The gene pool of a population can be described either by gene frequencies or by genotype frequencies. Let's imagine that there are N diploid individuals in a population that differ in one pair of alleles (A and a); D - means the number of homozygotes for the dominant allele (AA); P is the number of homozygotes for the recessive allele (aa); N - number of heterozygotes (Aa). Thus, there will be three types of individuals in the population, having genotypes AA, Aa, aa, respectively. Since each individual with the AA genotype has two A alleles, and each Aa individual has one A allele, the total number of A alleles will be 2D + H. Then p - the frequency of occurrence of the dominant allele A is equal to:
The frequency of the recessive allele (a) is usually denoted by q. The sum of the frequencies of genes A and a is equal to one, p+q=1, hence q=1-p. If a gene has only two alleles (A and a) with frequencies p and q, what are the frequencies of the three possible genotypes?
The question posed is answered by the Hardy-Weinberg law. At first glance, it may seem that individuals with a dominant phenotype will be found more often than those with a recessive phenotype. However, the 3:1 ratio is observed only in the offspring of two individuals heterozygous for the same alleles. Mendel's laws say nothing about the frequencies of genotypes and phenotypes in populations. They are discussed in the said law. It was formulated independently by the mathematician J. Hardy in England and the physician Wilhelm Weinberg in Germany. To understand the meaning of this law, suppose that males and females in a population interbreed randomly, or, what is the same thing, the gametes of males and females will combine randomly to form zygotes. In the zygote, the maternal and paternal chromosomes are combined, each of the homologous chromosomes carries one allele from a given pair. The formation of individuals with the AA genotype is determined by the probability of receiving the A allele from the mother and the A allele from the father, i.e. рхр = р2.
The emergence of genotype aa is similar, the frequency of occurrence of which is equal to q2. The Aa genotype can arise in two ways: the body receives allele A from the mother, allele a from the father, or, conversely, allele A from the father, allele a from the mother. The probability of both events is equal to pq, and the total probability of the occurrence of genotype Aa is equal to 2pq. Thus, the frequency of the three possible genotypes can be expressed by the equation: (p+q)2=p2+2pq+q2=1
It follows from the equation that if the crossing is random, then the frequencies of the genotypes are related to the frequencies of the alleles by simple ratios according to the Newton binomial formula.
Let's look at an example when the allele frequencies of a given gene in a population are 0.1A; 0.3a (the geometric expression of the Hardy-Weinberg law for this case is presented in Fig. 21). In the offspring, per 100 zygotes there will be 49 AA homozygotes, 9 aa homozygotes and 42 Aa heterozygotes, i.e. this corresponds to the genotype ratio we already know - p2(AA) : 2pq(Ad): q2(aa).
Interestingly, in the next generation, gametes with the A allele will appear with a frequency of 0.7 (0.49 from AA homozygotes + 0.21 from Aa heterozygotes). This ratio will continue in the future. The frequencies of genes, and accordingly genotypes, remain unchanged from generation to generation - this is one of the main provisions of the Hardy-Weinberg law. However, this law is probabilistic in nature and therefore is implemented in an infinitely large population. At the same time, gene frequencies remain unchanged if: there is unlimited panmixia; absent natural selection; new mutations of the same genes do not arise; there is no migration of individuals with other genotypes from neighboring populations.
Heredity and variability are properties of organisms. Genetics as a science
Heredity– the ability of organisms to transmit their characteristics and developmental characteristics to their offspring.
Variability– a variety of characteristics among representatives of a given species, as well as the ability of descendants to acquire differences from their parent forms.
Genetics– the science of the laws of heredity and variability.
2. Describe the contribution of scientists you know to the development of genetics as a science by filling out the table.
History of the development of genetics
3. What methods of genetics as a science do you know?
The main method of genetics is hybridological. This is the crossing of certain organisms and analysis of their offspring. This method was used by G. Mendel.
Genealogical - the study of genealogies. Allows you to determine patterns of inheritance of traits.
Twin - comparison of identical twins, allows you to study modification variability (determine the impact of the genotype and environment on the development of the child).
Cytogenetic - study under a microscope of the chromosome set - the number of chromosomes, features of their structure. Allows detection of chromosomal diseases.
4. What is the essence of the hybridological method of studying the inheritance of traits?
The hybridological method is one of the methods of genetics, a way of studying hereditary properties organism by crossing it with a related form and subsequent analysis of the characteristics of the offspring.
5. Why can peas be considered a successful object for genetic research?
Pea species differ from each other in a small number of clearly distinguishable characteristics. Peas are easy to grow; in the Czech Republic they reproduce several times a year. In addition, in nature, peas are self-pollinators, but in experiments, self-pollination is easily prevented, and the researcher can easily pollinate the plant with the same pollen from another plant.
6. Inheritance of which pairs of traits in peas was studied by G. Mendel?
Mendel used 22 pure pea lines. Plants of these lines had very pronounced differences from each other: the shape of the seeds (round - wrinkled); seed color (yellow – green); bean shape (smooth – wrinkled); arrangement of flowers on the stem (axillary - apical); plant height (normal – dwarf).
7. What is meant in genetics by a pure line?
A pure line in genetics is a group of organisms that have some characteristics that are completely transmitted to the offspring due to the genetic homogeneity of all individuals.
Patterns of inheritance. Monohybrid cross
1. Give definitions of concepts.
Allelic genes– genes responsible for the manifestation of one trait.
Homozygous organism– an organism containing two identical allelic genes.
Heterozygous organism– an organism containing two different allelic genes.
2. What is meant by monohybrid crossing?
Monohybrid crossing is the crossing of forms that differ from each other in one pair of alternative characters.
3. Formulate a rule for the uniformity of first-generation hybrids.
When crossing two homozygous organisms that differ from each other in one trait, all hybrids of the first generation will have the trait of one of the parents, and the generation for this trait will be uniform.
4. Formulate the splitting rule.
When two descendants (hybrids) of the first generation are crossed with each other, in the second generation a split is observed and individuals with recessive traits appear again; these individuals make up ¼ of the total number of descendants of the first generation.
5. Formulate the law of gamete purity.
When formed, each of them includes only one of the two “elements of heredity” responsible for a given trait.
