out horizontally, and the gametes of the mother's body are written down vertically.
Answer: six organisms out of sixteen are dominant in two traits - the probability of birth is 6/16 = 37.5%. Ten patients: 10/16 = 62.5%, of which two patients had two characteristics: 2/16 = 12.5%.
№4. Two recessive genes located in different areas X-chromosomes cause diseases such as hemophilia and muscular dystrophy in humans. Their dominant alleles control normal blood clotting and muscle tone.
U-chromosome does not contain allelic genes.
The bride’s mother suffers from dystrophy, but according to her pedigree she has normal blood clotting, and her father had hemophilia, but without any dystrophic signs.
The groom exhibits both diseases.
Analyze the offspring in this family.
Solution.
Answer: all children have the disease, 50% with hemophilia and 50% with dystrophy.
Inheritance of linked genes. The phenomenon of crossing over.
№1. The gene for human growth and the gene that determines the number of fingers on the limbs are in the same linkage group at a distance of 8 morganids.
Normal height and five fingers are recessive traits. Tall growth and polydactyly (six-fingered) occur in an autosomal dominant pattern.
The wife is of normal height and has five fingers on each hand. The husband is heterozygous for two pairs of alleles, and he inherited the tall gene from his father, and the six-fingered gene from his mother.
Determine the percentage of probable phenotypes in the offspring.
Solution.
Answer: 46% 46% 4% 4%
№2. Two genes that regulate metabolic reactions in the human body are linked to X-chromosome and are located from each other at a distance of 32 morganids. U-chromosome does not contain allelic genes.
Dominant genes control normal metabolism.
The effects of various mutagenic factors change the sequence of nucleotides in these areas X-chromosomes, which leads to deviations in the synthesis of substances and hereditary diseases of a recessive type.
Healthy parents give birth to a sick child with two mutant genes in the genotype.
What is the percentage chance of having your next child with a metabolic disorder?
Solution. According to the conditions of the problem in this family, the sick child is the son in X a U because A healthy father cannot make daughters sick.
The son received recessive genes from the mother, therefore, the mother’s genotype is heterozygous
We draw up a crossing scheme:
Answer: the probability of having sick children is 33%, of which 17% are sick with two metabolic diseases, 8% with one disease and 8% with another.
4 . 2. EPISTASE
Epistasis (from the Greek epístasis - stop, obstacle), the interaction of two non-allelic (i.e. belonging to different loci) genes, in which one gene, called epistatic or a suppressor gene, suppresses the action of another gene, called hypostatic. Suppressor genes are known in animals (mammals, birds, insects) and plants. They are usually designated I or Su in the case of the dominant state of genes and i or su for their recessive alleles (from the English words inhibitor or suppressor). During epistasis, an allele of one of the genes suppresses the effect of alleles of other genes, for example, A > B or B > A, a > B or b > A, etc.
Epistatic interaction of genes is opposite in nature to complementary. In the case of complementary interaction, one gene is complemented by another. The epistatic effect of genes is very similar in nature to the phenomenon of dominance; the only difference is that with dominance, the allele suppresses the manifestation of a recessive allele belonging to the same allelomorphic pair. During epistasis, the allele of one gene suppresses the manifestation of an allele from another allelomorphic pair, i.e., a non-allelic gene.
Phenotypically, epistasis is expressed in a deviation from the segregation expected with digenic inheritance, however, there is no violation of G. Mendel’s laws in this case, since the distribution of alleles of interacting genes fully complies with the law of independent combination of traits.
Currently, epistasis is divided into two types: dominant and recessive. The most well-known examples of gene interaction by type of epistasis are given in Table. 13.
Table 13
Splitting of characters during epistasis |
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Splitting in F 2 | ||
Inheritance of plumage color in chickens. |
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Inheritance of fruit color in pumpkins (white, yellow and |
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green). |
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Inheritance of oat grain color (black, gray and white). |
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Inheritance of the color of horses (gray, black and red). |
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Inheritance of coat color in Labrador dogs. |
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Inheritance of coat color in mice. |
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Double recessive epistasis (cryptomeria). |
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Dominant epistasis(A > B or B > A). Dominant epistasis is understood as the suppression by a dominant allele of one gene of the action of an allelic pair of another gene.
The epistatic system has been found in chickens. Some chicken breeds have white plumage (white Leghorn, white Plymouth rock, Wyandotte, etc.), while other breeds have colored plumage (Australorp, New Hampshire, striped Plymouth rock, etc.). The white plumage of different chicken breeds is determined by several different genes. For example, the dominant white color is determined by the CCII genes (white leghorns), and the recessive white color is determined by the ccii genes (white Wyandottes). GeneS determines the presence of a pigment precursor (chromogen), i.e., the color of the feather, its allele - the absence of the chromogen and, therefore, the uncolored feather of the bird. Gene I is a suppressor of the action of gene C, the allele i does not suppress its action. In the presence of even one dose of gene I in the bird’s genotype, the effect of color genes will not manifest itself. Therefore, when crossing white leghorns (CCII) with white windottes (ccii), the F1 hybrids turn out to be white (CCIi). When F 1 hybrids are crossed with each other in the second generation, a split in color occurs in the ratio of 13 white: 3 colored (Fig. 38).
F1: | ||
F2: | 9 С -I - : 3 ссI - : | l ccii: 3 C-ii |
painted |
Rice. 38. Inheritance of color in chickens through the interaction of two pairs of genes (epistasis): I – suppresses color, i – does not suppress color,
C – presence of pigment, c – absence of pigment
Thus, suppression of the action of the dominant allele of the gene that determines the development of color (C) by the dominant allele of another gene (I) causes cleavage in F2 according to the 13: 3 phenotype.
Dominant epistasis can also give another ratio of phenotypes in F 2, namely: 12: 3: 1. In this case, the recessive homozygote (aabb) is phenotypically distinguishable from one of the heterozygous classes A-bb or aaB-. This type of splitting of characters has been established for the color of fruits in pumpkins (Fig. 39). This plant has three known fruit colors: white, yellow and green. The dominant allele of gene A determines the yellow color of the pumpkin, the recessive allele determines the green color. The second gene B exhibits an epistatic effect - it suppresses the formation of both yellow and green pigments, making them white. Recessive allele - does not affect the color of pumpkin fruits. When crossing plants with white (AABB) and green (aabb) fruits, all offspring F 1 will be white. BF 2 splitting of features will correspond to the formula 12: 3: 1.
F1: A-B-
Rice. 39. Inheritance of fruit color traits in pumpkins
In the above example, the suppressor gene itself does not determine any qualitative reaction or synthetic process, but only suppresses the action of other genes.
