Czech explorer Gregor Mendel(1822-1884) considered founder of genetics, since he was the first, even before this science took shape, to formulate the basic laws of inheritance. Many scientists before Mendel, including the outstanding German hybridizer of the 18th century. I. Kelreuter noted that when crossing plants belonging to different varieties, great variability is observed in the hybrid offspring. However, no one was able to explain the complex splitting and, moreover, reduce it to precise formulas due to the lack of a scientific method of hybridological analysis.
It was thanks to the development of the hybridological method that Mendel managed to avoid the difficulties that had confused earlier researchers. G. Mendel reported on the results of his work in 1865 at a meeting of the Society of Natural Scientists in Brünn. The work itself, entitled “Experiments on Plant Hybrids,” was later published in the “Proceedings” of this society, but did not receive proper assessment from contemporaries and remained forgotten for 35 years.
As a monk, G. Mendel conducted his classical experiments on crossing different varieties of peas in the monastery garden in Brünn. He selected 22 pea varieties that had clear alternative differences in seven characteristics: seeds yellow and green, smooth and wrinkled, flowers red and white, plants tall and short, etc. An important condition of the hybridological method was the mandatory use of pure, i.e., as parents. forms that do not split according to the studied characteristics.
A successful choice of object played a major role in the success of Mendel's research. Peas are self-pollinators. To obtain first-generation hybrids, Mendel castrated the flowers of the mother plant (removed the anthers) and artificially pollinated the pistils with the pollen of the male parent. When obtaining second-generation hybrids, this procedure was no longer necessary: he simply left the F 1 hybrids to self-pollinate, which made the experiment less labor-intensive. Pea plants reproduced exclusively sexually, so that no deviations could distort the results of the experiment. And finally, in peas, Mendel discovered a sufficient number of pairs of brightly contrasting (alternative) and easily distinguishable pairs of characters for analysis.
Mendel began his analysis with the simplest type of crossing - monohybrid, in which the parent individuals differ in one pair of traits. The first pattern of inheritance discovered by Mendel was that all first-generation hybrids had the same phenotype and inherited the trait of one of the parents. Mendel called this trait dominant. An alternative trait of the other parent, which did not appear in hybrids, was called recessive. The discovered pattern was named I of Mendel's law, or the law of uniformity of hybrids of the 1st generation. During the analysis of the second generation, a second pattern was established: the splitting of hybrids into two phenotypic classes (with a dominant trait and with a recessive trait) in certain numerical ratios. By counting the number of individuals in each phenotypic class, Mendel established that splitting in a monohybrid cross corresponds to the formula 3: 1 (three plants with a dominant trait, one with a recessive trait). This pattern is called Mendel's II law, or law of segregation. Open patterns emerged in the analysis of all seven pairs of characteristics, on the basis of which the author came to the conclusion about their universality. When self-pollinating F 2 hybrids, Mendel obtained the following results. Plants with white flowers produced offspring with only white flowers. Plants with red flowers behaved differently. Only a third of them gave uniform offspring with red flowers. The offspring of the rest were split in the ratio of red and white colors in a ratio of 3: 1.
Below is a diagram of the inheritance of pea flower color, illustrating Mendel's I and II laws.
In an attempt to explain the cytological basis of open patterns, Mendel formulated the idea of discrete hereditary inclinations contained in gametes and determining the development of paired alternative characters. Each gamete carries one hereditary deposit, i.e. is “pure”. After fertilization, the zygote receives two hereditary deposits (one from the mother, the other from the father), which do not mix and later, when gametes are formed by the hybrid, they also end up in different gametes. This hypothesis of Mendel was called the rule of “purity of gametes.” The combination of hereditary inclinations in the zygote determines what character the hybrid will have. Mendel denoted the inclination that determines the development of a dominant trait with a capital letter ( A), and recessive is capitalized ( A). Combination AA And Ahh in the zygote determines the development of a dominant trait in the hybrid. A recessive trait appears only when combined ahh.
In 1902, V. Betson proposed to designate the phenomenon of paired characters by the term “allelomorphism”, and the characters themselves, accordingly, “allelomorphic”. According to his proposal, organisms containing the same hereditary inclinations began to be called homozygous, and those containing different inclinations - heterozygous. Later, the term “allelomorphism” was replaced by the shorter term “allelism” (Johansen, 1926), and the hereditary inclinations (genes) responsible for the development of alternative traits were called “allelic”.