6. Using generally accepted symbols, draw up a monohybrid crossing scheme.
Using this example, characterize the cytological basis of monohybrid crossing.
P is the parent generation, F1 is the first generation of descendants, F2 is the second generation of descendants, A is the gene responsible for the dominant trait, and A is the gene responsible for the recessive trait.
As a result of meiosis, the gametes of the parent individuals will contain one gene each, responsible for the inheritance of a certain trait (A or a). In the first generation, the somatic cells will be heterozygous (Aa), so half of the gametes of the first generation will contain the A gene, and the other half will contain the a gene. As a result of random combinations of gametes in the second generation, the following combinations will arise: AA, Aa, aA, aa. Individuals with the first three gene combinations will have the same phenotype (due to the presence of a dominant gene), while those with the fourth will have a different phenotype (recessive).
7. Solve the genetic problem of monohybrid crossing.
Task 1.
In watermelon, the green color of the fruit dominates over the striped color. By crossing a green-fruited variety with a striped-fruited one, first-generation hybrids with green-colored fruits were obtained. The hybrids were cross-pollinated and 172 second-generation hybrids were obtained. 1) How many types of gametes does a green-fruited plant produce? 2) How many F2 plants will be heterozygous? 3) How many different genotypes will there be in F2? 4) How many plants with striped fruit color will there be in F2? 5) How many homozygous plants with green fruit color will there be in F2?
Solution
A – green color, and – striped color.
Since when crossing plants with green and striped fruits, plants with green fruit were obtained, we can conclude that the parent individuals were homozygous (AA and aa) (according to Mendel’s rule of uniformity of first-generation hybrids).
Let's draw up a crossing diagram.
Answers:
1. 1 or 2 (in case of heterozygote)
2. 86
3. 3
4. 43
5. 43.
Task 2.
Long hair in cats is recessive to short hair. A long-haired cat crossed with a heterozygous short-haired cat produced 8 kittens. 1) How many types of gametes does a cat produce? 2) How many types of gametes does a cat produce? 3) How many phenotypically different kittens are there in the litter? 4) How many genotypically different kittens are there in the litter? 5) How many kittens in the litter have long hair?
Solution
A – short hair, and – long hair. Since the cat had long hair, she is homozygous, her genotype is aa. The cat has genotype Aa (heterozygous, short hair).
Let's draw up a crossing diagram.
Answers:
1. 2
2. 1
3. 4 with long and 4 with short
4. 4 with genotype Aa, and 4 with genotype aa
5. 4.
Multiple alleles. Analysis cross
1. Give definitions of concepts.
Phenotype– a set of all signs and properties of an organism that are revealed in the process of individual development in given conditions and are the result of the interaction of the genotype with a complex of factors of the internal and external environment.
Genotype- this is the totality of all the genes of an organism, which are its hereditary basis.
2. Why are the concepts of dominant and recessive genes relative?
The gene for any trait may have other “conditions” that cannot be called either dominant or recessive. This phenomenon can occur as a result of mutations and is called “multiple allelism.”
3. What is meant by multiple allelism?
Multiple allelism is the existence of more than two alleles of a given gene in a population.
4. Fill out the table.
Types of interaction of allelic genes
5. What is analytical crossing and what is its practical significance?
Test crossing is used to establish the genotype of individuals that do not differ in phenotype. In this case, the individual whose genotype needs to be established is crossed with an individual homozygous for the recessive gene (aa).
6. Solve the problem of analyzing crossing.
Task.
The white color of the phlox corolla dominates over the pink one. A plant with a white corolla was crossed with a plant with a pink color. 96 hybrid plants were obtained, of which 51 are white and 45 are pink. 1) What genotypes do the parent plants have? 2) How many types of gametes can a plant with a white corolla produce? 3) How many types of gametes can a plant with a pink corolla produce? 4) What phenotypic ratio can be expected in the F2 generation from crossing F1 hybrid plants with white flowers with each other?
Solution.
A – white color, and – pink color. The genotype of one plant A.. is white, the second aa is pink.
Since in the first generation there is a 1:1 split (51:45), the genotype of the first plant is Aa.
Let's draw up a crossing diagram.
Answers:
1. Aa and aa.
2. 2
3. 1
4. 3 with a white corolla: 1 with a pink corolla.
Dihybrid cross
1. Give definitions of concepts.
Dihybrid cross– crossing of individuals in which differences from each other in two characteristics are taken into account.
Punnett lattice is a table proposed by the English geneticist Reginald Punnett as a tool, which is a graphical record for determining the compatibility of alleles from parental genotypes.
2. What ratio of phenotypes is obtained by dihybrid crossing of diheterozygotes? Illustrate your answer by drawing a Punnett lattice.
A – Yellow color of seeds
a – Green color of seeds
B – Smooth seed shape
c – Wrinkled shape of seeds.
Yellow smooth (AABB) × Green wrinkled (AABB) =
R: AaBv×AaBv (diheterozygotes)
Gametes: AB, Av, aB, av.
F1 in the table:
Answer: 9 (yellow smooth):3 (green smooth):3 (yellow wrinkled):1 (green wrinkled).
3. Formulate the law of independent inheritance of characteristics.
In a dihybrid cross, the genes and traits for which these genes are responsible are inherited independently of each other.
4. Solve genetic problems for dihybrid crossing.
Task 1.
Black coloring in cats dominates over fawn, and short hair dominates over long hair. Purebred Persian cats (black longhaired) were crossed with Siamese cats (fawn shorthaired). The resulting hybrids were crossed with each other. What is the probability of getting a purebred Siamese kitten in F2; a kitten phenotypically similar to a Persian; long-haired fawn kitten (express in parts)?
Solution:
A – black color, and – fawn.
B – short hair, B – long hair.
Let's create a Punnett lattice.
Answer:
1) 1/16
2) 3/16
3) 1/16.
Task 2.