A slightly different mechanism of dominant epistasis is known for grain color in oats. In this case, the suppressor gene performs two functions: it ensures the manifestation of the trait and at the same time promotes
has an epistatic effect on another gene. In this crop, dominant genes were established that determine black (gene A) and gray (gene B) grain color. In addition, gene A exhibits an epistatic effect on gene B. When crossing parental forms of black-seeded (AABB) and white-seeded (aabb) in F 1, all offspring will be black-seeded (AaBb). Since gene A suppresses the expression of gene B, in F 1 all offspring will be black-seeded. BF 2 splitting will be 12: 3: 1:
F 2: | 9 A-B- | : 3 A-bb : | 3 aaB- | : 1 aabb |
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12 black: | |||||
Another example of dominant epistasis is the interaction of genes that determine coat color in horses. Gene B in the dominant state determines black color, and in the recessive state
- redhead. Gene C in the dominant state determines the gray color (causes early graying of horses). In addition, gene C additionally exhibits an epistatic effect on gene B, regardless of whether the latter is in a dominant or recessive state. As a result of the action of the suppressor gene C, the color of horses, regardless of the allelic state of the gene B, becomes gray. Therefore, from crossing gray horses of the BBCC genotype with red horses (bbcc) in F 1, gray offspring (BbCC) are born. When crossing horses of gray color with each other, splitting is observed in F 2
Lection 12:3:1.
F1: | ||||
F2: | 9 B-C- : | 3 bbС- | : 3 V-ss : | 1 aabb |
12 gray: 3 black: 1 red
Recessive epistasis. Recessive epistasis is understood as this type of interaction when the recessive allele of one gene, being in a homozygous state, does not allow it to manifest itself
dominant or recessive alleles of another gene: aa > B;
aa>bb or bb>A; bb> aa.
We have already had the opportunity to become acquainted with the 9:3:4 cleavage as a result of the complementary interaction of genes. But these same cases can also be considered as an example of recessive epistasis.
An example of recessive epistasis is the coat color of Labrador dogs. Coat pigmentation is provided by the B gene, which in the dominant state produces black color, and in the recessive state (b) produces brown color. There is also a gene E, which in a dominant state does not affect the manifestation of color, but being in a recessive state (cc) suppresses the synthesis of both black and brown pigments. Such dogs become white. Cleavage in
F 2 will be the following: | ||||
F2: 9 B-E- | vve- : | 3 V-ee | : 1 vwee |
|
brown | ||||
9 black: 3 brown: 4 white
When crossing black mice (AAcc) with albinos (aaCC), all individuals F 1 (AaCc) have agouti-type coloring (an example of the complementary action of genes), and in F 2 9 parts of all individuals turn out to be agouti (A - C -), 3 parts are black (A -ss) and 4 – albinos (aaS - iaass). These results can be explained by assuming that recessive epistasis of type aa >C - occurs. In this case, mice of genotype aaaC turn out to be white, because the gene in the homozygous state, causing the absence of pigment, thereby prevents the manifestation of the pigment distribution gene C.
An interesting example of recessive epistasis is the Bombay blood phenotype. In rare cases, residents of Bombay (India), whose genotype contains dominant alleles I A or I B, have blood type zero (O). It was found that this defect is caused by a recessive mutation in the H gene, which is not homologous to the A or B loci, which in the homozygous state leads to the formation of a defective H substance during the synthesis of A or B antigens, with which the glycosyltransferase enzyme cannot interact and, therefore, , normal antigens are not formed. The nature of the H-substance has not yet been studied
on the. Antigens A and B are found only in individuals with the HH or Hh genotype. When both parents have genotype I A I B Hh, different blood groups may occur among the offspring in the following ratio:
ni: 3A: 6AB: 3B: 4O.
In addition to the described cases of recessive epistasis, there are also those when the recessive allele of each gene in the homozygous state simultaneously reciprocally suppresses the action of another pair of genes, i.e. aa epistatizes over B -, abb over A -. This interaction of two recessive genes is called double recessive epistasis (cryptomerism). In this case, in a dihybrid cross, the phenotypic split will correspond to 9: 7, as in the case of complementary gene interaction. Consequently, the same relationship can be interpreted both as a complementary interaction and as epistating. In itself, a genetic analysis of inheritance during the interaction of genes without taking into account the biochemistry and physiology of the development of a trait in ontogenesis cannot reveal the nature of this interaction. But without genetic analysis it is impossible to understand the hereditary determination of the development of these characteristics.
4 . 3. POLYMERISM
In the types of gene interactions considered so far, we have dealt with alternative, i.e., qualitatively different, traits. However, such properties of organisms as height, weight, egg production of chickens, the amount of milk and its fat content in livestock, wool length in sheep, the amount of protein in the endosperm of corn and wheat grains, the content of vitamins in plants, the rate of biochemical reactions, the properties of the nervous activity of animals and etc., cannot be divided into clear phenotypic classes. Such characteristics must be assessed in quantitative terms, which is why they are most often called quantitative or dimensional.
The study of the inheritance of polymeric traits began in the first decade of our century. Thus, when crossing wheat plants with red and white (uncolored) grains, the Swedish geneticist G. Nilsson-Ehle in 1908 discovered the usual monohybrid split in F 2 in a ratio of 3: 1. However, when crossing some lines of wheat that differ in the same characteristics, In F 2, a split is observed in the ratio of 15/16 colored and 1/16 white (Fig. 40).
The color of grains from the first group varied from dark red to pale red. Genetic analysis of wheat plants in F 3 from seeds
F 2 that plants grown from white grains and from grains with the darkest (red) color do not further split. From grains with an intermediate type of color, plants developed that in subsequent generations gave rise to splitting according to the color of the grain.
A1 A1 A2 A2 | a1 a1 a2 a2 |
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F3:
A 1A 2♂
A1 a2
a1 A2
a1 a2
A1 a1 A2 a2
♀ A 1 A 2 | A1 a2 | a1 A2 | a1 a2 |
A1 A1 | A1 A1 | A1 a1 | A1 a1 |
A2 A2 | A2 a2 | A2 A2 | A2 a2 |
A1 A1 | A1 A1 | A1 a1 | A1 a1 |
A2 a2 | a2 a2 | A2 a2 | a2 a2 |
A1 a1 | A1 a1 | a1 a1 | a1 a1 |
A2 A2 | A2 a2 | A2 A2 | A2 a2 |
A1 a1 | A1 a1 | a1 a1 | a1 a1 |
A2 a2 | a2 a2 | A2 a2 | a2 a2 |
Rice. 40. Inheritance of grain color in Triticum through the interaction of two pairs of genes (polymerism).
Analysis of the nature of the splitting made it possible to establish that in this case the red color of the grains is determined by two dominant alleles of two different genes, and the combination of their recessive alleles in a homozygous state determines the absence of color. The intensity of grain color depends on the number of dominant genes present in the genotype.
Polymerism is inherent in genes that are represented by independent units, i.e. are non-allelic, but their products perform the same function (Fig. 41).