Hybridological analysis involves reciprocal crossing of parental forms, i.e. using the same individual first as the maternal parent (forward crossing) and then as the paternal parent (backcrossing). If both crosses produce the same results, corresponding to Mendel’s laws, then this indicates that the analyzed trait is determined by an autosomal gene. Otherwise, the trait is linked to sex, due to the localization of the gene on the sex chromosome.
Letter designations: P - parental individual, F - hybrid individual, ♀ and ♂ - female or male individual (or gamete),
capital letter (A) is a dominant hereditary disposition (gene), lowercase letter (a) is a recessive gene.
Among the second generation hybrids with yellow seed color there are both dominant homozygotes and heterozygotes. To determine the specific genotype of a hybrid, Mendel proposed crossing the hybrid with a homozygous recessive form. It is called analyzing. When crossing a heterozygote ( Ahh) with the analyzer line (aa), splitting is observed both by genotype and phenotype in a 1: 1 ratio.
If one of the parents is a homozygous recessive form, then the analyzing cross simultaneously becomes a backcross - a return crossing of the hybrid with the parent form. The offspring from such a cross are designated Fb.
The patterns Mendel discovered in his analysis of monohybrid crosses also appeared in dihybrid crosses in which the parents differed in two pairs of alternative traits (for example, yellow and green seed color, smooth and wrinkled shape). However, the number of phenotypic classes in F 2 doubled, and the phenotypic splitting formula was 9: 3: 3: 1 (for 9 individuals with two dominant traits, three individuals each with one dominant and one recessive trait, and one individual with two recessive traits ).
To facilitate the analysis of splitting in F 2, the English geneticist R. Punnett proposed a graphical representation of it in the form of a lattice, which began to be called after his name ( Punnett grid). On the left, vertically, it contains the female gametes of the F1 hybrid, and on the right - the male ones. The inner squares of the lattice contain the combinations of genes that arise when they merge, and the phenotype corresponding to each genotype. If the gametes are placed in a lattice in the sequence shown in the diagram, then in the lattice you can notice the order in the arrangement of genotypes: all homozygotes are located along one diagonal, and heterozygotes for two genes (diheterozygotes) are located along the other. All other cells are occupied by monoheterozygotes (heterozygotes for one gene).
The cleavage in F 2 can be represented using phenotypic radicals, i.e. indicating not the entire genotype, but only the genes that determine the phenotype. This entry looks like this:
The dashes in the radicals mean that the second allelic genes can be either dominant or recessive, and the phenotype will be the same.
Dihybrid crossing scheme
(Punnet grid)
| AB | Ab | aB | ab | |
| AB | AABB yellow Ch. |
AABb yellow Ch. |
AaBB yellow Ch. |
AaBb yellow Ch. |
| Ab | AABb yellow Ch. |
AAbb yellow wrinkle |
AaBb yellow Ch. |
Aabb yellow wrinkle |
| aB | AaBB yellow Ch. |
AaBb yellow Ch. |
aaBB green Ch. |
aaBb green Ch. |
| ab | AaBb yellow Ch. |
Aabb yellow wrinkle |
aaBb green Ch. |
aabb |
The total number of F2 genotypes in the Punnett lattice is 16, but there are 9 different ones, since some genotypes are repeated. The frequency of different genotypes is described by the rule:
In an F2 dihybrid cross, all homozygotes occur once, monoheterozygotes twice, and diheterozygotes four times. The Punnett grid contains 4 homozygotes, 8 monoheterozygotes and 4 diheterozygotes.
Segregation by genotype corresponds to the following formula:
1AABB: 2AABBb: 1AAbb: 2AaBB: 4AaBBb: 2Aabb: 1aaBB: 2aaBBb: 1aabb.
Abbreviated as 1:2:1:2:4:2:1:2:1.
Among the F 2 hybrids, only two genotypes repeat the genotypes of the parental forms: AABB And aabb; in the rest, recombination of parental genes occurred. It led to the emergence of two new phenotypic classes: yellow wrinkled seeds and green smooth ones.
Having analyzed the results of dihybrid crossing for each pair of characters separately, Mendel established the third pattern: the independent nature of inheritance of different pairs of characters ( Mendel's III law). Independence is expressed in the fact that splitting for each pair of characteristics corresponds to the monohybrid crossing formula 3: 1. Thus, a dihybrid crossing can be represented as two simultaneously occurring monohybrid ones.