In tomatoes, the round shape of the fruit dominates over the pear-shaped one, and the red color of the fruit dominates over the yellow one. By crossing a heterozygous plant with a red color and pear-shaped fruits and a yellow-fruited plant with rounded fruits, 120 plants were obtained. 1) How many types of gametes does a heterozygous plant with red fruit color and pear-shaped form form? 2) How many different phenotypes resulted from such a cross? 3) How many different genotypes resulted from this crossing? 4) How many plants were obtained with a red color and a rounded fruit shape? 5) How many plants were obtained with a yellow color and a rounded fruit shape?
Solution
A – round shape, and – pear shape.
B – red color, c – yellow color.
Let's determine the genotypes of the parents, the types of gametes and write down the crossing scheme.
Let's create a Punnett lattice.
Answer:
1. 2
2. 4
3. 4
4. 30
5. 30.
Chromosomal theory of heredity. Modern ideas about the gene and genome
1. Give definitions of concepts.
Crossing over– the process of exchange of sections of homologous chromosomes during conjugation in prophase I of meiosis.
Chromosome map- this is a diagram relative position and relative distances between genes on certain chromosomes that are in the same linkage group.
2. In what case does the law of independent inheritance of characteristics occur?
When crossing over, Morgan's law is violated, and the genes of one chromosome are not inherited linked, since some of them are replaced by allelic genes of the homologous chromosome.
3. Write the main provisions of T. Morgan’s chromosomal theory of heredity.
A gene is a section of a chromosome.
Allelic genes (genes responsible for one trait) are located in strictly defined places (loci) of homologous chromosomes.
Genes are located linearly on chromosomes, that is, one after another.
During the formation of gametes, conjugation occurs between homologous chromosomes, as a result of which they can exchange allelic genes, that is, crossing over can occur.
4. Formulate Morgan's law.
Genes located on the same chromosome during meiosis end up in one gamete, that is, they are inherited linked.
5. What determines the probability of divergence of two non-allelic genes during crossing over?
The probability of divergence of two non-allelic genes during crossing over depends on the distance between them in the chromosome.
6. What is the basis for compiling genetic maps of organisms?
Calculating the frequency of crossing over between any two genes on the same chromosome that are responsible for different traits makes it possible to accurately determine the distance between these genes, and therefore begin to build a genetic map, which is a diagram of the relative arrangement of genes that make up one chromosome.
7. Why are chromosome maps made?
Using genetic maps, you can find out the location of the genes of animals and plants and the information from them. This will help in the fight against various currently incurable diseases.
Hereditary and non-hereditary variability
1. Give definitions of concepts.
Reaction rate– the ability of a genotype to form different phenotypes in ontogenesis, depending on environmental conditions. It characterizes the share of participation of the environment in the implementation of the trait and determines the modification variability of the species.
Mutation- a persistent (that is, one that can be inherited by the descendants of a given cell or organism) transformation of the genotype, occurring under the influence of the external or internal environment.
2. Fill out the table.
3. What determines the limits of modification variability?
The limits of modification variability depend on the reaction norm, which is genetically determined and inherited.
4. What do combinative and mutational variability have in common and how do they differ?
General: both types of variability are caused by changes in the genetic material.
Differences: combinative variability occurs due to the recombination of genes during the fusion of gametes, and mutational variability is caused by the action of mutagens on the body.
5. Fill out the table.
Types of mutations
6. What is meant by mutagenic factors? Give relevant examples.
Mutagenic factors are influences that lead to the occurrence of mutations.
These may be physical effects: ionizing radiation and ultraviolet radiation, which damages DNA molecules; chemical substances, disrupting DNA structures and replication processes; viruses that insert their genes into the DNA of the host cell.
Inheritance of traits in humans. Hereditary diseases in humans
1. Give definitions of concepts.
Gene diseases– diseases caused by gene or chromosomal mutations.
Chromosomal diseases– diseases caused by changes in the number of chromosomes or their structure.
2. Fill out the table.
Inheritance of traits in humans
3. What is meant by sex-linked inheritance?
Sex-linked inheritance is the inheritance of traits whose genes are located on the sex chromosomes.
4. What traits in humans are inherited in a sex-linked manner?
Hemophilia and color blindness are inherited in humans in a gender-linked manner.
5. Solve genetic problems on the inheritance of traits in humans, including sex-linked inheritance.
Task 1.
In humans, the gene for long eyelashes is dominant over the gene for short eyelashes. A woman with long eyelashes, whose father had short eyelashes, married men with short eyelashes. 1) How many types of gametes does a woman produce? 2) How many types of gametes are produced in men? 3) What is the probability of having a child with long eyelashes in this family (in%)? 4) How many different genotypes and how many phenotypes can there be among the children of a given couple?
Solution
A – long eyelashes
a – short eyelashes.
The female is heterozygous (Aa), since the father had short eyelashes.
The man is homozygous (aa).
Answer:
1. 2
2. 1
3. 50
4. 2 genotypes (Aa) and 2 phenotypes (long and short eyelashes).
Task 2.
In humans, the free earlobe is dominant over the non-free earlobe, and the smooth chin is recessive to the chin with a triangular fossa. These traits are inherited independently. From the marriage of a man with a loose earlobe and a triangular dimple on the chin and a woman with a loose earlobe and a smooth chin, a son was born with a smooth chin and a loose earlobe. What is the probability of having a child in this family with a smooth chin and a loose earlobe; with a triangular dimple on the chin (%)?
Solution
A – free earlobe
a – non-free earlobe
B – triangular fossa
c – smooth chin.
Since the couple had a child with homozygous characteristics (aabv), the genotype of the mother is Aavv, and the genotype of the father is aaBv.
Let's write down the genotypes of the parents, types of gametes and the crossing scheme.
Let's create a Punnett lattice.
Answer:
1. 25
2. 50.
Task 3.
In humans, the gene causing hemophilia is recessive and is located on the X chromosome, while albinism is caused by an autosomal recessive gene. Parents normal according to these characteristics gave birth to an albino and hemophiliac son. 1) What is the probability that their next son will exhibit these two abnormal signs? 2) What is the probability of having healthy daughters?
Solution:
X° - presence of hemophilia (recessive), X - absence of hemophilia.
A – normal skin color
a – albino.
Genotypes of parents:
Mother - X°HAa
Father - HUAa.
Let's create a Punnett lattice.