Gene A 1
Such genes were called polymeric, and since they clearly influence the same trait, it was customary to designate them with one Latin letter indicating the index for different members: A 1, A 2, A 3, etc. Consequently, the original parental forms that gave 15:1 split in F 2 had genotypes A 1 A 1 A 2 A 2 ia 1 a 1 a 2 a 2 .
Obviously, in a trihybrid cross, if the F 1 hybrid has no number of polymer genes in the heterozygous state
two, but three A 1 a 1 A 2 a 2 A 3 a 3 or more, then the number |
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combinations of genotypes in F 2 increases. |
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In the experiment of G. Nilsson-Ehle, trihybrid |
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splitting in F 2 according to grain color genes |
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wheat gave the ratio: 63 plants |
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with red endosperm and 1 plant – with non- |
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painted. In F 2 all transitions were observed |
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dy from the intense coloring of grains with genoty- |
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pom A 1 A 1 A 2 A 2 A 3 A 3 until it is completely absent |
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u a 1 a 1 a 2 a 2 a 3 a 3 . At the same time, the frequencies of genotypes with |
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different quantities | dominant | |||||||||
distributed | next | |||||||||
1+6+15+20+15+6+1=64. | ||||||||||
Rice. 42. Distribution curves | In Fig. 42 shows the distribution curves |
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genotype frequencies in F2 | frequency limits | genotypes with different |
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cumulative polymer at | number of dominant genes cumulatively |
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crossing: | ||||||||||
1 – monohybrid; 2 – digi- | th action with their independent combination |
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bridnom; 3 – trihybrid | in mono-, di- and trihybrid crosses |
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vaniyah. From this comparison it is clear that the greater the number of dominant genes that determine a given trait, the greater the amplitude of variability.
When inheriting a quantitative trait, the offspring of a hybrid forms a continuous variation series according to the phenotypic manifestation of this trait.
The study of polymer genes is not only of theoretical, but also of great practical interest. It has been established that many economically valuable traits in animals and plants, such as milk production of livestock, egg production of chickens, animal weight and height, ear length, corn cob length, sugar content in beet roots, fertility and precocity of animals, length of the growing season in plants and many others are inherited according to the type of polymer.
The variability of a quantitative trait, in contrast to an alternative one, is assessed by the amplitude of its variation. The amplitude of variation of a trait itself is hereditarily determined and has adaptive significance in individual development. As an example of what has been said, let us cite the experiment of E. East on crossing two forms of corn - long-ear and short-ear. As can be seen from the results presented in Fig. 43, cobs along their length in the original corn lines No. 60 (short-cob) and No. 54 (long-cob), as well as in the hybrids of the first and second generations, are distributed with a certain pattern. It is easy to see that these two lines differ greatly from each other, but within each of them the length of the cobs varies slightly. This indicates that they are hereditarily relatively homogeneous. There is no variation in the size of the cobs in the parent forms.
In hybrid plants, the length of the ears appears to be intermediate, with little variability in the row. When F1 plants self-pollinate in the next generation (F2), the range of variability in cob length increases significantly. If we draw a distribution curve of classes according to the length of the cobs, plotting the size of the cobs on the abscissa and their number on the ordinate, it turns out to be similar to the distribution curve of polymeric dominant genes (see Fig. 42). Consequently, a continuous series of changes in the length of a corn cob can be represented as a series of genotypes during trihybrid crossing with a different number of dominant genes that determine a given quantitative trait.
F1:
F2:
Rice. 43. Inheritance and variability of cob length (in centimeters)
in Zea mays in F 1 and F 2
The fact that with a small number of second-generation plants studied, some of them reproduce the ear length characteristic of the parental forms may indicate the participation of a small number of polymer genes in determining the ear length in crossed forms. This assumption follows from the well-known formula 4n, which determines the number of possible combinations of gametes forming zygotes in F 2, depending on the number of gene pairs by which the original parental forms differed. The appearance in F2 of plants similar to the parental forms, with a sample size of 221 plants, indicates that the number of independently inherited genes determining ear length should not exceed three (43 = 64) or four (44 = 256).
The given examples of analysis of the inheritance of quantitative traits illustrate only one of the possible ways to study complex and fluctuating traits. Greater variability of a trait, first of all, indicates its complex genetic conditioning and, on the contrary, less variability of a trait indicates a smaller number of factors determining it.
As is known, dominance is the suppression of the action of one allele by another allele, representing one gene: A > a, B > b, C > c, etc.
But there is an interaction in which the allele of one of the genes suppresses the effect of alleles of other genes, for example A > B or B A, a > B or b > A, etc. This phenomenon of “dominance” between genes is called epistasis.
Epistatic interaction of genes is opposite in nature to complementary interaction.
Genes that suppress the action of other genes are called suppressors or inhibitors. They can be either dominant or recessive. Suppressor genes are known in animals (mammals, birds, insects) and plants. They are usually designated I or Su in the case of a dominant gene state and i or su for their recessive alleles (from the English words inhibitor or suppressor).
Currently, epistasis is divided into two types: dominant and recessive.
Under dominant epistasis understand the suppression by a dominant allele of one gene of the action of an allelic pair of another gene. Of the many examples of dominant epistasis established both in animals and plants, we will cite only a few.
Some breeds of chickens have white plumage (white Leghorn, white Primutrock, etc.), while other breeds have colored plumage (Australorp, New Hampshire, striped Plymouthrock, etc.). The white plumage of different chicken breeds is determined by several different genes. For example, the dominant white color is determined by the CCII genes (white Leghorns), and the recessive white color is determined by the ccii genes (white Sussex, white Minor, white Plymouth Rock). Gene C determines the presence of a precursor to pigment (chromogen), i.e., the color of the feather, its allele c - the absence of a chromogen and, therefore, the uncolored feather of the bird. Gene I is a suppressor of the action of gene C, allele i does not suppress its action. In the presence of even one dose of gene I in the bird's genotype, the effect of color genes will not manifest itself. Therefore, when white CCII Leghorns are crossed with colored CCii breeds, the white CCIi color will generally dominate. When white ccii Plymouth rocks are crossed with colored CCii breeds, the hybrids end up being colored Ccii. Therefore, in Leghorns white color is dominant, and in Plymouth Rocks it is recessive.
If white Leghorns CCII are crossed with recessive white Plymouth Rocks ccii, then in the first generation the chickens also turn out to be white CcIi. When F 1 hybrids are crossed with each other in the second generation, a split in color occurs in the ratio of 13/16 white: 3/16 colored.
How can the resulting relationship be explained? First of all, this ratio indicates splitting into two genes; this splitting can be represented as 9(C-I-) + 3(ccI-) + 1(ccii) = 13 and 3(C-ii), which corresponds to formula 9: 3:3:1.
