Material on genetics for the exam. Problems for monohybrid and dihybrid crosses. Task design scheme
The sixth building of the Unified State Examination in Biology is tasks. For people who are just starting out in biology or exam preparation in particular, they are terrifying. Very in vain. One has only to figure out how everything will become simple and easy. 🙂
Refers to the basic level, with a correct answer, you can get 1 primary point.
To successfully complete this task, you should know the following topics given in the codifier:
Topics in the codifier for task No. 6
Genetics, its tasks. Heredity and variability are properties of organisms. Methods of genetics. Basic genetic concepts and symbolism. Chromosomal theory of heredity. Modern ideas about the gene and genome
Patterns of heredity, their cytological basis. Patterns of inheritance established by G. Mendel, their cytological foundations (mono- and dihybrid crossing). Laws of T. Morgan: linked inheritance of traits, violation of the linkage of genes. Sex genetics. Inheritance of sex-linked traits. Interaction of genes. The genotype as an integral system. Human genetics. Methods for studying human genetics. Solution of genetic problems. Drawing up cross-breeding schemes.
"I will solve the exam" divides tasks into two large groups: monohybrid crossing and dihybrid crossing.
Before solving problems, we suggest compiling a small glossary of terms and concepts in order to understand what is required of us.
Theory for crossbreeding tasks
There are two types of traits: recessive and dominant.
« Dominant trait overrides recessive trait' is a fixed phrase. What does it mean to suppress? This means that in the choice between a dominant and a recessive trait, the dominant one will necessarily appear. Anyway. A dominant trait is indicated by a capital letter, and a recessive trait is indicated by a small letter. Everything is logical. In order for a recessive trait to appear in the offspring, it is necessary that the gene carries a recessive trait from both the female and the male.
For clarity, let's imagine a sign, for example, the color of a kitten's coat. Suppose we have two options for the development of events:
- Black wool
- White wool
Black wool dominates over white. In general, tasks always indicate what dominates what, applicants are not required to know everything, especially from genetics.
Black wool would then be denoted by a capital letter. The most commonly used are A, B, C, and so on alphabetically. White wool, respectively, in small letters.
A black wool.
a white wool.
If the fusion of gametes results in combinations: AA, Aa, aA, then this means that the wool of the descendants of the first generation will be black.
If, when the gametes are fused, the combination aa is obtained, then the wool will be white.
About what gametes the parents have will be said in the condition of the problem.
Gametes, or sex cells, are reproductive cells that have a haploid (single) set of chromosomes and are involved, in particular, in sexual reproduction.
Zygote A diploid cell resulting from fertilization.
Heterozygous - two genes that determine one trait are the same (AA or aa)
Homozygous - two genes that determine one trait are different (Aa)
Dihybrid cross- crossing organisms that differ in two pairs of alternative traits.
monohybrid cross- crossing, in which the crossed organisms differ in only one trait.
Analyzing cross- crossing a hybrid individual with an individual homozygous for recessive alleles.
Gregor Mendel - "father" of genetics
So, how to distinguish between these types of crossing:
With monohybrid crossing, we are talking about one trait: color, size, shape.
In a dihybrid cross, we are talking about a pair of traits.
With analyzing crosses, one individual can be absolutely any, but the other gametes must carry exclusively recessive traits.
alleles- different forms of the same gene located in the same regions of homologous chromosomes.
It doesn't sound very clear. Let's figure it out:
1 gene carries 1 trait.
1 allele carries one trait value (it can be dominant or recessive).
Genotype is the totality of the genes of an organism.
Phenotype- a set of characteristics inherent in an individual at a certain stage of development.
Problems are often asked to indicate the percentage of individuals with a particular genotype or phenotype, or to indicate the splitting by genotype or phenotype. If we simplify the definition of the phenotype, then the phenotype is the external manifestation of traits from the genotype.
In addition to any concepts, you need to know the laws of Gregor Mendel - the father of genetics.
Gregor Mendel crossed peas with fruits that differed in color and smoothness of the skin. Thanks to his observations, three laws of genetics appeared:
I. The law of uniformity of hybrids of the first generation:
With monohybrid crossing of different homozygotes, all descendants of the first generation will be the same in phenotype.
II. splitting law
When crossing the offspring of the first generation, a splitting of 3:1 in phenotype and 1:2:1 in genotype is observed.
III. Law of independent splitting
When dihybrid crossing of two different homozygotes in the second generation, phenotypic splitting is observed in a ratio of 9:3:3:1.
When the skill of solving genetic problems is obtained, the question may arise: why should I know Mendel's laws, if I can perfectly solve the problem and find splitting in particular cases? Attention answer: in some tasks it may be necessary to indicate by what law splitting occurred, but this applies more to tasks with a detailed answer.
Having been savvy in theory, you can finally move on to tasks. 😉
Analysis of typical tasks No. 6 USE in biology
Types of gametes in an individual
How many types of gametes are formed in an individual with the aabb genotype?