Answer: the probability of showing signs of albinism and hemophilia (genotype X°Uaa) in the next son is 6.25%. The probability of having healthy daughters is (XXAA genotype) – 6.25%.
Task 4.
Hypertension in humans is determined by a dominant autosomal gene, while optic atrophy is caused by a sex-linked recessive gene. A woman with optic atrophy married a man with hypertension whose father also had hypertension and whose mother was healthy. 1) What is the probability that a child in this family will suffer from both anomalies (in%)? 2) What is the probability of having a healthy child (in%)?
Solution.
X° - presence of atrophy (recessive), X - absence of atrophy.
A – hypertension
a – no hypertension.
Genotypes of parents:
Mother - Х°Х°аa (since she is sick with atrophy and without hypertension)
Father - HUAa (since he is not sick with atrophy and his father had hypertension, and his mother is healthy).
Let's create a Punnett lattice.
Answer:
1. 25
2.0 (only 25% of daughters will not have these deficiencies, but they will be carriers of atrophy and without hypertension).
TASKS AND METHODS OF GENETICS.
Genetics – a fairly young science.OIts founder is the Austrian naturalist Gregor Mendel (1822–1884). In 1865, at a meeting of the Society of Natural History Lovers in the city of Brno (Czech Republic), G. Mendel examined the mechanism for preserving the adaptive characteristics of a species over a series of generations. In 1866, he published his work “Experiments on Plant Hybrids,” but this publication did not attract the attention of his contemporaries. In the spring of 1900, three botanists - G.de Friz in Holland, K. Chermak in Austria and K. Correns in Germany - independently of each other, at completely different sites, opened important pattern inheritance of traits in the offspring of hybrids. But it turned out that they simply “rediscovered” the patterns of inheritance considered by G. Mendel in 1865. Nevertheless, the official date of birth of genetics is still considered to be 1900.
Genetics – the science of heredity and variability of living organisms.Heredity – this is the body’s ability to transmit its characteristics and developmental features to subsequent generations. The elementary units of heredity are genes located on chromosomes. The transmission of traits by inheritance occurs during the process of reproduction. During sexual reproduction, the inheritance of traits and developmental characteristics occurs through germ cells. In asexual reproduction, inheritance occurs through vegetative cells and spores, which contain the material basis of heredity. Character traits species, breeds, varieties are preserved from generation to generation by plants, animals, microorganisms due to heredity. But with sexual reproduction, there is less similarity between parents and the new generation, since variability takes place.
Variability – this is the property of an organism to acquire new characteristics in the process of individual development. Variation provides material for the activity of selection and the process of evolution. Due to variability, individuals of the same species differ from each other. The appearance of new characteristics in individuals of the same species depends on changes in the material basis of the organism’s heredity and on external conditions affecting the organism.
The set of all hereditary characteristics of an organism (genes) is calledgenotype .
The set of signs of an organism (external and internal) that appear in the process of life activity is calledphenotype . Therefore, the phenotype is determined by the genotype, but external conditions the existence of organisms in which the genotype is realized can largely determine the manifestation of certain traits. Individuals of the same species having the same genotype may differ from each other depending on the conditions of existence and development. We can conclude that the phenotype develops through the interaction of the genotype and environmental conditions.
The main task of genetics is the study of such important problems as storage, transmission, implementation and variability of hereditary information. The following methods are used to solve these problems.
The most widely used method in genetics is the hybridological method of studying heredity.
Main features of the hybridological method:
1) G. Mendel did not take into account the entire diverse complex of traits in parents and their descendants, but isolated and analyzed inheritance according to individual traits;
2) an accurate quantitative accounting of the inheritance of each trait in a number of successive generations was carried out;
3) G. Mendel traced the nature of the offspring of each hybrid separately.
For his research, G. Mendel chose peas, since this plant has many well-differentiated characteristics (shape of seeds, color of seeds and flowers); Peas are characterized by self-pollination, which allowed Mendel to analyze the offspring of each individual separately. For crossing, G. Mendel selected plants that had pairs of alternative (mutually exclusive) traits.
Let's look at monohybrid crosses in detail. Classic example A monohybrid cross is the crossing of pea plants that have yellow and green seeds.
When crossing a pea variety that has yellow seeds with a plant that has green seeds, all the offspring of the first generation turned out to have yellow seeds. When pea plants with wrinkled and smooth seeds were crossed, all the offspring ended up with smooth seeds. The pattern discovered by G. Mendel was called the rule of uniformity of first generation hybrids, orlaw of dominance (I Mendel's law).
The trait that appears in the first generation is calleddominant (yellow color of seeds, smooth surface of seeds), and the trait is not manifested (suppressed trait) -recessive (green color, wrinkled surface of seeds). Mendel introduced genetic symbolism to record the results of crossing; P – parents; – female; – male; x – crossing sign; G – gametes; F – offspring; hybrids of the first, second and subsequent generations are designated by the letter F with a number below - F 1 , F 2 , F 3 ...; the letters of the Latin alphabet A, a, B, b, C, c, D, d... denote individual hereditary traits, while dominant traits are denoted by capital letters A, B, C, D,..., and recessive traits - a, b, respectively , s, d…
When drawing up a crossing scheme, it is necessary to remember that each somatic cell has a diploid set of chromosomes. All chromosomes are paired. Genes that determine the alternative development of the same trait and are located in identical regions of homologous chromosomes are calledallelic genes or alleles. There are always two allelic genes in a zygote, and the genotypic formula for any trait must be written in two letters. If any pair of alleles is represented by two dominant (AA) or two recessive (aa) genes, such an organism is calledhomozygous. If in the same allele one gene is dominant and the other is recessive, then such an organism is calledheterozygous (Ah).
Genetic recording is carried out as follows:
From the above crossbreeding scheme we see that the first generation hybrids are uniform in their dominant character. This pattern, as already mentioned, is known asMendel's first law: When crossing two homozygous organisms that differ from each other in an alternative version of the same trait, all hybrids of the first generation will be uniform in both phenotype and genotype, and will carry the characteristics of both parents in the genotype.
Mendel's experiments showed that a dominant gene manifests itself in both homozygous and heterozygous states, and a recessive gene manifests itself only in a homozygous state.