As was established later, the independent type of inheritance is due to the localization of genes in different pairs of homologous chromosomes. The cytological basis of Mendelian segregation is the behavior of chromosomes during cell division and the subsequent fusion of gametes during fertilization. In prophase I of the reduction division of meiosis, homologous chromosomes conjugate, and then in anaphase I they diverge to different poles, due to which allelic genes cannot enter the same gamete. When they diverge, non-homologous chromosomes freely combine with each other and move to the poles in different combinations. This determines the genetic heterogeneity of germ cells, and after their fusion during the process of fertilization, the genetic heterogeneity of zygotes, and as a consequence, the genotypic and phenotypic diversity of the offspring.
Independent inheritance of different pairs of traits makes it easy to calculate segregation formulas in di- and polyhybrid crosses, since they are based on simple monohybrid cross formulas. When calculating, the law of probability is used (the probability of the occurrence of two or more phenomena at the same time is equal to the product of their probabilities). A dihybrid cross can be decomposed into two, and a trihybrid cross into three independent monohybrid crosses, in each of which the probability of the manifestation of two different traits in F 2 is equal to 3: 1. Therefore, the formula for splitting the phenotype in F 2 dihybrid cross will be:
(3: 1) 2 = 9: 3: 3: 1,
trihybrid (3: 1) 3 = 27: 9: 9: 9: 3: 3: 3: 1, etc.
The number of phenotypes in an F2 polyhybrid cross is equal to 2 n, where n is the number of pairs of characteristics in which the parent individuals differ.
Formulas for calculating other characteristics of hybrids are presented in Table 1.
Table 1. Quantitative patterns of segregation in hybrid offspring
for various types of crossings
| Quantitative characteristics | Type of crossing | ||
| monohybrid | dihybrid | polyhybrid | |
| Number of gamete types formed by hybrid F 1 | 2 | 2 2 | 2 n |
| Number of gamete combinations during the formation of F 2 | 4 | 4 2 | 4n |
| Number of phenotypes F 2 | 2 | 2 2 | 2 n |
| Number of genotypes F 2 | 3 | 3 2 | 3 |
|
Phenotype splitting in F 2 |
3: 1 | (3: 1) 2 | (3:1)n |
| Segregation by genotype in F 2 | 1: 2: 1 | (1: 2: 1) 2 | (1:2:1)n |
The manifestation of the patterns of inheritance discovered by Mendel is possible only under certain conditions (independent of the experimenter). They are:
- Equally probable formation by hybridomas of all varieties of gametes.
- All possible combinations of gametes during the process of fertilization.
- Equal viability of all varieties of zygotes.
If these conditions are not met, then the nature of segregation in the hybrid offspring changes.
The first condition may be violated due to the non-viability of one or another type of gamete, possibly due to various reasons, for example, the negative effect of another gene manifested at the gametic level.
The second condition is violated in the case of selective fertilization, in which there is a preferential fusion of certain types of gametes. Moreover, a gamete with the same gene can behave differently during the process of fertilization, depending on whether it is female or male.
The third condition is usually violated if the dominant gene has a lethal effect in the homozygous state. In this case, in F 2 monohybrid crossing as a result of the death of dominant homozygotes AA instead of a 3:1 split, a 2:1 split is observed. Examples of such genes are: the gene for platinum fur color in foxes, the gene for gray coat color in Shirazi sheep. (More details in the next lecture.)
The reason for deviation from Mendelian segregation formulas can also be incomplete manifestation of the trait. The degree of manifestation of the action of genes in the phenotype is denoted by the term expressivity. For some genes it is unstable and highly dependent on external conditions. An example is the recessive gene for black body color in Drosophila (mutation ebony), the expressivity of which depends on temperature, as a result of which individuals heterozygous for this gene can have a dark color.
Mendel's discovery of the laws of inheritance was more than three decades ahead of the development of genetics. The work “Experience with Plant Hybrids” published by the author was not understood and appreciated by his contemporaries, including Charles Darwin. The main reason for this is that at the time of the publication of Mendel’s work, chromosomes had not yet been discovered and the process of cell division, which, as mentioned above, constituted the cytological basis of Mendelian patterns, had not yet been described. In addition, Mendel himself doubted the universality of the patterns he discovered when, on the advice of K. Nägeli, he began to check the results obtained on another object - the hawkweed. Not knowing that the hawksbill reproduces parthenogenetically and, therefore, it is impossible to obtain hybrids from it, Mendel was completely discouraged by the results of the experiments, which did not fit into the framework of his laws. Under the influence of failure, he abandoned his research.