We have two pairs of allelic chromosomes:
First couple: aa
Second pair: bb
These are all homozygotes. You can make only one combination: ab.
Types of gametes when crossing
How many types of gametes are formed in diheterozygous pea plants during dihybrid crossing (genes do not form a linkage group)? Write down a number for your answer.
Since the plants are diheterozygous, this means that, according to both traits, they have one allele dominant, and the second recessive.
We get the genotypes AaBb and AaBb.
Gametes in tasks are denoted by the letter G, moreover, without commas, in circles, gametes of one individual are indicated first, then a semicolon (;) is put, gametes of another individual are written, also in circles.
Crossing is indicated by an "x".
Let's write out the gametes, for this we will sort through all the combinations:
The gametes of the first and second individuals turned out to be the same, so their genotype was also the same. So, we got 4 different types of gametes:
Calculation of the proportion of diheterozygotes
When crossing individuals with AaBb genotypes with AaBb (genes are not linked), the proportion (%) of heterozygotes for both alleles (diheterozygotes) in the offspring will be ....
Let's create a Punnett lattice. To do this, we write the gametes of one individual in a column, the gametes of the other in a row, we get a table:
Let's find the diheterozygotes in the table:
Total zygotes: 16
Diheterozygotes: 4
Let's calculate the percentage: =
Application of Mendel's laws
The rule of uniformity of the first generation will appear if the genotype of one of the parents is aabb, and the other is
According to the rule of uniformity, monohybrid homozygotes should be crossed, one with a dominant trait, and the second with a recessive trait. Hence, the genotype of the other individual must be AABB.
Answer: AABB.
Phenotype ratio
The genotype of one of the parents will be AaBb if, during analyzing dihybrid crossing and independent inheritance of traits, splitting in the phenotype in the offspring is observed in the ratio. Write down the answer in the form of a sequence of numbers showing the ratio of the resulting phenotypes, in descending order.
Analyzing dihybrid cross, which means that the second individual has a recessive dihomozygote: aabb.
Here you can do without the Punnett lattice.
Generations are denoted by the letter F.
F1: AaBb; abb; aaBb; aabb
All four variants of phenotypes are different, so they relate to each other as 1:1:1:1.
Answer: 1111
Genotype ratio
What is the ratio of genotypes in offspring obtained from crossing individuals with AaBb x AABB genotypes?
AaBb x AABB
F1: AaBb; abb; aaBb; aabb
All 4 genotypes are different.
Inheritance of certain traits or diseases
What is the probability of the birth of healthy boys in a family where the mother is healthy and the father is sick with hypertrichosis, a disease caused by the presence of a gene linked to the Y chromosome?
If the trait is linked to the Y chromosome, then it does not affect the X chromosome.
The female sex is homozygous: XX, and the male is heterozygous XY.
Solving problems with sex chromosomes practically does not differ from solving problems with autosomes.
Let's make a gene and trait table, which should also be compiled for problems about autosomal chromosomes, if the traits are indicated and this is important.
The letter above the Y indicates that the gene is linked to that chromosome. Traits are dominant and recessive, they are indicated by capital and small letters, they can refer to both the H-chromosome and the Y-chromosome, depending on the task.
♀XX x XY a
F1: XX-girl, healthy
XY a - boy, sick
Boys born to this couple will be 100% sick, so 0% healthy.
Blood types
What blood type according to the ABO system does a person with the genotype I B I 0 have? Write down a number for your answer.
Let's use the table: 
In our genotype, agglutinogens B and 0 are recorded. This pair gives the third blood group.
Working with a schema
Based on the pedigree shown in the figure, determine the probability (in percent) of the birth of parents 1 and 2 of a child with a trait marked in black, with the complete dominance of this trait. Write your answer as a number.
So, we are learning to analyze such schemes.
We see that the sign is manifested in both men and women, which means that it is not sex-linked.
It manifests itself in every generation, which means it is dominant.
Since one of the children of the couple did not show the trait, then the parents are heterozygotes.
F1: AA- appears
Aa- shows up
Aa- shows up
aa - does not appear
3 - appears from 4
Basic terms of genetics
- Gene- This is a section of the DNA molecule that carries information about the primary structure of one protein. A gene is a structural and functional unit of heredity.
- Allelic genes (alleles)- different variants of the same gene encoding an alternative manifestation of the same trait. Alternative signs - signs that cannot be in the body at the same time.
- Homozygous organism- an organism that does not give splitting for one reason or another. Its allelic genes equally affect the development of this trait.
- heterozygous organism- an organism that gives splitting according to one or another feature. Its allelic genes affect the development of this trait in different ways.
- dominant gene is responsible for the development of a trait that manifests itself in a heterozygous organism.
- recessive gene is responsible for the trait, the development of which is suppressed by the dominant gene. A recessive trait appears in a homozygous organism containing two recessive genes.