Then G. Mendel crossed the first generation hybrids with each other and obtained the following results: out of 8023 pea seeds obtained in the second generation, there were 6022 yellow and 2001 green. The same ratio was obtained with other variants of crossing between first-generation hybrids. Based on this, Mendel came to the conclusion that in the second generation there is a splitting of characters in a ratio of 3: 1, that is, 75% of individuals carry dominant characters, and 25% have recessive ones.
Let's make a genetic record of this cross:
We see that according to the phenotype there was a splitting of 3: 1, and according to the genotype 1AA: 2Aa: 1aa.
This pattern is calledrules for splitting second generation hydrides, orII Mendel's law , which is formulated as follows:When crossing two heterozygous individuals (Aa hybrids) having a pair of alternative variants of one trait, the offspring are split according to this trait in a ratio of 3: 1 in phenotype and 1: 2: 1 in genotype.
Recording crosses can be done in another way, using the so-called Punnett grid, which was proposed by the English geneticist Punnett. The principle of constructing a grid is simple: the gametes of a female individual are recorded along the horizontal line at the top, and the gametes of a male individual are recorded along the vertical line on the left, and at the intersection of the vertical and horizontal lines the genotype and phenotype of the descendants are determined.
Let's make a genetic record of the considered examples using the Punnett lattice.
Typically, genetic recording using the Punnett grid is used in the analysis of more complex crosses. For us, it allows us to easily figure out why the offspring of the first generation are uniform, while splitting occurred in the second generation.
Genes are known to be located on chromosomes. In the example considered, a pea plant with yellow seeds carries a pair of yellow alleles in some pair of homologous chromosomes. As a result of meiosis, homologous chromosomes diverge into different gametes, and with them allelic genes (yellow seed color). Since both allelic genes are the same in a homozygote (AA), all gametes carry this gene. It’s the same with plants that have green seeds (aa): both allelic genes are the same, therefore the gametes carry the same gene. We conclude: a homozygous individual always produces one type of gametes.
Thus, if the maternal individual (AA) gives one type of gametes (A) and the paternal individual (aa) gives one type of gametes (a), then only one combination of gametes is possible - Aa, that is, all hybrids of the first generation are uniform and are heterozygous for this characteristic (seed color). Phenotypically, all plants will have yellow seeds. When crossing two heterozygotes (Aa), each individual produces gametes with a dominant gene (A) and a recessive gene (a) in equal numbers, from which four possible combinations of zygotes should be expected. An egg with gene (A) can be fertilized by a sperm with either a dominant gene (A) or a recessive gene (a); in the same way, an egg with gene (a) can be fertilized with equal probability by a sperm with gene (A) and gene (a). As a result, four zygotes are formed: AA: Aa: aA: aa. We see that according to the phenotype we received 3 individuals with dominant traits and one individual with recessive traits, that is, a ratio of 3: 1. According to the genotype, the ratio is 1AA: 2Aa: 1aa.OIt follows from this that if in the future we obtain offspring from each group of individuals of the second generation through self-pollination, then homozygous individuals AA and aa will produce only uniform offspring, without splitting, and the offspring of individuals with the genotype Aa (heterozygous) will continue to split further.
G. Mendel explained this by the fact that gametes are genetically pure, that is, they carry only one gene from an allelic pair. Based on this conclusion, G. Mendel formulatedlaw of gamete purity: The pairs of alternative characteristics found in each organism do not mix and, when gametes are formed, one by one they pass into them in their pure form.
(Definition from the book: T. L. Bogdanova. Biology. Tasks and exercises. M., graduate School, 1991)
TASKS AND METHODS OF GENETICS.
MENDEL'S FIRST AND SECOND LAWS
Task No. 1.
How many types of gametes are formed by individuals with genotype BB; with the BB genotype; with genotype bb?
D ano:Genotypes:
1) Vv
2) BB
3) cc
Solution:
The number of expected gamete types is determined by the formula:X = 2 n , where n is the number of pairs of alternative characteristics of the organism being studied, andX – number of gamete types.
Find:
number of gamete types – ?
1) Вв – genotype of an individual.
One pair of alternative features.
We determine the number of gamete combinations:X = 2 1 , from hereX = 2 (V, v).
2) BB – genotype of the individual; there are no alternative signs.
X = 2 0 , from hereX = 1 (V).
3) bb – genotype of the individual; there are no alternative signs.
Let us determine the number of gamete combinations:X = 2 0 , from hereX = 1 (c).
ANSWER: 2 types of gametes; 1 type of gametes; 1 type of gametes.
Task No. 2.
How many types of gametes does an individual produce: a) homozygous for a recessive gene? b) homozygous for a dominant gene? c) heterozygous?
D ano:Genotypes:
1) aa
2) AA
3) Aa
Solution:
a) aa – genotype of the individual
X = 2 0 = 1 (a)
b) AA – genotype of the individual
X = 2 0 = 1 (A)
c) Aa – genotype of the individual
X = 2 1 = 2 (A, a)
Find:
X - ?
ANSWER: a) 1 type of gametes; b) 1 type of gametes; c) 2 types of gametes.
Task No. 3.
The smooth coloring of watermelons is inherited as a recessive trait. What offspring will result from crossing two heterozygous plants with striped fruits?
Task No. 4.
Find possible variants of gametes for organisms with the following genotypes: AA, BB, Cs, DD.
D ano:Genotypes:
AA, BB, SS, DD
Solution:
1) AA is a homozygous organism that produces one type of gametes: A.
2) BB is a heterozygous organism that produces two types of gametes: B and c.
3) CC is a heterozygous organism that produces two types of gametes: C and s.
Find:
possible variants of gametes – ?
4) DD is a homozygous organism that produces one type of gametes: D.
ANSWER: 1) A; 2) In, in; 3) C, s; 4) D.
Task No. 3.
Determine the genotypes and phenotypes of offspring from the marriage of brown-eyed heterozygous parents.
Note: if in the task we're talking about about people, then the following designations for parents are introduced: ○ – women; □ – men.
ANSWER: 1АА: 2Аа: 1аа; 3 children with brown and one with blue eyes.
Task No. 4.