Obviously, in this case, the color of Leghorns is not due to the presence of special white color genes, but to the action of the color suppressor gene (I-). Then the genotype of homozygous white leghorns should be CCII, where I is the color suppressor gene, and C is the color gene. White Plymouth Rocks genotype must be homozygous for two recessive factors ccii, where c is the absence of color and i is the absence of color suppression. Due to epistatation I > C, hybrid chickens of the first generation CcIi should be white. In F 2, all chickens with genotypes 9/16 C-I-, 3/1 6 ccI- and 1/1 6 ccii should also be white, and only chickens of one phenotypic class 3/1 6 (C-ii) are colored, because it contains the color gene and does not have its suppressor.
Thus, suppression of the action of the dominant allele of the gene that determines color development by the dominant allele of another gene (suppressor) causes phenotypic cleavage in F2 in a ratio of 13: 3.
Dominant epistasis can also give another ratio when splitting in F 2 according to phenotype, namely 12: 3: 1. In this case, a form homozygous for both recessive factors aabb will be phenotypically distinguishable from forms with dominant alleles of two genes A-B- and forms with one of them: aaB- and A-bb. This decoupling has been established for the inheritance of fruit color in pumpkins, peels in onions, and other characteristics. In this case, the dominant inhibitor also takes part in the cleavage.
We analyzed the interaction of only two genes. In fact, many genes interact via epistasis. Suppressor genes usually do not themselves determine any qualitative reaction or synthetic process, but only suppress the action of other genes. However, when we say that a suppressor gene does not have its own qualitative effect on a trait, this applies only to this trait. In fact, the inhibitor, by suppressing, for example, pigment formation, can have a pleiotropic effect on other properties and characteristics.
Under recessive epistasis understand this type of interaction when the recessive allele of one gene, being in a homozygous state, does not allow dominant or recessive alleles of other genes to manifest themselves: aa > B or aa > bb.
We have already had the opportunity to become acquainted with the splitting 9:3:4 as a result. But these same cases can also be considered as an example of recessive epistasis.
When crossing black mice (AAbb) with white mice (aaBB), all individuals F 1 (AaBb) have agouti-type coloring, and in F 2 9/16 of all individuals are agouti (A-B-), 3/16 black (A-bb ) and 4/16 white (aaB- and aabb). These results can be explained by assuming that recessive epistasis of type aaB- occurs. In this case, mice of the aaB- genotype turn out to be white because gene a in the homozygous state, causing the absence of pigment, thereby prevents the manifestation of the gene that distributes pigment B.
In addition to the described cases of single recessive epistasis, there are also those when the recessive allele of each gene in a homozygous state simultaneously reciprocally suppresses the action of the dominant alleles of each gene, i.e. aa epistatizes over B-, and bb over A-. This interaction of two suppressive recessive genes is called double recessive epistasis. In this case, in a dihybrid cross, the phenotypic split will correspond to 9: 7, as in the case of complementary gene interaction.
Consequently, the same relationship can be interpreted both as a complementary interaction and as epistating. In itself, a genetic analysis of inheritance during the interaction of genes without taking into account the biochemistry and physiology of the development of a trait in ontogenesis cannot reveal the nature of this interaction. But without genetic analysis it is impossible to understand the hereditary determination of the development of these characteristics.
Now let us turn to the problem of interaction of non-allelic genes. If the development of a trait is controlled by more than one pair of genes, then this means that it is under polygenic control. Several main types of gene interaction have been established: complementarity, epistasis, polymery and pleiotropy.
The first case of non-allelic interaction was described as an example of deviation from Mendel's laws by English scientists W. Betson and R. Punnett in 1904 when studying the inheritance of comb shape in chickens. Different chicken breeds have different comb shapes. Wyandottes have a low, regular, papillary crest known as a “rose crest.” Brahmas and some fighting chickens have a narrow and high crest with three longitudinal elevations - “pea-shaped”. Leghorns have a simple or leaf-shaped crest consisting of a single vertical plate. Hybridological analysis showed that the simple crest behaves as a completely recessive trait in relation to the rose and pisiform. The splitting in F 2 corresponds to formula 3: 1. When crossing races with a rose-shaped and pea-shaped crest, the first generation hybrids develop a completely new form of crest, reminiscent of half a walnut kernel, and therefore the crest was called “nut-shaped.” When analyzing the second generation, it was found that the ratio of different comb forms in F 2 corresponds to the formula 9: 3: 3: 1, which indicated the dihybrid nature of the crossing. A crossing scheme was developed to explain the mechanism of inheritance of this trait.
Two non-allelic genes take part in determining the shape of the comb in chickens. The dominant R gene controls the development of the rose crest, and the dominant P gene controls the development of the pisiform crest. The combination of recessive alleles of these rrpp genes causes the development of a simple comb. The nut-shaped comb develops when both dominant genes are present in the genotype.
The inheritance of comb shape in chickens can be attributed to the complementary interaction of non-allelic genes. Complementary, or additional, genes are those that, when acted together in a genotype in a homo- or heterozygous state, determine the development of a new trait. The action of each gene individually reproduces the trait of one of the parents.
Diagram illustrating the interaction of non-allelic genes,
determining the shape of the comb in chickens
The inheritance of genes that determine the shape of the comb in chickens fully fits into the dihybrid crossing scheme, since they behave independently during distribution. The difference from a conventional dihybrid cross appears only at the phenotypic level and boils down to the following:
- F 1 hybrids are not similar to either parent and have a new trait;
- In F 2, two new phenotypic classes appear that result from the interaction of either dominant (nut comb) or recessive (simple comb) alleles of two independent genes.
Mechanism complementary interaction studied in detail using the example of the inheritance of eye color in Drosophila. The red color of the eyes in wild-type flies is determined by the simultaneous synthesis of two pigments - brown and bright red, each of which is controlled by a dominant gene. Mutations affecting the structure of these genes block the synthesis of either one or the other pigment. So, a recessive mutation brown(the gene is located on chromosome 2) blocks the synthesis of bright red pigment, and therefore homozygotes for this mutation have brown eyes. Recessive mutation scarlet(the gene is located on the 3rd chromosome) disrupts the synthesis of brown pigment, and therefore homozygotes stst have bright red eyes. When both mutant genes are simultaneously present in the genotype in a homozygous state, both pigments are not produced and the flies have white eyes.
In the described examples of complementary interaction of non-allelic genes, the phenotypic splitting formula in F 2 corresponds to 9: 3: 3: 1. Such splitting is observed if the interacting genes individually have different phenotypic manifestations and it does not coincide with the phenotype of a homozygous recessive. If this condition is not met, other phenotype relationships take place in F2.
For example, when crossing two varieties of shaped pumpkin with a spherical fruit shape, the first generation hybrids have a new characteristic - flat or disc-shaped fruits. When crossing hybrids with each other in F 2, a splitting is observed in the ratio of 9 disc-shaped: 6 spherical: 1 elongated.
Analysis of the diagram shows that two non-allelic genes with the same phenotypic manifestation (spherical shape) take part in determining the shape of the fruit. The interaction of dominant alleles of these genes gives a disc-shaped shape, the interaction of recessive alleles gives an elongated shape.