- Genotype- a set of genes in the diploid set of an organism. The set of genes in a haploid set of chromosomes is called a genome.
- Phenotype- the totality of all the characteristics of an organism.
When solving problems in genetics, it is necessary:
- Determine the types of crossing and interactions of allelic and non-alelic genes (determine the nature of crossing).
- Determine the dominant and recessive trait(s) by the condition of the problem, drawing, scheme or by the results of crossing F 1 and F 2.
- Enter the letter designations of the dominant (capital letter) and recessive (capital letter) traits, if they are not given in the problem statement.
- Write down the phenotypes and genotypes of the parent forms.
- Write down the phenotypes and genotypes of the offspring.
- Draw up a crossbreeding scheme, be sure to indicate the gametes that form the parental forms.
- Write down the answer.
When solving problems on the interaction of non-allelic genes, it is necessary:
- Make a brief note of the task.
- Conduct an analysis of each feature separately, making an appropriate entry for each feature.
- Apply monohybrid cross formulas, if none of them are suitable, then
- Add the weight of the numbers in the offspring, divide the sum by 16, find one part and express all the numbers in parts.
- Based on the fact that the splitting in F 2 of dihybrid crossing goes according to the formula 9A_B_: 3A_bb: 3 aaB_: l aabb, find the Fr genotypes
- By F 2 find F genotypes
- F to find the genotypes of the parents.
Formulas for determining the nature of crossing:
where n is the number of alleles, pairs of traits
- Splitting by genotype - (3:1) n
- Splitting by phenotype - (1:2:1) n
- Number of gamete types - 2 n
- Number of phenotypic classes - 2 n
- Number of genotypic classes - 3 n
- The number of possible combinations, combinations of gametes - 4 n
Basic rules for solving genetic problems:
- If, when two phenotypically identical individuals are crossed, their offspring show splitting of traits, then these individuals are heterozygous.
- If, as a result of crossing individuals that differ phenotypically in one pair of traits, offspring are obtained in which splitting is observed in the same pair of traits, then one of the parental individuals was heterozygous, and the other was homozygous for a recessive trait.
- If, when crossing phenotypically identical (one pair of traits) individuals in the first generation of hybrids, the traits are split into three phenotypic groups in a 1:2:1 ratio, then this indicates incomplete dominance and that the parental individuals are heterozygous.
- If, when two phenotypically identical individuals are crossed, the characters in the offspring are split in a ratio of 9:3:3:1, then the original individuals were diheterozygous.
Having worked through these topics, you should be able to:
- Give definitions: gene, dominant trait; recessive trait; allele; homologous chromosomes; monohybrid crossing, crossing over, homozygous and heterozygous organism, independent distribution, complete and incomplete dominance, genotype, phenotype.
- Using the Punnett lattice, illustrate crossings for one or two traits and indicate what numerical ratios of genotypes and phenotypes should be expected in the offspring from these crossings.
- Outline the rules of inheritance, segregation, and independent distribution of traits, the discovery of which was Mendel's contribution to genetics.
- Explain how mutations can affect the protein encoded by a particular gene.
- Specify the possible genotypes of people with blood groups A; AT; AB; O.
- Give examples of polygenic traits.
- Indicate the chromosomal mechanism of sex determination and the types of inheritance of sex-linked genes in mammals, use this information in solving problems.
- Explain the difference between sex-linked and sex-dependent traits; give examples.
- Explain how human genetic diseases such as hemophilia, color blindness, sickle cell anemia are inherited.
- Name the features of plant and animal breeding methods.
- Indicate the main directions of biotechnology.
- To be able to solve the simplest genetic problems using this algorithm:
Problem solving algorithm
- Determine the dominant and recessive trait based on the results of crossing the first generation (F1) and the second (F2) (according to the condition of the problem). Enter the letter designations: A - dominant and - recessive.
- Write down the genotype of an individual with a recessive trait or an individual with a genotype known by the condition of the problem and gametes.
- Write down the genotype of F1 hybrids.
- Make a diagram of the second crossing. Write the gametes of the F1 hybrids in the Punnett grid horizontally and vertically.
- Write down the genotypes of the offspring in the gamete crossing cells. Determine the ratio of phenotypes in F1.
Task design scheme.
Letter designations:
a) dominant trait _______________
b) recessive trait _______________
Gametes
F1(first generation genotype)
| gametes | ||
| ? | ? |
Punnett lattice
| F2 |
|
Phenotype ratio in F2: _____________________________
Answer:_________________________
Examples of solving problems for monohybrid crossing.
Task."There are two children in the Ivanov family: a brown-eyed daughter and a blue-eyed son. The mother of these children is blue-eyed, but her parents had brown eyes. How is eye color inherited in humans? What are the genotypes of all family members? Eye color is a monogenic autosomal trait."
The eye color trait is controlled by one gene (by condition). The mother of these children is blue-eyed, and her parents had brown eyes. This is possible only in the THAT case if both parents were heterozygous, therefore, brown eyes dominate over blue ones. Thus, grandmother, grandfather, father and daughter had the genotype (Aa), and mother and son - aa.
Task."A rooster with a pink comb is crossed with two hens that also have a pink comb. The first gave 14 chickens, all with a pink comb, and the second - 9 chickens, of which 7 with a pink comb and 2 with a leaf comb. The shape of the comb is a monogenic autosomal trait. What are genotypes of all three parents?
Before determining the genotypes of the parents, it is necessary to find out the nature of the inheritance of the comb shape in chickens. When a rooster was crossed with a second hen, 2 chickens with a leaf-shaped comb appeared. This is possible when the parents are heterozygous, therefore, it can be assumed that the pink-shaped comb in chickens dominates over the leaf-shaped one. Thus, the genotypes of the rooster and the second hen are Aa.
When the same rooster was crossed with the first hen, no splitting was observed, therefore, the first hen was homozygous - AA.
Task."In the family of brown-eyed right-handed parents, fraternal twins were born, one of which is brown-eyed left-handed, and the other blue-eyed right-handed. What is the probability of the birth of the next child, similar to their parents?"
The birth of a blue-eyed child in brown-eyed parents indicates the recessiveness of the blue color of the eyes, respectively, the birth of a left-handed child in right-handed parents indicates the recessiveness of the better possession 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 | AaVv x AaVv |
| F, | A_vv, aaB_ |
A_vv - phenotypic radical, which shows that this child is left-handed with brown eyes. The genotype of this child can be - Aavv, AAvv.
Further solution of this problem is carried out in the traditional way, by constructing the Punnett lattice.
| AB | Av | aB | Av | |
| AB | AABB | AAVv | AaBB | AaVv |
| Av | AAVv | AAvv | AaVv | aww |
| aB | AaBB | AaVv | aaBB | AaVv |
| aw | AaVv | aww | aawww | aww |
Underlined are 9 variants of descendants that we are interested in. There are 16 possible options, so the probability of having a child similar to their parents is 9/16.
Ivanova T.V., Kalinova G.S., Myagkova A.N. "General Biology". Moscow, "Enlightenment", 2000
- Topic 10. "Monohybrid and dihybrid crossing." §23-24 pp. 63-67
- Topic 11. "Genetics of sex." §28-29 pp. 71-85
- Topic 12. "Mutational and modification variability." §30-31 pp. 85-90
- Topic 13. "Selection." §32-34 pp. 90-97
Municipal State Educational Institution Lyceum No. 4
Rossosh Rossoshsky municipal district of the Voronezh region.
Methodological development in biology to help students passing the exam.
"Tasks in genetics" for grade 11
biology teacher of the highest qualification category
2016
I. Tasks for monohybrid crossing (complete and incomplete dominance)
1. In rabbits, gray coat color dominates over black. A homozygous gray rabbit was crossed with a black rabbit. Determine the phenotypes and genotypes of rabbits?
2. In guinea pigs, black coat color is dominant over white. We crossed two heterozygous males and a female. What will be the first generation hybrids? What law is manifested in this inheritance? 
3. When crossing two white pumpkins in the first generation, ¾ of the plants were white and ¼ were yellow. What are the genotypes of the parents if white is dominant over yellow? 
4. Black cow Nochka brought a red calf. A black calf was born to the red cow Zorka. These cows are from the same herd with one bull. What are the genotypes of all animals? Consider different options. ( The black gene is dominant.)
5. How many dwarf pea plants can be expected when sowing 1200 seeds obtained by self-pollination of tall heterozygous pea plants? (Seed germination is 80%).
6. Watermelon fruit may be green or striped. All watermelons obtained from crossing plants with green and striped fruits had only the green color of the fruit rind. What color of watermelon fruits can be in F 2?
7. In snapdragons, plants with wide leaves, when crossed with each other, always give offspring with the same leaves, and when an narrow-leaved plant is crossed with a broad-leaved plant, plants with leaves of intermediate width appear. What will be the offspring from crossing two individuals with leaves of intermediate width. 
1. In humans, glaucoma is inherited as an autosomal recessive trait (a), and Marfan syndrome, accompanied by an anomaly in the development of connective tissue, is inherited as an autosomal dominant trait (B). The genes are located in different pairs of autosomes. One of the spouses suffers from glaucoma and had no ancestors with Marfan's syndrome, and the second is diheterozygous according to these characteristics. Determine the genotypes of the parents, the possible genotypes and phenotypes of the children, the probability of having a healthy child. Make a scheme for solving the problem. What law of heredity is manifested in this case?
2. We crossed undersized (dwarf) tomato plants with ribbed fruits and plants of normal height with smooth fruits. In the offspring, two phenotypic groups of plants were obtained: undersized with smooth fruits and normal height with smooth fruits. When crossing undersized tomato plants with ribbed fruits with plants having a normal stem height and ribbed fruits, all offspring had a normal stem height and ribbed fruits. Create crossover patterns. Determine the genotypes of parents and offspring of tomato plants in two crosses. What law of heredity is manifested in this case? 
3. When crossing horned red cows with polled black bulls, calves of two phenotypic groups were born: horned black and polled black. With further crossing of these same horned red cows with other polled black bulls, the offspring were individuals of polled red and polled black. Write schemes for solving the problem. Determine the genotypes of the parents and offspring in two crosses. What law of heredity is manifested in this case? 
4. Tomato fruits can be red and yellow, naked and pubescent. It is known that the genes for yellow color and pubescence are recessive. Of the tomatoes harvested on the collective farm, there were 36 tons of red hairless and 12 tons of red hairy ones. How many (approximately) yellow fluffy tomatoes can be in a collective farm crop if the source material was heterozygous?
5. When crossing a motley crested (B) hen with the same rooster, eight chickens were obtained: four motley crested chickens, two white (a) crested and two black crested. Make a scheme for solving the problem. Determine the genotypes of parents, offspring, explain the nature of the inheritance of traits and the appearance of individuals with variegated color. What laws of heredity are manifested in this case? 
6. In mice, black coat color is dominant over brown (a). The long tail is determined by the dominant gene (B) and develops only in the homozygous state, in heterozygotes a short tail develops. Recessive genes that determine the length of the tail in the homozygous state cause the death of embryos. The genes of the two traits are not linked. When a female mouse with black hair and a short tail was crossed with a male with brown hair and a short tail, the following offspring were obtained: black mice with a long tail, black mice with a short tail, brown mice with a long tail, and brown mice with a short tail. Make a scheme for solving the problem. Determine the genotypes of parents and offspring, the ratio of phenotypes and genotypes of offspring, the probability of death of the embryos. What law of heredity is manifested in this case? Justify the answer. 
7. A strawberry plant was crossed with mustachioed white-fruited plants with beardless red-fruited plants (B). All hybrids turned out to be mustachioed pink-fruited. When analyzing the crossing of F 1 hybrids, phenotypic splitting occurred in the offspring. Make a scheme for solving the problem. Determine the genotypes of the parental individuals, first-generation hybrids, as well as the genotypes and phenotypes of the offspring in analyzing crosses (F 2). Determine the nature of the inheritance of the color trait of the fetus. What laws of heredity are manifested in these cases? 
8. When crossing phlox plants with white flowers and a funnel-shaped corolla with a plant with cream flowers and flat corollas, 78 descendants were obtained, among which 38 form white flowers with flat corollas, and 40 - cream flowers with flat corollas. When crossing phloxes with white flowers and funnel-shaped corollas with a plant with cream flowers and flat corollas, phloxes of two phenotypic groups were obtained: white with funnel-shaped corollas and white with flat corollas. Draw two crosses. Determine the genotypes of the parents and offspring in two crosses. What law of heredity is manifested in this case?
9. There are two types of hereditary blindness, each of which is determined by recessive alleles of genes (a or b). Both alleles are on different pairs of homologous chromosomes. What is the probability of the birth of a blind grandson in a family in which maternal and paternal grandmothers are dihomozygous and suffer from various types of blindness, and both grandfathers see well (do not have recessive genes). Make a scheme for solving the problem. Determine the genotypes and phenotypes of grandparents, their children and possible grandchildren. 
III. Tasks for the inheritance of blood groups of the AB0 system
The boy has group I, his sister has group IV. What are the genotypes of the parents?
The father has IV blood type, the mother has I. Can a child inherit the blood type of his mother?
Two children were mixed up in the maternity hospital. The first pair of parents has I and II blood groups, the second pair - II and IV. One child has the II group, and the second - the I group. Identify the parents of both children.
Is it possible to transfuse blood to a child from a mother if her blood type is AB, and her father has 00? Explain the answer.
5. Blood group and Rh factor - autosomal unlinked traits. The blood group is controlled by three alleles of one gene - i 0, I A, I B. Alleles I A and I B are dominant with respect to the allele i 0. The first group (0) is determined by recessive genes i 0, the second group (A) is determined by the dominant allele I A, the third group (B) is determined by the dominant allele I B, and the fourth (AB) is determined by two dominant allele I A I B . The positive Rh factor R dominates the negative r. The father has the third blood type and positive Rh (diheterozygote), the mother has the second group and positive Rh (dihomozygous). Determine the genotypes of the parents. What blood type and Rh factor can children in this family have, what are their possible genotypes and phenotype ratio? Make a scheme for solving the problem. What law of heredity is manifested in this case?
IV
1. In canaries, the presence of a crest is a dominant autosomal trait (A); the sex-linked gene X B determines the green color of the plumage, and X b - brown. Birds are homogametic male and heterogametic female. A crested green female was crossed with a male without a crest and green plumage (heterozygous). The offspring included chicks crested green, without crest green, crested brown and without crest brown. Make a scheme for solving the problem. Determine the genotypes of parents and offspring, their gender. What laws of heredity are manifested in this case?
2. The body color of Drosophila is determined by an autosomal gene. The eye color gene is located on the X chromosome. Drosophila is heterogametic in males. A female with a gray body and red eyes was crossed with a male with a black body and white eyes. All offspring had a gray body and red eyes. The males obtained in F1 were crossed with the parent female. Make a scheme for solving the problem. Determine the genotypes of the parents and females F 1, the genotypes and phenotypes of the offspring in F 2 . What proportion of females from the total number of offspring in the second crossing is phenotypically similar to the parental female? Specify their genotypes. 
3. In humans, the gene for brown eyes dominates over blue eyes (A), and the gene for color blindness is recessive (color blindness - d) and is linked to the X chromosome. A brown-eyed woman with normal vision, whose father had blue eyes and suffered from color blindness, marries a blue-eyed man with normal vision. Make a scheme for solving the problem. Determine the genotypes of the parents and possible offspring, the probability of having color-blind children with brown eyes in this family, and their gender. Explain the answer.
4. In humans, the inheritance of albinism is not sex-linked (A is the presence of melanin in skin cells, a is albinism), and hemophilia is sex-linked (X n is normal blood clotting, X h is hemophilia). Determine the genotypes of the parents, as well as the possible genotypes, sex and phenotypes of children from the marriage of a diheterozygous woman for both alleles and an albino man with normal blood clotting. Make a scheme for solving the problem. Explain the answer.
1. When corn plants with smooth colored grains were crossed with a plant producing wrinkled uncolored seeds, in the first generation all plants produced smooth colored grains. When analyzing crosses of hybrids from F 1, there were four phenotypic groups in the offspring: 1200 smooth colored, 1215 wrinkled uncolored, 309 smooth uncolored, 315 wrinkled colored. Make a scheme for solving the problem. Determine the genotypes of the parents and offspring in two crosses. Explain the formation of four phenotypic groups in the second cross.
2. When crossing a diheterozygous corn plant with a colored seed and a starchy endosperm and a plant with an uncolored seed and a waxy endosperm, the offspring obtained splitting by phenotype: 9 plants with a colored seed and a starchy endosperm; 42 - with colored seed and waxy endosperm; 44 - with uncolored seed and starchy endosperm; 10 - with uncolored seed and waxy endosperm. Make a scheme for solving the problem. Determine the genotypes of the original individuals, the genotypes of the offspring. Explain the formation of four phenotypic groups. 
3. A diheterozygous pea plant with smooth seeds and tendrils was crossed with a plant with wrinkled seeds without tendrils. It is known that both dominant genes (smooth seeds and the presence of antennae) are localized on the same chromosome; crossing over does not occur. Make a scheme for solving the problem. Determine the genotypes of the parents, the phenotypes and genotypes of the offspring, the ratio of individuals with different genotypes and phenotypes. What law is manifested in this case? 
VI. Pedigree tasks
1. Based on the pedigree shown in the figure, determine and explain the nature of the inheritance of the trait highlighted in black (dominant or recessive, sex-linked or not. Determine the genotypes of the offspring 1,3,4,5,6,7. Determine the probability of birth from parents 1 ,2 next child with a trait highlighted in black on the pedigree. 
2. According to the pedigree shown in the figure, determine and justify the genotypes of parents, offspring, indicated in the diagram by the numbers 1,6,7. Set the probability of giving birth to a child with an inherited trait in woman number 6, if this trait has never manifested itself in her husband's family.
Based on the pedigree shown in the figure, determine and explain the nature of the inheritance of the trait highlighted in black. Determine the genotypes of parents, offspring1,6 and explain the formation of their genotypes.
According to the pedigree shown in the figure, determine the nature of the inheritance of the trait (dominant or recessive, linked or not linked to sex), highlighted in black, the genotypes of parents and children in the first generation. Indicate which of them is a carrier of the gene, the trait of which is highlighted in black.
Tasks for the interaction of genes
1. complementarity
In parrots, feather color is determined by two pairs of genes. The combination of two dominant genes determines the green color. Individuals recessive for both pairs of genes are white. The combination of the dominant gene A and the recessive gene b determines the yellow color, and the combination of the recessive gene a with the dominant gene B determines the blue color. 
When two green individuals were crossed, they received parrots of all colors. Determine the genotypes of parents and offspring.
2. epistasis
The coat color of rabbits (as opposed to albinism) is determined by the dominant gene. The color of the color is controlled by another gene located on a different chromosome, and the gray color dominates over black (in albino rabbits, color color genes do not show themselves). 
What characteristics will the hybrid forms obtained from crossing a gray rabbit born from an albino rabbit with an albino carrying the black color gene have?
3. pleiotropy.
One of the breeds of chickens is distinguished by shortened legs. This feature is dominant. The gene controlling it simultaneously causes the shortening of the beak. At the same time, in homozygous chickens, the beak is so small that they are not able to break through the egg shell and die without hatching from the egg. 3,000 chicks have been produced in the incubator of a farm that breeds only short-legged hens. How many of them are short-legged?
4. polymer
The son of a white woman and a black man married a white woman. Can this couple have a child darker than their father?
ANSWERS:
I. Tasks for monohybrid crossing (complete and incomplete dominance)
1. A- gray color
a- black color R: AA and aa.
F 1: Ah, all grey.
2. A - black flower a - white color
R: Aa and aa.
F 1: AA 2Aa, aa. splitting law.
3. A - white color
a- yellow
Both parents have the Aa genotype.
F 1 splitting by genotype 1aa 2Aa 1 aa
4. Night-Aa, her calf-aa. Dawn - aa, her calf - Aa, bull - Aa.
5. A- high
a-dwarf
F 1 1AA 2Aa 1aa
240 dwarf plants.
6.A- green color
a striped color
F2 1AA 2Aa 1aa
F 2 - splitting 3 green: 1 striped
7. A- wide leaves
a narrow leaves
Aa - intermediate leaf width
F 1: splitting by phenotype: 1 - wide leaves, 2 - intermediate leaf width, 1 - narrow leaves.
Genotype: AA 2Aa aa
II. Tasks for dihybrid crossing1. A - normal, a - glaucoma.
B - Marfan's syndrome, b - normal.
One of the spouses suffers from glaucoma and had no ancestors with Marfan syndrome: aabb. The second spouse is diheterozygous: AaBb.
| norm. | norm | glaucoma | glaucoma |
Probability of having a healthy baby (normal/normal) = 1/4 (25%). In this case, the third law of Mendel (the law of independent inheritance) is manifested.
a- dwarfism
B - smooth
in-ribbed
first cross - P: aabb and AaBB, got F 1 - aaBb and AaBb
second - P: aabb and AAbb, got F 1 - Aabb.
4.R-AaBv and aavv.F1: 9 red heads. 3 red. Op., 3 females, 1 female op.4 tons op.
5. In this case, intermediate inheritance of color appears. AA - black, Aa - motley, aa - white. the parents of both the hen and the rooster have AaBB genotypes. And the gametes form the same: AB, aB. when they merge, the formation of genotypes occurs -AABB - black crested, AaBB - mottled crested, aaBB - white crested. ratio -1/2/1.
color gene:
A - black
a- brown
tail length gene:
B - long
c- short
vv - lethal
Вв - shortened
Decision:
1) AaBv x aaBv
black short x brown short
gametes - AB, AB, AB, AB AB AB
AaBB AABB AABB AABB AABB AABB AABB
h.d.h.u. k.d. to y. h. y letal k. y lethal
black with a long tail - 1/8
black with shortened - 2/8
brown with a long tail - 1/8
brown with shortened - 2/
lethal - 2/8. Law of Independent Succession
7. A- mustachioed
a- beardless
B - red
BB- pink
1) first crossing:
R AAvv * aaVV
mustache white used cr.
2) analyzing cross:
Aaaa * aaa
G AB AB AB AB AB
F 2 AaBb - mustachioed pink-fruited; Aabb - mustachioed white-fruited;
aaBb - beardless rose-fruited; aabb - beardless white-fruited;
3) the nature of the inheritance of the color trait of the fetus - incomplete dominance. In the first crossing - the law of uniformity of hybrids, in the second (analyzing) - independent inheritance of traits.
8. A - flat beaters,
a - funnel-shaped corollas.
B - white flowers,
b - cream flowers
first cross:
R aaVv x Aavv
G ab ab av av
F 1 AaBb aavb
second cross:
R aaBB x Aawb
G aB Av av
F 1 AaBb aaBv
In this case, Medel's third law is manifested - the law of independent inheritance.
9. A - normal, a - blindness No. 1.
B - normal, b - blindness No. 2.
Maternal grandmother AAbb, paternal grandmother aaBB. Grandpas - AABB.
| |
Probability of having a blind grandchild 0%
Tasks for the inheritance of blood groups of the AB0 system
1. Boy-j0j0. Sister - JAJB
P J A j 0 and J A J B
2. Father - J A J B
Mother-j 0 j 0.
No, because children can have either the 2nd or 3rd blood group.
3. first pair of parents:
P: j 0 j 0 x J A j 0 or j 0 j 0 x J A J A
G: j 0 J A , j 0 j0 J A
F: J A j0 (2 gr) , j 0 j 0 (1 gr.) or J A j 0 (2 gr)
second pair of parents
P: J A J A x J A J B or J A j 0 x J A J B
G: J A ; J A , J B J A j 0 J A , J B
F: J A J A (2) J A J B (4) J A J A (2) J A J B (4) J A j 0(2) J in j 0(3 gr.)
In the first pair of parents, the son has 1 gr. and he received gene 0 from both parents. The second couple are the parents of a boy with the 2nd blood group.
This problem can be solved orally, because a child with 1 gr. blood cannot be born to a couple in which there is a person with 4 blood group
4. It is impossible, because in children, blood types are possible: A0 (II) or B0 (III), therefore, the blood of the fourth group, which the mother has, cannot be transfused.
5. Digeterozygous father I B i 0 Rr, dihomozygous mother I A I A RR.
| IV group | IV group | II group | II group |
Children in this family may have IV or II blood group, all Rh-positive. The proportion of children with IV blood group is 2/4 (50%). The law of independent inheritance appears (Mendel's third law).
IV. Sex-linked and autosomal inheritance tasks
1. A - the presence of a crest, a - no crest.
X B - green plumage, X b - brown plumage.
A_X B Y-crested green female
aaX B X b - male without crest with green plumage (heterozygous)
Among the offspring were chicks without crest - aa. They got one gene a from their mother, one from their father. Therefore, the mother must have the gene a, hence the mother of Aa.
P AaX B Y x aaX B X b
| AaX V X V | aaX V X V | AaX B Y | aaX B Y |
|
| AaX B Xb | aaX B X b | AaX b Y | aaX b Y |
In this case, the law of independent inheritance (Mendel's third law) and sex-linked inheritance appeared.
2. A - gray body
a black body
X B red eyes
X in - white eyes
R 1 AAX B X B * aaX in Y
gray body black body
red eyes white ch.
G AX B aX to aY
F 1 AaX B X b AaX B Y
P 2 AAX B X B * AaX B Y
G AX B AX B aX B AY aY
F 2 AAX B X B AaX B X B AAX B Y AaX B Y
F 2 all offspring have a gray body and red eyes.
Sex Ratio-50% : 50:%
3. A - brown eyes,
and blue eyes.
X D - normal vision,
X d - color blindness.
A_X D X _ brown-eyed woman with normal vision
aaX d Y is the woman's father, he could give his daughter only aX d, therefore, the brown-eyed woman is AaX D X d.
AaX D Y. - woman's husband
P AaX D X d x aaX D Y
G AX D AX d aX D ax d aX D aY
F 1 AAX D X D AaX D X d aaX D X D aaX D X d AAX D Y Aa X d Y aaX D XY aaX d Y
The probability of having a color-blind child with brown eyes is 1/8, (12.5%), it is a boy.
4. A - normal, a - albinism.
X H - normal, X h - hemophilia.
Woman AaX H X h , man aaX H Y
G AX H AX h aX H aX h aX H aY
F1 AaX H X H AaX H Y AaX H X h AaX h Y aaX H X H aaX H Y aaX H X h aaX h Y
Ph.D. Ph.D. Ph.D. c.g. g.n. g.n. g.n. y..y.
Splitting by eye color - 1:1 by blood clotting - all daughters are healthy, boys - 1:1.
V. Problems on linked inheritance1 . A - smooth grains,
a - wrinkled grains.
B - colored grains,
b - uncolored grains.
R AABB x AABB
Since uniformity was obtained in the first generation (Mendel's first law), therefore, homozygotes were crossed, in F1 a diheterozygote was obtained, carrying dominant traits.
Analyzing cross:
| normal gametes | |||||||
| recombinant gametes | |||||||
| smooth | wrinkled. | smooth | wrinkled. |
Since in the second generation an unequal number of phenotypic groups turned out, therefore, linked inheritance took place. Those phenotypic groups that are presented in large numbers are not crossovers, but groups that are presented in small numbers are crossovers formed from recombinant gametes in which linkage was broken due to crossing over in meiosis.
2. A- colored seed a- not colored seed B- starchy endosperm b- waxy endosperm seed, starch endosperm 42- Aavv- okr. seed, wax endosperm 44- aaBv- unstained seed, starchy endosperm 10- aavb- unstained seed waxy endosperm The presence in the offspring of two groups (42 - with colored waxy endosperm; 44 - with unstained waxy endosperm) in approximately equal proportions - the result of linked inheritance of alleles A and B , a and B among themselves. The other two phenotypic groups are formed as a result of crossing over.
A - smooth seeds,
a - wrinkled seeds
B - the presence of antennae,
b - without antennae
| smooth | wrinkled. |
||
If crossing over does not occur, then only two types of gametes are formed in the diheterozygous parent (full linkage).
- A gray body
F 1 AaBv all gray with normal wings. law of uniformity)
P AaVv x AaVv
T.K. there is no expected Mendelian splitting, which means that a crossing-over has occurred:
linked
F 2 AABB AABB AABB AABB