A person’s ability to use predominantly the right hand dominates the ability to use predominantly the left hand. A right-handed man whose mother was left-handed married a right-handed woman who had three siblings, two of whom were left-handed.Odetermine the possible genotypes of a woman and the probability that children born from this marriage will be left-handed.
Answer: if a woman is homozygous, then the probability of being born left-handed will be 0, if she is heterozygous, then 25% will be born left-handed.
Task No. 5.
When crossing heterozygous red-fruited tomatoes with yellow-fruited ones, 352 plants with red fruits were obtained.Osteel plants had yellow fruits. Determine how many plants were yellow in color?
Answer:352 plants.
Task No. 6.
Myoplegia (periodic paralysis) is inherited as a dominant trait. Determine the probability of having children with anomalies in a family where the father is heterozygous and the mother does not suffer from myoplegia.
Answer: the probability of having children with anomalies will be 50%.
Task No. 7.
In tomatoes, the gene that determines the red color of fruits is dominant to the gene for yellow color. 3021 tomato bushes obtained from hybrid seeds were yellow in color, and 9114 were red.
Question: a) how many heterozygous plants are there among the hybrids? b) does this characteristic (fruit color) belong to Mendelians?
2) Let’s count the number of heterozygous plants, constituting 2/3 of the number of all red-fruited plants:
(9114: 3) · 2 = 6742 plants.
3) The trait “fruit color” is Mendelian, since the ratio of bushes with yellow and red fruits is 1: 3, that is, it obeys Mendel’s second law.
Answer: a) 6742 plants; b) applies.
Task No. 8.
Gene for black body color cattle dominates the red gene. What kind of offspring can be expected from crossing: a) two heterozygous individuals? b) a red bull and a hybrid cow?
Answer: a) 75% black calves, 25% red calves;
b) 50% black calves, 50% red calves.
DIHYBRID CROSSING.
MENDEL'S THIRD LAW
Dihybrid called a crossing in which individuals differ in two pairs of alleles. G. Mendel crossed two varieties of peas - with smooth yellow seeds and with green wrinkled ones. The first generation hybrids all had smooth yellow seeds, since in the first generation only dominant traits always appear. When crossing the first generation hybrids with each other, a split was discovered: 315 yellow smooth seeds, 101 yellow wrinkled seeds, 108 green smooth seeds, 32 green wrinkled seeds.
Let us analyze the results of crossing: as we see, when crossing homozygous forms, the first generation hybrids are uniform, as in monohybrid crossing; in the second generation, the characteristics are split and four groups of individuals different in phenotype are formed (yellow smooth, yellow wrinkled, green smooth, green wrinkled), and the ratio of their phenotypes is 9: 3: 3: 1. We see that with dihybrid crossing there is an increase the number of phenotypes is doubled compared to a monohybrid cross. If we look at the ratio of each trait separately, we will see that it is 3: 1, as with monohybrid crossing, that is, splitting for each trait occurs independently. Based on this, G. Mendel formulatedIII law – the law of independent inheritance of characteristics: splitting for each pair of characteristics occurs independently of other pairs of characteristics.
Let's consider the cytological foundations of Mendel's III law. Two homozygous pea plants with genotypesAABB (yellow smooth seeds) andaaww (green wrinkled seeds) form one type of gametes -AB Andaw . As a result of their crossing, the offspring will be uniform -AaVv (yellow smooth seeds). When hybrids of the first generation are crossed with each other, each plant forms four types of gametes, and from each pair of allelic genes only one gets into the gamete. As a result of random divergence of chromosomes in the first division of meiosis, the geneA can get into the same gamete with the geneIN , forming a type of gametesAB , or with the geneV (forming the second type of gametes -Av ), and the geneA can combine with a geneIN (forming the third type of gametes -aB ) or with a geneV (forming the fourth type of gametes -aw ):
During fertilization, each of the four types of gametes(AB, Av, aV, av) one organism can meet with any of the gametes(AB, Av, aV, av) another organism. All possible combinations of male and female gametes can be easily established using a Punnett grid.
By genotype: 1ААВВ: 2ААВв: 2АаВВ: 4АаВв: 1АABB: 2AaBB: 1aaBB: 2 aaBB: 1aaBB.
As already mentioned, the resulting hybrids in the second generation have the following ratio of phenotypes: 9 parts - yellow smooth, 3 - green smooth, 3 - yellow wrinkled and 1 part green wrinkled. In this case, the characteristics are inherited independently of each other and for each of them the usual splitting is observed - 3: 1.
Thus, in a dihybrid cross, each pair of characters, when split in the offspring, behaves in the same way as in a monohybrid cross, that is, independently of the other pair of characters. Such splitting is obtained with complete dominance for each pair of characteristics. In polyhybrid crossing (parental forms differ in several or many traits), the splitting for each trait is the same.
PROBLEMS FOR DIHYBRID
AND ANALYZING CROSSING
Task No. 1.
Write the possible types of gametes produced by organisms with the following genotypes: a) AABB, b) SsDD, c) EeFf;
d) ddhh (genes are inherited independently).
ANSWER: a) AB; b) SD, SD; c) Ef, Ef, eF, ef; d) dh.
Task No. 2.
Normal growth in oats dominates over gigantism, and early ripening dominates over late ripening. The genes for both traits are located on different pairs of chromosomes. What traits will hybrids obtained from crossing parents heterozygous for both traits have? What is the phenotype of the parents?
ANSWER: P – normal, early ripening oat plants;
F 1 : 9 parts of normal growth, early ripening; 3 – normal growth, late ripening; 3 – giant early ripening; 1 part giant late ripening.
Task No. 3.
In Drosophila, gray body color and the presence of bristles are dominant characters that are inherited independently. What offspring should be expected from crossing a yellow female without bristles with a male heterozygous for both traits?
ANSWER: 25% – gray, without bristles; 25% – gray, with bristles; 25% – yellow, with bristles; 25% – yellow, without bristles.
Task No. 4.
When a black rooster without a crest was crossed with a brown crested hen, all the offspring turned out to be black and crested. Determine the genotypes of parents and offspring. What traits are dominant? What percentage of brown, crestless chickens will result from crossing first-generation hybrids with each other?