Recognition came to Mendel at the very beginning of the twentieth century, when in 1900 three researchers - G. de Vries, K. Correns and E. Cermak - independently published the results of their studies, reproducing Mendel's experiments, and confirmed the correctness of his conclusions . Since by this time mitosis, almost completely meiosis (its full description was completed in 1905), as well as the process of fertilization, had been completely described, scientists were able to connect the behavior of Mendelian hereditary factors with the behavior of chromosomes during cell division. The rediscovery of Mendel's laws became the starting point for the development of genetics.
The first decade of the twentieth century. became the period of the triumphal march of Mendelism. The patterns discovered by Mendel were confirmed in the study of various characteristics in both plant and animal objects. The idea of the universality of Mendel's laws arose. At the same time, facts began to accumulate that did not fit within the framework of these laws. But it was the hybridological method that made it possible to clarify the nature of these deviations and confirm the correctness of Mendel’s conclusions.
All pairs of characters that were used by Mendel were inherited according to the type of complete dominance. In this case, the recessive gene in the heterozygote has no effect, and the phenotype of the heterozygote is determined solely by the dominant gene. However, a large number of traits in plants and animals are inherited according to the type of incomplete dominance. In this case, the F 1 hybrid does not completely reproduce the trait of one or the other parent. The expression of the trait is intermediate, with a greater or lesser deviation in one direction or the other.
An example of incomplete dominance can be the intermediate pink color of flowers in night beauty hybrids obtained by crossing plants with a dominant red and recessive white color (see diagram).
Scheme of incomplete dominance in the inheritance of flower color in the night beauty
As can be seen from the diagram, the law of uniformity of first-generation hybrids applies in crossing. All hybrids have the same color - pink - as a result of incomplete dominance of the gene A. In the second generation, different genotypes have the same frequency as in Mendel’s experiment, and only the phenotypic segregation formula changes. It coincides with the formula for segregation by genotype - 1: 2: 1, since each genotype has its own characteristic. This circumstance facilitates the analysis, since there is no need for analytical crossing.
There is another type of behavior of allelic genes in a heterozygote. It is called codominance and is described in the study of the inheritance of blood groups in humans and a number of domestic animals. In this case, a hybrid whose genotype contains both allelic genes exhibits both alternative traits equally. Codominance is observed when inheriting blood groups of the A, B, 0 system in humans. In people with a group AB(IV group) there are two different antigens in the blood, the synthesis of which is controlled by two allelic genes.
129. Analyze the pattern of inheritance of traits presented in the textbook during dihybrid crossing. Fill in the blanks in the proposed algorithm for solving the dihybrid crossing problem
1. Write down the object of study and the designation of genes
3. Let's create a Punnett lattice
130. Based on the analysis of the results obtained in task 129, answer the questions
1) How many types of gametes does a parent plant with smooth yellow seeds produce? - 2. With green wrinkled seeds? - 2
2) What is the probability (%) of producing F1 plants with yellow seeds as a result of the first cross? 50 . With green seeds? 50
3) What is the probability (%) of the appearance of F1 plants with yellow smooth seeds as a result of the first cross? 25. With yellow wrinkled ones? 25. With green smooth ones? 25. With green wrinkled ones? 25
4) How many different genotypes can there be among the first generation hybrids? 2
5) How many different phenotypes can there be among the first generation hybrids? 2
6) How many types of gametes does the F1 plant with yellow smooth seeds produce? 4
7) What is the probability (%) of the appearance of F2 plants with yellow seeds as a result of self-pollination? 50 . With green seeds? - 50
8) What is the probability (%) of the appearance of F2 plants with yellow smooth seeds as a result of crossing? 25. With yellow wrinkled ones? 25. With green smooth ones? 25. With green wrinkled ones? 25
9) How many different genotypes can there be among the second generation hybrids? - 9
10) How many different phenotypes can there be among second generation hybrids? - 4
131. Solve the problem
In humans, right-handedness dominates over left-handedness, and brown eye color dominates over blue. A brown-eyed, right-handed man, whose mother was a blue-eyed left-hander, and a blue-eyed, right-handed woman, whose father was left-handed, marry. 1) How many different phenotypes can their children have? 2) How many different genotypes can there be among their children? 3) What is the probability (%) that this couple will have a left-handed child?