Another example of complementary interaction is provided by the inheritance of coat color in mice. Wild gray coloration is determined by the interaction of two dominant genes. Gene A is responsible for the presence of pigment, and the gene IN- for its uneven distribution. If the genotype contains only the gene A (A-bb), then the mice are uniformly colored black. If only the gene is present IN (aaB-), then the pigment is not produced and the mice turn out to be uncolored, just like a homozygous recessive aabb. This action of genes leads to the fact that in F2 the phenotypic splitting corresponds to the formula 9: 3: 4.
F 2
| AB | Ab | aB | ab | |
| AB | AABB ser. |
AABb ser. |
AaBB ser. |
AaBb ser. |
| Ab | AABb ser. |
AAbb black |
AaBb ser. |
Aabb black |
| aB | AaBB ser. |
AaBb ser. |
aaBB white |
aaBb white |
| ab | AaBb ser. |
Aabb black |
aaBb white |
aabb |
Complementary interactions have also been described in the inheritance of flower color in sweet peas. Most of the varieties of this plant have purple flowers with violet wings, which are characteristic of the wild Sicilian race, but there are also varieties with a white color. By crossing plants with purple flowers with plants with white flowers, Betson and Punnett found that the purple color of flowers completely dominates over white, and in F 2 a ratio of 3: 1 is observed. But in one case, the crossing of two white plants produced offspring consisting only of plants with colored flowers. Self-pollination of F 1 plants produced offspring consisting of two phenotypic classes: with colored and uncolored flowers in a ratio of 9/16: 7/16.
The results obtained are explained by the complementary interaction of two pairs of non-allelic genes, the dominant alleles of which ( WITH And R) individually are not capable of ensuring the development of purple coloration, as well as their recessive alleles ( ssrr). Coloring appears only if both dominant genes are present in the genotype, the interaction of which ensures the synthesis of pigment.
purple
F 2
| C.P. | Cp | cP | cp | |
| C.P. | CCPP purple |
CCPp purple |
CCPP purple |
CCPP purple |
| Cp | CCPp purple |
CCpp white |
CCPP purple |
Ccpp white |
| cP | CCPP purple |
CCPP purple |
ccPP white |
ccPp white |
| cp | CCPP purple |
Ccpp white |
ccPp white |
In the example given, the splitting formula in F 2 is 9: 7 due to the absence of the dominant alleles of both genes having their own phenotypic manifestation. However, the same result is obtained if the interacting dominant genes have the same phenotypic manifestation. For example, when crossing two varieties of corn with purple grains in F 1, all hybrids have yellow grains, and in F 2 a split of 9/16 yellow is observed. : 7/16 viol.
Epistasis- another type of non-allelic interaction, in which the action of one gene is suppressed by another non-allelic gene. A gene that prevents the expression of another gene is called epistatic, or suppressor, and one whose action is suppressed is called hypostatic. Both a dominant and a recessive gene can act as an epistatic gene (dominant and recessive epistasis, respectively).
An example of dominant epistasis is the inheritance of coat color in horses and fruit color in pumpkins. The pattern of inheritance of these two traits is absolutely the same.
F 2
| C.B. | Cb | cB | cb | |
| C.B. | CCBB ser. |
CCBB ser. |
CCBB ser. |
CcBb ser. |
| Cb | CCBb ser. |
CCbb ser. |
CcBb ser. |
CCBB ser. |
| cB | CCBB ser. |
CcBb ser. |
ccBB black |
ccBb black |
| cb | CcBb ser. |
CCBB ser. |
ccBb black |
ccbb red |
The diagram shows that the dominant gene for gray color WITH is epistatic to the dominant gene IN, which causes the black color. In the presence of a gene WITH gene IN does not exhibit its effect, and therefore F 1 hybrids carry a trait determined by the epistatic gene. In F 2, the class with both dominant genes merges in phenotype (gray color) with the class in which only the epistatic gene is represented (12/16). Black coloration appears in 3/16 hybrid offspring whose genotype lacks the epistatic gene. In the case of a homozygous recessive, the absence of a suppressor gene allows the recessive gene c to appear, which causes the development of a red color.
Dominant epistasis has also been described in the inheritance of feather color in chickens. The white color of the plumage in Leghorn chickens dominates over the colored ones of black, speckled and other colored breeds. However, the white coloration of other breeds (for example, Plymouth Rocks) is recessive to the colored plumage. Crosses between individuals with dominant white coloration and individuals with recessive white coloration in F 1 produce white offspring. In F2, a splitting ratio of 13:3 is observed.
Analysis of the diagram shows that two pairs of non-allelic genes take part in determining feather color in chickens. Dominant gene of one pair ( I) is epistatic in relation to the dominant gene of the other pair, causing the development of color ( C). In this regard, only those individuals whose genotype contains the gene have colored plumage WITH, but lacks an epistatic gene I. In recessive homozygotes ccii there is no epistatic gene, but they do not have the gene that ensures the production of pigment ( C), which is why they are white.
As an example recessive epistasis we can consider the situation with the albinism gene in animals (see above for the scheme of inheritance of coat color in mice). The presence in the genotype of two alleles of the albinism gene ( ahh) does not allow the dominant color gene to appear ( B) - genotypes aaB-.
Polymer type of interaction was first established by G. Nielsen-Ehle while studying the inheritance of grain color in wheat. When crossing a red-grain wheat variety with a white-grain one in the first generation, the hybrids were colored, but the color was pink. In the second generation, only 1/16 of the offspring had red grain color and 1/16 had white grain; the rest had an intermediate color with varying degrees of severity of the trait (from pale pink to dark pink). Analysis of segregation in F2 showed that two pairs of non-allelic genes are involved in determining the color of grain, the effect of which is summed up. The degree of severity of the red color depends on the number of dominant genes in the genotype.
Polymer genes are usually designated by the same letters with the addition of indices, in accordance with the number of non-allelic genes.
The effect of dominant genes in a given cross is additive, since the addition of any of them enhances the development of the trait.
F 2
| A 1 A 2 | A 1 a 2 | a 1 A 2 | a 1 a 2 | |
| A 1 A 2 | A 1 A 1 A 2 A 2 red |
A 1 A 1 A 2 Aa 2 bright pink |
A 1 a 1 A 2 A 2 bright pink |
A 1 a 1 A 2 a 2 pink |
| A 1 a 2 | A 1 A 1 A 2 a 2 bright pink |
A 1 A 1 a 2 a 2 pink |
A 1 a 1 A 2 a 2 pink |
A 1 a 1 a 2 a 2 pale pink. |
| a 1 A 2 | A 1 a 1 A 2 A 2 bright pink |
A 1 a 1 A 2 a 2 pink |
a 1 a 1 A 2 A 2 pink |
a 1 a 1 A 2 a 2 pale pink. |
| a 1 a 2 | A 1 a 1 A 2 a 2 pink |
A 1 a 1 a 2 a 2 pale pink. |
a 1 a 1 A 2 a 2 pale pink. |
a 1 a 1 a 2 a 2 |
The type of polymerization described, in which the degree of development of a trait depends on the dose of the dominant gene, is called cumulative. This type of inheritance is common for quantitative traits, which include coloration, because its intensity is determined by the amount of pigment produced. If we do not take into account the degree of expression of color, then the ratio of painted and uncolored plants in F2 corresponds to the formula 15: 1.