3) Let’s determine the percentage of brown chicks without a crest:
ANSWER: R 1 : ahbb, ahhcc; F 1 : AaBv;
R 2 : AaBv, AaBv;
F 2 : 1ААВВ: 2ААВв: 2АаВВ: 4АаВв: 1АABB: 2AaBB: 1aaBB: 2aaBB: 1aaBB; dominant features - black plumage and the presence of a crest; There will be 6% brown chicks without crest.
Task No. 5.
A pumpkin that had yellow disc-shaped fruits was crossed with a pumpkin that had white spherical fruits. All hybrids from this crossing had a white color and disc-shaped fruits. What signs dominate? What are the genotypes of parents and offspring?
ANSWER: R: aabb, ahhcc; F 1 : AaBv; The dominant features are white color and disc-shaped fruit.
Task No. 6.
Polydactyly (polydactyly) and the absence of small molars are transmitted as dominant characters. The genes for these traits are located in different pairs of chromosomes. What is the probability of having children without anomalies in a family where both parents suffer from both diseases and are heterozygous for these gene pairs?
Answer: the probability of having children without anomalies is 1/16.
Task No. 7.
In humans, some forms of myopia dominate over normal vision, and brown eye color dominates over blue. What kind of offspring can be expected from the marriage of a nearsighted, brown-eyed man with a blue-eyed woman with normal vision? Determine all possible genotypes of parents and offspring.
25% – normal vision brown eyes
25% – normal vision blue eyes
ANSWER: 1) R: aavv,aaww; F 1 : AaBv;
2) R: aavv,AAbb; F 1 : AaVv, aaVV;
3) R: aavv,aavV; F 1 : AaVv,Aaww;
4) R: aavv,AAVV; F 1 : AaVv,AAww, aaWw, aww.
Task No. 8.
Some forms of cataracts and deaf-muteness in humans are transmitted as recessive unlinked traits.
Questions:
1. What is the probability of having children with two anomalies in a family where both parents are heterozygous for two pairs of genes?
2. What is the probability of having children with two anomalies in a family where one of the parents suffers from cataracts and deaf-muteness, and the second spouse is heterozygous for these characteristics?
Answer: in the first case, the probability of having children with two anomalies will be 1/16, or 6%, in the second -¼, or 25%.
Task No. 9.
Glaucoma (eye disease) has two forms: one form is determined by a dominant gene, and the other by a recessive gene. Genes are located on different chromosomes. What is the probability of having a sick child in the family:
a) where both spouses suffer from different forms of glaucoma and are homozygous for both pairs of genes;
b) where both spouses are heterozygous for both pairs of genes?
ANSWER: a) 100% of sick children;
b) 13/16 sick children, or 81%.
Task No. 10.
In snapdragons, the red color of the flower is not completely dominant over the white. The hybrid plant is pink in color. The normal flower shape is completely dominant over the pyloric one. What offspring will result from crossing two diheterozygous plants?
ANSWER: 3/16 – red flowers of normal shape;
6/16 – pink flowers of normal shape;
1/16 – red flowers of pyloric form;
2/16 – pink pyloric flowers;
3/16 – white flowers of normal shape;
1/16 – white pyloric flowers.
Hardy–Weinberg law
We will consider Mendelian populations:
– individuals are diploid;
– reproduce sexually;
the population size is infinitely large; and panmictic populations, where random free crossing of individuals occurs in the absence of selection.
Consider in a population one autosomal gene represented by two alleles A And A.
Let us introduce the following notation:
N – total number of individuals in the population
D – number of dominant homozygotes ( AA)
H – number of heterozygotes ( Ahh)
R – number of recessive homozygotes ( aa)
Then: D + H + R = N.
Since individuals are diploid, the number of all alleles for the gene in question will be 2N.
Total number of alleles A And A:
A= 2D + H;
A= H + 2R.
Let us denote the proportion (or frequency) of the allele A through p, and allele A– through g, then:
Since a gene can be represented by alleles A or A and no others, then p + g = 1.
The state of population equilibrium was described by a mathematical formula in 1908, independently of each other, by mathematician J. Hardy in England and physician W. Weinberg in Germany (Hardy-Weinberg law).
If p is the gene frequency A, and g is the gene frequency A, using the Punnett lattice, we can represent in a generalized form the nature of the distribution of alleles in the population:
The ratio of genotypes in the described population:
p2 AA: 2pg Ahh: g 2 ah.
The Hardy–Weinberg law in its simplest form:
p2 AA+ 2pg Ahh+ g 2 ahh = 1.
Problem No. 36
The population contains 400 individuals, of which with genotypes AA – 20, Ahh– 120 and ahh– 260 individuals. Determine Gene Frequencies A And A.
Given: |
Solution: |
|
| N=400 D=20 H=120 R = 260 p – ? g – ? |
 |
Answer: gene frequency A– 0.2; gene A – 0,8.
Problem No. 37
In Shorthorn cattle, the red color is dominant over the white color. Hybrids from crossing red and white - roan color. The shorthorn breeding area has 4,169 red, 3,780 roan and 756 white recorded animals. Determine the frequency of genes for red and white coloration of cattle in this area.
Answer: frequency of the red color gene – 0.7; white – 0.3.
Problem No. 38
In a sample of 84,000 rye plants, 210 plants turned out to be albino, because... their recessive genes are in a homozygous state. Determine Allele Frequencies A And A, as well as the frequency of heterozygous plants.
Answer: gene frequency A And A – 0,5.
Problem No. 40
Allele frequencies p = 0.8 and g = 0.2 are known in the population. Determine genotype frequencies.
Answer: allele frequency A– 0.45; allele A – 0,55.
Problem No. 42
In a herd of cattle, 49% of animals are red (recessive) and 51% are black (dominant). What is the percentage of homo- and heterozygous animals in this herd?
Answer: in the population 81% of individuals with the genotype AA, 18% with genotype Ahh and 1% with genotype ahh.
Entertaining genetic problems
Problem No. 44. "A Tale of Dragons"
The researcher had 4 dragons: fire-breathing and non-fire-breathing females, fire-breathing and non-fire-breathing males. To determine the fire-breathing ability of these dragons, all kinds of crossings were carried out.
1. Fire-breathing parents - all offspring are fire-breathing.
2. Non-fire-breathing parents - all offspring are non-fire-breathing.
3. A fire-breathing male and a non-fire-breathing female - the offspring contain approximately equal numbers of fire-breathing and non-fire-breathing dragonets.
4. A non-fire-breathing male and a fire-breathing female - all offspring are non-fire-breathing.
Assuming that the trait is determined by an autosomal gene, identify the dominant allele and record the genotypes of the parents.
Solution:
By crossing No. 4 we determine: A- non-fire-breathing, A– fire-breathing; fire-breathing – ♀ ahh and ♂ ahh; non-fire-breathing male – ♂ AA;
by crossing No. 3: non-fire-breathing female – ♀ Ahh.
Problem No. 45. “Consultant of the company “Cocktail”.
Imagine that you are a consultant for a small company “Cocktail”, which literally translated from English means “rooster tail”. The company breeds exotic breeds of roosters for their tail feathers, which are readily purchased by hat store owners all over the world. Feather length is determined by the gene A(long) and A(short), color: IN– black, b– red, width: WITH– wide, With– narrow. The genes are not linked. There are many different roosters and chickens on the farm with all possible genotypes, the data of which is entered into the computer. Next year, increased demand for hats with long black narrow feathers is expected. What crosses should be made to get the maximum number of birds with fashionable feathers in the offspring? It is not worth crossing pairs with absolutely identical genotypes and phenotypes.
Solution:
F 1: A*IN*cc
1. R: ♀ AABBss × ♂ aabbss
2. R: ♀ AABBss × ♂ AAbbss
3. R: ♀ AAbbss × ♂ aaBBSS etc.
Problem No. 46. "Smuggler".
IN small state Foxland has been breeding foxes for several centuries. Fur is exported, and money from its sale forms the basis of the country's economy. Silver foxes are especially prized. They are considered a national treasure and transporting them across borders is strictly prohibited. A cunning smuggler who did well in school wants to deceive customs. He knows the basics of genetics and suggests that the silver coloration of foxes is determined by two recessive alleles of the coat color gene. Foxes with at least one dominant allele are red. What needs to be done to get silver foxes in the smuggler’s homeland without violating the laws of Foxland?
Solution:
Conduct test crossings and find out which red foxes are heterozygous for color alleles, transport them across the border;
in the smuggler's homeland, cross them with each other, and 1/4 of the descendants will be silver in color.
Problem No. 47. "Will Prince Uno's wedding be upset?"
The only crown prince, Uno, is about to marry the beautiful princess Beatrice. Uno's parents learned that there were cases of hemophilia in Beatrice's family. Beatrice has no brothers or sisters. Aunt Beatrice has two healthy, strong sons. Uncle Beatrice spends his days hunting and feels great. The second uncle died as a boy from loss of blood, the cause of which was a deep scratch. Uncles, aunts and Beatrice's mother are children of the same parents. How likely is it that the disease can be transmitted through Beatrice to the royal family of her fiancé?
Solution:
By constructing a putative family tree, it can be proven that the hemophilia gene was on one of Beatrice's grandmother's X chromosomes; Beatrice's mother could receive it with a probability of 0.5; Beatrice herself – with a probability of 0.25.
Problem No. 48. "Royal dynasties".
Let's assume that Emperor Alexander I had a rare mutation on his Y chromosome. Could this mutation be present in:
a) Ivan the Terrible;
b) Peter I;
c) Catherine II;
d) Nicholas II?
Solution:
Let’s immediately cross out Catherine II, due to her being female.
Let's cross out Ivan the Terrible too - he is a representative of the Rurikovich family and did not belong to the Romanov dynasty.
Answer: Nicholas II could have had the mutation.
Problem No. 49. “Leafing through the novel “War and Peace.”
Let's assume that Prince Nikolai Andreevich Bolkonsky had a rare mutation on the X chromosome. Pierre Bezukhov had the same mutation. What is the likelihood of this mutation occurring in:
a) Natasha Rostova;
b) son of Natasha Rostova;
c) son Nikolai Rostov;
d) the author of “War and Peace”?
Answer:
Andrei Bolkonsky did not receive an X chromosome from his father. His wife was not a relative of either the Bolkonskys or the Bezukhovs. Consequently, Prince Andrei’s son does not have the mutation.
Natasha Rostova married Pierre Bezukhov. Pierre passed on the X chromosome to his daughters, but not to his sons. Consequently, Natasha Rostova’s daughters received the mutation, but her sons did not.
The son of Nikolai Rostov received his X chromosome from his mother, the daughter of the old Prince Bolkonsky (out of Princess Marya’s 2 chromosomes, there was a mutation in only one, therefore, she passed the X chromosome to her son with a probability of 50%).
Lev Nikolaevich: the action of the novel ends several years before Tolstoy’s birth; the author himself does not appear on the pages of the novel. But the writer’s father was a retired officer Count Nikolai Ilyich Tolstoy, and his mother was nee Volkonskaya, i.e. The writer's parents were the prototypes of Nikolai Rostov and his wife, nee Maria Bolkonskaya. Their future son Leo will receive a mutation with a 50% probability.
Problem No. 50. "The dispute between Bender and Panikovsky."
Bender and Panikovsky had a dispute: how is color inherited in budgies? Bender believes that the color of parrots is determined by one gene that has 3 alleles: S o– recessive in relation to the other two, C g And S are codominant, therefore in parrots with the genotype S o S o – White color, C g C g And C g C o– blue, S f S f And S f S o– yellow and S g S w- green color. And Panikovsky believes that color is formed under the influence of two interacting genes A And IN. Therefore, parrots with the genotype A*B*– green, A*bb- blue, aaB*- yellow, aabb– white.
They compiled 3 genealogies:
1. P: W × B |
F1: G, B F1: B F1: G, F, G, G, F, F, F, G, F |
Which genealogies could have been compiled by Bender, which by Panikovsky?
Answer: pedigrees 1 and 2 could have been compiled by Panikovsky, and pedigree 3 by Bender.
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