A - right-handedness, a - left-handedness
B - brown eyes, B - blue eyes
The genotype of men is AaBv, women - Aavv
R ♀AaVv × ♂AaVv Gametes A, a, B, c A, a, B, c F1 AB: AB: AB: AB
132. Solve the problem
The black coat color and floppy ear in dogs are dominant over the brown coat and erect ear. Purebred black dogs with floppy ears were crossed with dogs with brown coat coloring and erect ears. Hybrids crossed with each other. 1) What proportion of F2 puppies should be phenotypically similar to the F1 hybrid? 2) What proportion of F2 hybrids should be completely homozygous? 3) What proportion of F2 puppies should have a genotype similar to that of F1 hybrids?
A - black, a - brown
B - hanging ear, B - standing ear
We will determine the genotypes of the parents, the types of gametes and write down the crossing scheme
R black, floppy ears
× brown, erect ears
Gametes A, B a,c F1 black, floppy ears AaVv
133. Solve the problem
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. 1) What is the probability (%) of getting a purebred Siamese kitten in F2? 2) What is the probability (%) of getting a kitten in F2 that is phenotypically similar to a Persian? 3) What is the probability (%) of getting a long-haired fawn kitten in F2?
A - black color, a - fawn
B - short hair, B - long hair
We will determine the genotypes of the parents, the types of gametes and write down the crossing scheme
134. Finish the sentence
Linked genes are genes that are located on the same chromosome and are inherited linked
135. Below are the discoveries that belong to G. Mendel and T. Morgan. Distribute the following discoveries according to their affiliation with scientists:
a) the independence of hereditary characteristics has been established
b) it has been established that chromosomes are carriers of hereditary traits
c) a statement on the linkage of genes in a chromosome was formulated
d) quantitative patterns of inheritance of traits have been identified
e) the nature of the manifestation of the symptom has been established
f) the mechanism for determining sex in animals has been established (males and females differ in the set of chromosomes)
Reginald Punnett (1875-1967) as a tool, which is a graphical notation for determining the compatibility of alleles from parental genotypes. Along one side of the square are female gametes, along the other - male ones. This makes it easier and more visual to present the genotypes obtained by crossing parental gametes.Discoveries of G. Mendel - a, b, d, d
T. Morgan's discoveries - c, e
Monohybrid cross
In this example, both organisms have the Bb genotype. They can produce gametes containing the B or b allele (the first means dominance, the second means recessive). The probability of a descendant with the BB genotype is 25%, Bb - 50%, bb - 25%.
| Maternal | |||
|---|---|---|---|
| B | b | ||
| Paternal | B | BB | Bb |
| b | Bb | bb | |
Phenotypes are obtained in a 3:1 combination. A classic example is the color of a rat's fur: for example, B - black fur, b - white. In this case, 75% of the offspring will have a black coat (BB or Bb), while only 25% will have a white coat (bb).
Dihybrid cross
The following example illustrates a dihybrid cross between heterozygous pea plants. A represents the dominant allele for shape (round peas), a represents the recessive allele (wrinkled peas). B represents the dominant allele for color (yellow peas), b is the recessive allele (green). If each plant has the genotype AaBb, then, since the alleles for shape and color are independent, there can be four types of gametes in all possible combinations: AB, Ab, aB and ab.
| AB | Ab | aB | ab | |
|---|---|---|---|---|
| AB | AABB | AABb | AaBB | AaBb |
| Ab | AABb | AAbb | AaBb | Aabb |
| aB | AaBB | AaBb | aaBB | aaBb |
| ab | AaBb | Aabb | aaBb | aabb |
Makes 9 round yellow peas, 3 round green peas, 3 wrinkled yellow peas, 1 wrinkled green pea. Phenotypes in a dihybrid cross are combined in a ratio of 9:3:3:1.
Purpose of work: development of skills to use the Punnett grid, determine the type of gametes and genotypes of offspring.
Equipment: cards with tasks for students, collections of problems on genetics for schoolchildren.
Progress
An exercise in using the Punnett grid to indicate gamete type and genotypes.
The Punnett grid looks like a two-dimensional table, where the gametes of one parent are written in the upper part, and the gametes of the second parent are written vertically in the left part. And in the cells of the table at the intersection of rows and columns, the genotypes of the offspring are recorded in the form of combinations of these gametes. Thus, it becomes very easy to determine the probabilities for each genotype in a given cross.
Solving monohybrid crossing problems.
To solve this type of problem, use the following solution algorithm:
1. Read the problem statement carefully, think about every word.
2. Write a short note of the task. Don't forget to put the appropriate symbols.
3. Write down the crossing scheme. Remember that the solution to the problem depends on correct recording.
4. Using the crossing scheme, you can immediately answer the question of the problem, knowing Mendel’s first and second laws. Do it. If this is not possible, crossbreed.