However, in some cases the polymer is not accompanied by a cumulative effect. An example is the inheritance of seed shape in the shepherd's purse. Crossing two races, one of which has triangular fruits, and the other ovoid, produces in the first generation hybrids with a triangular fruit shape, and in the second generation, splitting is observed according to these two characteristics in a ratio of 15 triangles. : 1 eggs.
This case of inheritance differs from the previous one only at the phenotypic level: the absence of a cumulative effect with an increase in the dose of dominant genes determines the same expression of the trait (triangular shape of the fruit) regardless of their number in the genotype.
The interaction of non-allelic genes also includes the phenomenon pleiotropy— multiple actions of a gene, its influence on the development of several traits. The pleiotropic effect of genes is the result of a serious metabolic disorder caused by the mutant structure of a given gene.
For example, Irish Dexter cows differ from the Kerry breed, which is similar in origin, by having shorter legs and heads, but at the same time by better meat qualities and fattening ability. When crossing cows and bulls of the Dexter breed, 25% of the calves have characteristics of the Kerry breed, 50% are similar to the Dexter breed, and in the remaining 25% of cases, miscarriages of ugly bulldog-shaped calves are observed. Genetic analysis made it possible to establish that the cause of death of part of the offspring is the transition to a homozygous state of a dominant mutation that causes underdevelopment of the pituitary gland. In a heterozygote, this gene leads to the appearance of dominant traits of short legs, short heads and an increased ability to store fat. In a homozygote, this gene has a lethal effect, i.e. in relation to the death of offspring, it behaves like a recessive gene.
The lethal effect upon transition to a homozygous state is characteristic of many pleiotropic mutations. Thus, in foxes, dominant genes that control platinum and white-faced fur colors, which do not have a lethal effect in heterozygotes, cause the death of homozygous embryos at an early stage of development. A similar situation occurs when inheriting gray coat color in Shirazi sheep and underdevelopment of scales in mirror carp. The lethal effect of mutations leads to the fact that animals of these breeds can only be heterozygous and, during intrabreed crossings, produce a split in the ratio of 2 mutants: 1 normal.
F 1
F 1: 2 boards. : 1 black
However, most lethal genes are recessive, and individuals heterozygous for them have a normal phenotype. The presence of such genes in parents can be judged by the appearance in the offspring of homozygous freaks, abortions and stillborns. Most often, this is observed in closely related crosses, where the parents have similar genotypes, and the chances of harmful mutations passing into a homozygous state are quite high.
Drosophila has pleiotropic genes with a lethal effect. So, dominant genes Curly- upturned wings, Star- starry eyes, Notch- the jagged edge of the wing and a number of others in a homozygous state cause the death of flies in the early stages of development.
Known recessive mutation white, first discovered and studied by T. Morgan, also has a pleiotropic effect. In the homozygous state, this gene blocks the synthesis of eye pigments (white eyes), reduces the viability and fertility of flies and modifies the shape of the testes in males.
In humans, an example of pleiotropy is Marfan disease (spider finger syndrome, or arachnodactyly), which is caused by a dominant gene that causes increased finger growth. At the same time, it detects abnormalities of the lens of the eye and heart defects. The disease occurs against the background of increased intelligence, which is why it is called the disease of great people. A. Lincoln and N. Paganini suffered from it.
The pleiotropic effect of a gene appears to underlie correlative variation, in which a change in one trait entails a change in others.
The interaction of non-allelic genes should also include the influence of modifier genes that weaken or enhance the function of the main structural gene that controls the development of a trait. In Drosophila, modifier genes are known that modify the process of wing venation. At least three modifier genes are known that affect the amount of red pigment in the hair of cattle, as a result of which the coat color of different breeds ranges from cherry to fawn. In humans, modifier genes change eye color, increasing or decreasing its intensity. Their action explains the different eye colors in one person.
The existence of the phenomenon of gene interaction led to the emergence of such concepts as “genotypic environment” and “gene balance”. The genotypic environment means the environment into which the newly emerging mutation falls, i.e. the entire complex of genes present in a given genotype. The concept of “gene balance” refers to the relationship and interaction between genes that influence the development of a trait. Genes are usually designated by the name of the trait that arises during mutation. In fact, the manifestation of this trait is often the result of a dysfunction of the gene under the influence of other genes (suppressors, modifiers, etc.). The more complex the genetic control of a trait, the more genes are involved in its development, the higher the hereditary variability, since a mutation of any gene disrupts the gene balance and leads to a change in the trait. Consequently, for the normal development of an individual, not only the presence of genes in the genotype is necessary, but also the implementation of the entire complex of inter-allelic and non-allelic interactions.
Problem book on genetics 9th grade Monohybrid cross
PROBLEM ON GENETICS 9TH GRADE.
Monohybrid crossing.
Task No. 1
Task No. 2
1.What is the dominant coat color in rabbits?
2. What are the genotypes of parents and first-generation hybrids based on coat color?
3. What genetic patterns appear during such hybridization?
Problem No. 3
1. Which tomato fruit shape (spherical or pear-shaped) is dominant?
2. What are the genotypes of the parents and 1st and 2nd generation hybrids?
3. What genetic patterns discovered by Mendel appear during such hybridization?
Task No. 4 When crossing an ermine rooster and a hen, 35 chickens were obtained. Of these, 17 are ermine, 10 are black, 8 are white. How is ermine coloring inherited? What kind of parents should you take to get only ermine chickens?
Problem No. 5 At the school site, tomatoes that had a rounded fruit shape were crossed with ribbed tomatoes. F1 had round fruits. What fruits will the hybrids have from crossing the resulting offspring with a parent variety that has round fruits?
Note: varieties are homozygous. Write the resulting hybrids in genetic and chromosomal expression.
Problem No. 6 A tall plant was pollinated with a homozygous organism having normal stem growth. The offspring produced 20 plants of normal growth and 10 plants of tall growth. What split does this cross correspond to - 3:1 or 1:1?
Problem No. 7
Problem No. 8 Sasha and Pasha have gray eyes, and their sister Masha has green eyes. The mother of these children is grey-eyed, although both her parents had green eyes. The gene responsible for eye color is located on a non-sex chromosome (autosome). Determine the genotypes of parents and children. Draw up a crossing diagram.