5. If you come to a conclusion based on Mendel’s laws, you must check your statement. Do the crossbreeding.
Problem 1. In a tomato, the smooth skin of the fruit dominates over the pubescent skin. A homozygous form with smooth fruits is crossed with a plant with pubescent fruits. In F1 we got 54 plants, in F2 - 736. Some questions: How many types of gametes can a plant with pubescent fruits produce? How many F1 plants can be homozygous? How many F2 plants can have smooth fruits? How many F2 plants can have pubescent fruits? How many different genotypes can be formed in F2?
A – smooth skin of the fruit, a – pubescent skin of the fruit
Solution: 1. Write down the crossing scheme. The problem says that a homozygous plant with smooth seeds is crossed, which means its genotype is AA, and that of a pubescent plant is aa.
3. We carry out a crossbreeding analysis. In F2, splitting occurred: by genotype – 1 (AA) : 2 (Aa) : 1 (aa); by phenotype 3 (yellow-seeded plants): 1 (green-seeded plants).
4. Answer the questions of the task.
1) Plants with pubescent fruits produce one type of gametes, since its genotype is homozygous for a recessive trait.
2) All F1 plants are heterozygous. Therefore, the number of homozygous plants with pubescent fruits in F1 is 0.
3) In F2 – 736 plants. Plants with smooth fruits have genotypes AA and Aa. They make up 3/4 of the total number of plants - 736: 4 * 3 = 552.
4) Plants with pubescent fruits make up? of the total number in F2, i.e. 736: 4 = 184.
5) In F2, segregation by genotype occurred in a ratio of 1: 2: 1, i.e. F2 has 3 different genotypes.
Answers: 1) 1; 20; 3) 552; 4) 184; 5) 3.
Problem 2. The black color of the bristles in pigs dominates over the red one. What kind of offspring should be expected from crossing a black pig with the FF genotype and a black boar with the Ff genotype?
F – black color of the bristles, f – red color of the bristles.
Answer: all offspring have black stubble.
Task 3. Normal hearing in humans is determined by the dominant S gene, and hereditary deafness is determined by the recessive s gene. The marriage of a deaf woman with a normal man gave birth to a deaf child. Determine the genotypes of the parents.
S – normal hearing, s – hereditary deafness.
P deaf x normal
The child has a recessive trait, which means his genotype is ss. In the child’s genotype, one allele came from the mother’s body, and the second from the father’s. The mother exhibited a recessive trait due to the condition. Therefore, her genotype is ss. The father has normal hearing, which means that one allele is dominant and the other is recessive, which he passed on to the child. If the father was homozygous for this trait, then the child would be born with normal hearing, but heterozygous (a carrier of the deafness gene).
Answer: the genotypes of the parents are ss and ss.
Problem 4. From crossing a polled bull of the Aishir breed with a horned cow, 18 calves (all polled) were obtained in F1, and all 95 calves in F2. What is the number of polled calves in F2?
Sign: presence of horns. D – polled, d – horned.
95 3/4 = 71.5 = 72 polled calves
Answer: 72 polled calves in F2.
3. Solving problems of dihybrid crossing.
To successfully solve dihybrid crossing problems, you must be able to:
a) record the genotypes of the parental individuals. Remember, they are written in four letters: two pairs of alleles (two pairs of traits).
b) record the gametes that the parent produces. Remember, gametes are written in two letters: one allele (gene) from each pair of traits.
For example, AABB; AaBB; aaBB; AaVv
Gametes AB AB aB aB AB aB Av aw
1. Carefully read the conditions of the task (maybe even several times), find out all the points that cause you doubts.
2. Designate pairs of genes with certain letters (you can take any letter of the Latin alphabet).
3. Briefly write down the condition of the problem.
4. Write down the crossing scheme.
5. If both crossed individuals produce more than one type of gamete, construct a Pönnett lattice or solve the problem algebraically.
6. Record the results of the cross.
Give the answer to the question posed in the problem. If necessary, analyze the results obtained.
Problem 1. “In a family of brown-eyed, right-handed parents, fraternal twins were born, one of whom is brown-eyed, left-handed, and the other blue-eyed, right-handed. What is the probability of the next child being born similar to his parents?”
The birth of a blue-eyed child to brown-eyed parents indicates the recessiveness of blue eye color, respectively, the birth of a left-handed child to right-handed parents indicates the recessivity of better control of the left hand compared to the right. Let's introduce allele designations: A - brown eyes, a - blue eyes, B - right-handed, c - left-handed. Let's determine the genotypes of parents and children:
R AaBv x AaBv
F1 A_bb, aaB_
A_вв is a phenotypic radical, which shows that this child is left-handed with brown eyes. The genotype of this child may be Aavv, AAvv. If we look for an answer to the problem, then we need a phenotypic radical - A_B_ - a brown-eyed, right-handed child.
9 variants of descendants that interest us are underlined.
Answer. There are 16 possible options, so the probability of having a child similar to their parents is 56.25%.
Problem 2. A woman with brown eyes and red hair married a man with non-red hair and blue eyes. It is known that the woman’s father had brown eyes, and her mother had blue eyes, and both had red hair. The man's father did not have red hair and blue eyes, his mother had brown eyes and red hair. What are the genotypes of all these people? What kind of eyes and hair might the children of these spouses have?
We will denote the allelic gene responsible for the manifestation of brown eye color as A (it is well known to everyone that brown eye color dominates over blue color), and the allelic gene for blue eyes, accordingly, will be a. The same letter of the alphabet is required, since this is one sign - eye color. We will denote the allelic gene for non-red hair (hair color is the second characteristic being studied) as B, since it dominates over the allele responsible for the manifestation of red hair color - b. We can first write down the genotype of a woman with brown eyes and red hair incompletely, but A-bb. But since it is said that her father was brown-eyed with red hair, that is, with the genotype A-bb, and her mother was blue-eyed and also with red hair (aabb), then the second allele of a woman with A could only be a, that is, her the genotype will be Aabb. The genotype of a blue-eyed man with non-red hair can first be written as follows: aaB-. But since his mother had red hair, that is, bb, then the second allelic gene for B in a man could only be b. Thus, the man’s genotype will be written aaBb. The genotypes of his parents: father - aaB-; mothers - A-bb.
Р ♀Ааbb × ♂aaBb
Children from the marriage of the analyzed spouses will have equally likely genotypes or phenotypes: brown-eyed not red-haired, brown-eyed red-haired, blue-eyed not red-haired, blue-eyed red-haired in a ratio of 1:1:1:1.
Problem 3. Parents with a loose earlobe and a triangular dimple on the chin gave birth to a child with a fused earlobe and a smooth chin. Determine the genotypes of the parents, the first child, phenotypes and genotypes of other possible offspring. Make a diagram for solving the problem. Traits are inherited independently.
A – free earlobe, and – fused earlobe;
B – triangular dimple on the beard, c – smooth chin.
This leaves 16 possible genotypes with which children would be born. The genotype of the born child - the one indicated in the problem will be 1/16 or 6.25%, its genotype (aabv), and the remaining 9/16 or 56.25% - free lobe and triangular fossa, 3/16 or 18.75 % - free lobe and smooth chin, 3/16 or 18.75% - fused lobe and triangular chin.
4. Solving problems on analyzing crossing.
To successfully solve this type of problem, you must remember: analyzing crossing is crossing an individual of an unknown genotype with an individual homozygous for recessive alleles (traits).
When solving problems, use the following solution algorithm:
Read the problem carefully and label the alleles (traits, genes, genotypes) with the appropriate letters.
2. Briefly write down the condition of the problem.
3. Make a guess about the genotype of an unknown individual using Mendel's laws.
4. Confirm your assumption with a crossing diagram.
5. Draw a conclusion based on the final result of the crossing.
Problem 1. “A rooster with a rose-shaped comb was crossed with two hens, also having a rose-shaped comb. The first gave 14 chickens, all with a rose-shaped comb, and the second gave 9 chickens, of which 7 with a rose-shaped and 2 with a leaf-shaped comb. The shape of the comb is monogenic autosomal trait. What are the genotypes of all three parents?"
Solution. Before determining the genotypes of the parents, it is necessary to find out the nature of inheritance of the comb shape in chickens. When a rooster was crossed with a second hen, 2 chicks with leaf combs were produced. This is possible if the parents are heterozygous; therefore, it can be assumed that the rose-shaped comb in chickens is dominant over the leaf-shaped one. Thus, the genotypes of the rooster and the second hen are Aa.
When crossing the same rooster with the first hen, no splitting was observed, therefore, the first hen was homozygous - AA.