Problem No. 9 Mother is brunette; the father is blond, there were no brunettes in his pedigree. Three children were born: two blonde daughters and a brunette son. The gene for this trait is located on the autosome. Analyze the genotypes of offspring and parents.
Problem No. 10 A person develops a disease called sickle cell anemia. This disease is expressed in the fact that red blood cells are not round in shape, but sickle-shaped, as a result of which less oxygen is transported. Sickle cell anemia is inherited as an incompletely dominant trait, and the homozygous state of the gene leads to the death of the body in childhood. In the family, both spouses have signs of anemia. What is the percentage chance of them having a healthy child?
Problem No. 11 A blue-eyed man, both of whose parents had brown eyes, married a brown-eyed woman whose father had brown eyes and whose mother had blue eyes. From this marriage one blue-eyed son was born. Determine the genotypes of each of the individuals mentioned and draw a diagram of their pedigree.
Problem No. 12 In humans, the gene for polydactyly (multi-fingered) dominates the normal structure of the hand. The wife has a normal hand, the husband is heterozygous for the polydactyly gene. Determine the probability of having a multi-fingered child in this family.
Problem No. 13 Red-fruited strawberry plants, when crossed with each other, always produce offspring with red berries, and white-fruited strawberry plants - with white berries. As a result of crossing both varieties with each other, pink berries are obtained. What offspring arises when hybrid strawberry plants with pink berries are crossed with each other? What kind of offspring will you get if you pollinate red strawberries with pollen from hybrid strawberries with pink berries?
Problem No. 14 In tomatoes, the skin of the fruit may be smooth or drooping. One variety has smooth fruits, the other drooping. When crossed, F1 has smooth fruits, F2 has 174 plants with drooping fruits and 520 with smooth fruits. How is prolapse inherited? What will happen in both Fv?
Problem No. 15 One child in the family was born healthy, and the second had a severe hereditary disease and died immediately after birth. What is the probability that the next child in this family will be healthy? One pair of autosomal genes is considered.
Problem No. 16 A woman with red hair, whose mother and father have red hair, and whose brother has red hair, married a red-haired man, whose mother also has red hair, and whose father also has red hair. From this marriage a boy with red hair and a girl with red hair were born. Determine the genotypes of all mentioned individuals and draw up a pedigree chart for this family.
Problem No. 17 When crossing a gray rabbit, both of whose parents were gray, with a gray rabbit, whose parents were also gray, several black rabbits (grandchildren) were born. Determine the genotypes of each of the individuals mentioned and draw up a pedigree chart.
Problem No. 18 What kind of offspring F 1
And F 2
What will be the phenotype and genotype obtained by crossing red-fruited strawberries with white-fruited ones? (None of these fruit traits in strawberries gives complete dominance, which is why hybrids have pink fruit.
Problem No. 19 When black rabbits were crossed with each other, the offspring were black and white rabbits. Draw up a crossing scheme if it is known that one pair of autosomal genes is responsible for coat color.
Problem No. 20 In what ratio will the offspring split according to the phenotype obtained from crossing a heterozygous gray Karakul ram with the same sheep, if homozygotes for dominant genes are lethal? Why?
Problem No. 21 Crossed a motley rooster and a hen. We received 26 motley, 12 black and 13 white chickens. How is plumage color inherited in chickens?
Problem No. 22 In medicine, the distinction between the four human blood groups is of great importance. Blood type is a hereditary trait that depends on one gene. This gene has not two, but three alleles, designated by the symbols A, IN, 0 . Persons with genotype 00 have the first blood group, with genotypes AA or A0- second, BB or B0- third, AB– fourth (we can say that alleles A And IN dominate the allele 0 , whereas they do not suppress each other). What blood types are possible in children if their mother has the second blood group, and their father has the first?
Problem No. 23 Can a child be born with the first blood group and Rh(-) factor if: the mother has the second blood group Rh(+) factor, and the father has the third blood group Rh(+) factor?
1) What happens if the Rh(-) factor of the fetus enters the mother’s blood through the placenta? 2) Why can a second child from these parents be stillborn?
Problem No. 24 The boy has blood group 1, and his sister has blood group 4. What can be said about the blood types of their parents?
Problem No. 25 The woman has blood group 2, the child has blood group 1. Blood. Can a person with blood type 4 be the father of a child?
Problem No. 26 The woman has blood group 4, the child has blood group 2. Blood. The woman claims that the child’s father is D.I (a person with blood type 4). What will they answer in court?
Problem No. 27 Is it possible to transfuse the blood of parents to their children if they have blood types 2 and 4?
Problem No. 28 Can children only have blood type 2 if both parents have the same blood type?
Dihybrid crossing independent inheritance of genes
№1. Gene mutations that cause shortened limbs (A) and longhair (V) in sheep, are transmitted to the next generation in a recessive manner. Their dominant alleles form normal limbs (A) and short hair (IN). The genes are not linked.
The farm bred rams and sheep with dominant traits and produced 2,336 lambs. Of these, 425 are long-haired with normal limbs and 143 are long-haired with short limbs.
Determine the number of short-haired lambs and how many of them have normal limbs?
Solution. We determine the genotypes of parents based on recessive offspring. According to the rule of “purity of gametes,” in the offspring for each trait one gene is from the paternal organism, the other gene is from the maternal organism, therefore, the genotypes of the parents are diheterozygous.
1). Find the number of long-haired lambs: 425 + 143 = 568.
2). We find the number of shorthairs: 2336 – 568 = 1768.
3). We determine the number of shorthairs with normal limbs:
1768 ---------- 12 o'clock
x ----------- 9 hours x = 1326.
№2. A person has a gene for black skin color (IN) completely dominates the European skin gene (V), and the disease sickle cell anemia is manifested by an incompletely dominant gene (A), and allelic genes are in a homozygous state (AA) lead to the destruction of red blood cells, and the organism becomes unviable.
The genes for both traits are located on different chromosomes.
A pure Negroid woman gave birth to two mulattoes from a white man. One child had no signs of anemia, and the second died of anemia.
What is the probability of having the next child without signs of anemia?
Solution. We draw up a crossing scheme:
Answer: The probability of having a healthy child in this family is 1/4 = 25%
№3. Recessive genes (A) And (With) determine the manifestation of human diseases such as deafness and albinism. Their dominant alleles control the inheritance of normal hearing (A) and synthesis of melanin pigment (WITH).
The genes are not linked.
Parents have normal hearing; brunette mother, albino father. Three identical twins were born with two characteristics.
What is the probability that the next child in this family will have both diseases?
Solution.
According to the rule of “purity of gametes,” it was determined that the parents are diheterozygous:
Answer: The probability of having a child with both diseases is 1/8 = 12.5%
№4. Two pairs of autosomal genes exhibiting independent inheritance are being studied.
A rooster with a rose comb and feathered legs is bred with two hens with a rose comb and feathered legs.
From the first hen, chickens with feathered legs were obtained, some of them had a rose-shaped comb, and the other part had a simple comb.
The chickens from the second hen had a pink comb, and some of them had feathered legs and some had unfeathered legs.
Determine the genotypes of a rooster and two hens.
Solution.
According to the conditions of the problem, both parents have the same phenotypes, and in the offspring from two crosses, splitting occurred for each trait. According to G. Mendel's law, only heterozygous organisms can produce “splitting” in their offspring. We draw up two crossing schemes.
Interaction of nonallelic genes
№1. Two pairs of non-allelic unlinked genes that determine fur color in ermine are being studied.
Dominant gene of one pair (A) determines the color black, and its recessive allele (A)- blue color.
Dominant gene of the other pair (IN) contributes to the manifestation of pigmentation in the body, its recessive allele (V) does not synthesize pigment.
When black individuals were crossed with each other, the offspring turned out to be individuals with blue fur, black and albinos.
Analyze the genotypes of the parents and the theoretical ratio in the offspring.
Solution.
Answer: 9 black, 3 albino, 4 blue.
№2. The inheritance of plumage color in chickens is determined by two pairs of non-allelic unlinked genes located on the autosome.
Dominant gene of one pair (A) determines the synthesis of the melanin pigment, which ensures the presence of color. Recessive gene (A) does not lead to the synthesis of pigment and the chickens turn out to be white (feather albinism).
Dominant gene of the other pair (IN) suppresses the action of the genes of the first pair, as a result of which pigment synthesis does not occur, and chickens also become albinos. Its recessive allele (V) does not have a downward effect.
Two organisms heterozygous for two pairs of alleles are crossed.
Determine the ratio of chickens with colored plumage and albinos in the offspring.
Solution.
Answer: 13 white, 3 painted.
№3.
In oats, grain color is determined by two pairs of non-allelic unlinked genes.
One dominant gene (A) determines black color, another dominant gene (IN)- grey colour. The black gene suppresses the gray gene.
Both recessive alleles determine the white color of the grains.
When pollination of diheterozygous organisms occurred, the offspring produced plants with black, gray and white grains.
Determine the genotypes of parental organisms and the phenotypic ratio in the offspring.
Solution.
Answer: 12 black, 3 gray, 1 white.
Inheritance of genes located on sex chromosomes
№1. Normal blood clotting gene (A) in humans it is inherited in a dominant manner and is linked to X-chromosome. Recessive mutation of this gene (A) leads to hemophilia - inability to clot blood.
U-chromosome does not have an allelic gene.
Determine the percentage probability of having healthy children in a young family if the bride has normal blood clotting, although her sister has signs of hemophilia. The groom's mother suffers from this disease, and his father is healthy.
Solution. 1) Determine the genotype of the bride. According to the conditions of the problem, the bride's sister has a recessive genotype X A X A , which means both sisters receive the hemophilia gene (from their father). Therefore, a healthy bride is heterozygous.
2) Determine the genotype of the groom. Mother of the groom with signs of hemophilia X A X A , therefore, according to the chromosomal theory of sex, she passes the recessive gene to her son X A U .
Answer: the phenotypic ratio is 1:1, 50% of children are healthy.
№2. One pair of allelic genes is studied in X-chromosome that regulates color vision in humans.
Normal color vision is a dominant trait, while color blindness is a recessive trait.
Analyze the genotype of the maternal organism.
It is known that the mother has two sons, one of them has a sick wife and a healthy child. The second family has a daughter with signs of color blindness and a son whose color vision is normal.
Solution. 1) Determine the genotype of the first son. According to the conditions of the problem, he has a sick wife and a healthy child - it can only be a daughter X A X A . The daughter received the recessive gene from her mother, and the dominant gene from her father, therefore, the genotype of the male body is dominant (X A U) .
2) Determine the genotype of the second son. His daughter is sick X A X A , which means that she received one of the recessive alleles from her father, therefore the genotype of the male body is recessive (X A U - ) .
3) We determine the genotype of the mother’s organism based on her sons:
Answer: mother's genotype is heterozygous X A X A .
№3. Albinism in humans is determined by a recessive gene (A) located on the autosome, and one form of diabetes is determined by a recessive gene (V) , linked to the genitals X -chromosome.
Dominant genes are responsible for pigmentation (A) and normal metabolism (IN) .
U -chromosome does not contain genes.
The couple have dark hair color. Both mothers suffered from diabetes, and fathers were healthy.
One child was born with two symptoms.
Determine the percentage probability of having healthy and sick children in this family.
Solution. Applying the rule of “gamete purity” we determine the genotypes of the parents by hair color - heterozygous genotypes Ahh.
According to the chromosomal theory of sex, it was determined that the father has diabetes X V U - and the mother is healthy X IN X V .
We make a Punnett grid - the gametes of the father's body are written out horizontally, and the gametes of the mother's body are written down vertically.
Answer: six organisms out of sixteen are dominant in two traits - the probability of birth is 6/16 = 37.5%. Ten patients: 10/16 = 62.5%, of which two patients had two characteristics: 2/16 = 12.5%.
№4. Two recessive genes located in different areas X-chromosomes cause diseases such as hemophilia and muscular dystrophy in humans. Their dominant alleles control normal blood clotting and muscle tone.
U
The bride’s mother suffers from dystrophy, but according to her pedigree she has normal blood clotting, and her father had hemophilia, but without any dystrophic signs.
The groom exhibits both diseases.
Analyze the offspring in this family.
Solution.
Answer: all children have the disease, 50% with hemophilia and 50% with dystrophy.
Inheritance of linked genes. The phenomenon of crossing over.
№1. The gene for human growth and the gene that determines the number of fingers on the limbs are in the same linkage group at a distance of 8 morganids.
Normal height and five fingers are recessive traits. Tall growth and polydactyly (six-fingered) occur in an autosomal dominant pattern.
The wife is of normal height and has five fingers on each hand. The husband is heterozygous for two pairs of alleles, and he inherited the tall gene from his father, and the six-fingered gene from his mother.
Determine the percentage of probable phenotypes in the offspring.
Solution.
Answer: 46% 46% 4% 4%
№2. Two genes that regulate metabolic reactions in the human body are linked to X-chromosome and are located from each other at a distance of 32 morganids. U-chromosome does not contain allelic genes.
Dominant genes control normal metabolism.
The effects of various mutagenic factors change the sequence of nucleotides in these areas X-chromosomes, which leads to deviations in the synthesis of substances and hereditary diseases of a recessive type.
Healthy parents give birth to a sick child with two mutant genes in the genotype.
What is the percentage chance of having your next child with a metabolic disorder?
Solution. According to the conditions of the problem in this family, the sick child is the son V X A U because A healthy father cannot make daughters sick.
The son received recessive genes from the mother, therefore, the mother’s genotype is heterozygous
We draw up a crossing scheme:
Answer: the probability of having sick children is 33%, of which 17% are sick with two metabolic diseases, 8% with one disease and 8% with another.