Problem 2. The black color of the fur of minks dominates over the blue. How to prove the purebred nature of two black minks purchased by a fur farm?
A - black color, a - blue color
Genotype P-?
Solution: Both heterozygous and homozygous individuals can have black coloration. But purebred individuals are homozygous and do not produce segregation in the offspring. In order to determine the genotype of an individual with dominant traits, an analytical cross is carried out - crossed with an individual homozygous for recessive traits.
If the individual under study is homozygous (pure-bred) (AA), then the offspring from such a cross will have a black color and genotype Aa:
If the individual under study is heterozygous (Aa), then it produces two types of gametes and 50% of the offspring will have a black color and genotype Aa, and 50% will have a blue color and genotype aa.
Task 3. Parents have I and II blood groups. What blood types should we expect from our offspring? Let's write a crossing diagram.
Blood type is controlled by an autosomal gene. The locus of this gene is designated by the letter I, and its alleles are designated as A, B and 0. Moreover, alleles A and B are dominant, and allele 0 is recessive. Of the three alleles in the genotype, there can be only two.
1 var. P♀ I(A)I(A) x ♂ I(0)I(0)
2 var. P♀ I(A)I(0) x I(0)I(0)
G I(A), I(0) and I(0)
F1 I(A)I(0) and I(0)I(0)
When crossing, if one parent has the genotype I(A)I(A), all children will have only the second blood group. If one parent has I(A)I(0), two options are possible - I(A)I(A) - the second and I(0)I(0) - the first blood group.
Problem 4. In dogs, black coat color dominates over brown, and short hair dominates over long hair. A hunter has purchased a black dog with short hair and wants to make sure that he does not carry the brown and long hair alleles. Which female should be selected for crossing in order to check the genotype of the purchased dog?
Q – black coat color, q – brown coat color.
W – short wool, w – long wool.
P ♀ __ __ x ♂ Q_ W_
R♀ - ? F1 - ?
Solution. To determine the genotype of a dog, it is necessary to cross it with a female homozygous for recessive alleles, i.e. carry out an analytical cross. If the offspring produces puppies with brown and long hair, the genotype of the crossed dog will be heterozygous for both pairs of traits, or for one of the pairs of traits.
P ♀ qq ww x ♂ QqWw
G qw QW Qw qW qw
Answer. Phenotype splitting in F1:
P black shorthair = 0.25; 25%;
P black longhair =0.25; 25%;
P brown shorthair = 0.25; 25%;
P brown longhair = 0.25; 25%.
Another crossing option.
P ♀ qq ww x ♂ QqWW
P brown shorthair = 0.5; 50%.
This splitting of traits in puppies confirms that the dog being crossed was heterozygous in coat color.
Third crossing option.
P ♀ qq ww x ♂ QQWw
Answer: Phenotype splitting in F1:
P black shorthair =0.5; 50%;
P black longhair = 0.5; 50%.
This crossbreeding result confirms that the dog being crossed was heterozygous for coat length.
(1875-1967) as a tool, which is a graphical notation for determining the compatibility of alleles from parental genotypes. Along one side of the square are female gametes, along the other - male ones. This makes it easier and more visual to present the genotypes obtained by crossing parental gametes.
Phenotypes are obtained in a 3:1 combination. A classic example is the color of a rat's fur: for example, B - black fur, b - white. In this case, 75% of the offspring will have a black coat (BB or Bb), while only 25% will have a white coat (bb).
Dihybrid cross
The following example illustrates a dihybrid cross between heterozygous pea plants. A represents the dominant allele for shape (round peas), a represents the recessive allele (wrinkled peas). B represents the dominant allele for color (yellow peas), b is the recessive allele (green). If each plant has the genotype AaBb, then, since the alleles for shape and color are independent, there can be four types of gametes in all possible combinations: AB, Ab, aB and ab.
| AB | Ab | aB | ab | |
|---|---|---|---|---|
| AB | AABB | AABb | AaBB | AaBb |
| Ab | AABb | AAbb | AaBb | Aabb |
| aB | AaBB | AaBb | aaBB | aaBb |
| ab | AaBb | Aabb | aaBb | aabb |
Makes 9 round yellow peas, 3 round green peas, 3 wrinkled yellow peas, 1 wrinkled green pea. Phenotypes in a dihybrid cross are combined in a ratio of 9:3:3:1.
Tree method
There is an alternative, tree-based method, but it does not display gamete genotypes correctly:
It is advantageous to use when crossing homozygous organisms:
