How to solve a biology task. Solving genetic problems

with examples

explanations and solutions

to solve problems in genetics, it is necessary:

introduce symbols for signs. The dominant gene denotes capital letter, and the recessive gene is capitalized;

find out what kind of crossing in this case takes place (mono-, di- or polyhybrid);

write down the solution to the problem schematically and draw a conclusion about the probability of the requested event in% or fractions of one.

Monohybrid crossing.

One pair of alternative (mutually exclusive) traits is taken into account, therefore, one pair of allelic genes.

Problem number 1

In tomatoes, the gene for the red color of the fruit dominates the gene for the yellow color. What color will be the fruits of plants obtained from crossing heterozygous red-fruited plants with yellow-fruited plants? What are their genotypes?

Solution to problem number 1

The gene responsible for the red color of the fruits is dominant, we denote A, the gene for the yellow color of the fruits is recessive, we denote - a. The problem of monohybrid crossing, since in the condition of the plant, they differ in 1 pair of mutually exclusive signs (fruit color). One of the parental individuals is heterozygous, therefore, this individual carries the genes of one allelic pair in a different state, one gene is dominant, the other is recessive (Aa). Such an individual produces two types of gametes (A, a). The second individual is yellow-fruited, both genes are in the same state (aa), which means that the individual is homozygous and gives one type of gametes (a).

Knowing the genotypes of the parents, we will write down the solution to the problem and answer the question posed.

A - red fruits

a - yellow fruits

P - parents

D - gametes

F 1 - first generation

x - cross sign

P Aa x aa

red yellow

Answer: F 1 hybrids show phenotypic cleavage in a ratio of 1: 1 - 50% red-fruited, 50% yellow-fruited tomatoes. By genotype, splitting in a 1: 1 ratio - 50% heterozygous individuals (Aa), 50% homozygous individuals (aa).

Tasks for an independent solution.

3Assack number 2.

In Drosophila, gray body color dominates over black. From the crossing of flies with a gray body and flies with a black body, hybrids F 1 were obtained, which subsequently, when crossed among themselves, gave 192 individuals of the next generation.

1. How many gamete types are formed in the F I hybrid?

2. How many different phenotypes are there among the offspring of F 2?

3. How many different genotypes are there among the offspring of F 2?

4. How many homozygous gray flies are in F 2 (theoretically)?

5. How many blacks are there in F 2 (theoretically)?

Problem number 3.

In humans, the gene for long eyelashes dominates over the gene for short eyelashes. A woman with long eyelashes, whose father had short eyelashes, marries a man with short eyelashes.

1. How many types of gametes do a man have?

2. How many types of gametes does a woman have?

3. How many different genotypes can there be among the children of this married couple?

4. What is the probability that a child in this family will be born with long eyelashes?

5. What is the probability that a child in this family will be born with short eyelashes?

Problem number 4.

In humans, the gene for the early development of hypertension dominates over the gene that determines the normal development of the trait. In the family, both spouses suffer from hypertension with an early onset, their only daughter has a normal blood pressure... She is married and has two children. One of the daughter's children has normal blood pressure, while the other developed hypertension early.

1. How many different genotypes can there be among the children of the above spouses?

2. How many types of gametes does a daughter develop?

3. How many types of gametes does the husband of the daughter develop?

4. What is the probability that the daughter of these spouses will have a child with hypertension?

5. How many different genotypes can there be among grandchildren from a daughter?

Toolkit

in biology

Solving genetic problems

(for students in grades 9-11)

Biology teacher , chemistry

I. Explanatory note

II. Terminology

IV Examples of solving genetic problems

Monohybrid crossing

Dihybrid crossing

Vi. Genetic tasks

Explanatory note

The "Genetics" section of the school biology course is one of the most difficult for students to understand. Knowledge of the terminology of modern genetics, as well as solving problems of different levels of complexity, can facilitate the assimilation of this section.

At the moment, most of the textbooks used to study the sections of genetics in high school general education schools contain little training tasks on genetics. As a rule, they are not enough for the successful development of skills in solving genetic problems for mono-, di- and sex-linked inheritance of traits.

Solving genetic problems develops in schoolchildren logical thinking and allows them to understand deeper educational material, enables teachers to effectively monitor the level of student achievement.

The manual provides the basic terminology necessary for understanding and successfully solving genetic problems, generally accepted conventions, and also provides approximate algorithms for solving problems on different types inheritance.

For each task, the approximate number of points that a student can earn in case of successful completion of the task is given. It will also help to control the knowledge of students with at different levels readiness, that is, to differentiate the assessment of students' knowledge.

This teaching aid compiled to help biology teachers, high school students and applicants.

Terminology

Alternative signs - mutually exclusive, contrasting

Analyzing cross – crossbreedingthe individual whose genotype is neededestablish with an individual homozygous for recesssive gene;

Autosome- not related to sex chromosomes in diploid cells. In humans, the diploid chromosome set (karyotype) is represented by 22 pairs chromosomes(autosomes) and in one pair genitalchromosomes(gonos).

Mendel's second law (splitting rule)- when crossing two descendants (hybrids) of the first generation with each other, splitting is observed in the second generation and individuals with recessive traits appear again; these individuals make up one fourth ofallthe number of offspring of the second generation (splitting by genotype 1: 2: 1, by phenotype 3: 1);

Gamete - germ cell of a plant or animal organism carrying one gene from an allelic pair

Gene- a section of a DNA molecule (in some cases, RNA), in which information about the biosynthesis of one polypeptide chain with a specific amino acid sequence is encoded;

Genome- a set of genes contained in a haploid set of chromosomes of organisms of one biological species;

Genotype - localized in haploid set of chromosomes this organism... Unlike the concepts of genome and gene pool, it characterizes an individual, not a species (yet the difference between genotype and genome is the inclusion of non-coding sequences in the concept of "genome" that are not included in the concept of "genotype"). Together with environmental factors, it determines the phenotype of the organism;

Heterozygous organisms- organisms containing various allelic genes;

Homozygous organisms- organisms containing two identical allelic genes;

Homologous chromosomes- paired chromosomes, identical in shape, size and set of genes;

Dihybrid crossing -crossing of organisms that differ in two ways;

Morgan's Law -genes located on the same chromosome, during meiosis, fall into one gamete, that is, are inherited linked;

The law of gamete purity -during the formation of gametes, only one of the two allelic genes gets into each of them, it is called the law of gamete purity

Karyotype- a set of features (number, size, shape, etc.) complete set chromosomes, inherent in cells of a given biological species (species karyotype), a given organism (individual karyotype) or line (clone) of cells. The visual representation of a complete chromosome set (karyogram) is sometimes also called a karyotype.

Codominating -the type of interaction of allelic genes, in which in the offspringsigns of the genes of both parents appear;

TOcomplementary, or additional, interaction of genes - as a result of which new signs appear;

Locus - the part of the chromosome in which the gene is located;

Monohybrid crossing -crossing of organisms that differ in one trait (only one trait is taken into account);

Incomplete dominance -incomplete suppression by the dominant gene of the recessive allelic pair. In this case, intermediate traits appear, and the trait in homozygous individuals will not be the same as in heterozygous ones;

Mendel's first law (law uniformity of first generation hybrids)- when crossing two homozygous organisms that differ from each other by 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.

Pleiotropy (multiple gene action) - -this is the interaction of genes in whichone gene affects several traits at once;

Polymeraction of genes -this is such an interaction of genes when the more dominant genes in the genotype of those pairs that affect this quantitative trait, the more it manifests itself;

Polyhybrid crossing -crossing of organisms that differ in several ways;

Sex-linked inheritance – inheritance of a gene located on the sex chromosome.

Mendel's third law (the law of independent inheritance of traits) -with dihybrid crossing, the genes and traits for which these genes are responsible are combined and inherited independently of each other (the ratio of these phenotypic variants is as follows: 9: W: Z: 1);

Phenotype - the totality of all external and internal signs any organism;

Clean lines- organisms that do not interbreed with other varieties, homozygous organisms;

Epistasis- this is such an interaction of genes when one of them suppresses the manifestations of the other, non-allelic to it.

3.1. Symbols,

adopted in solving genetic problems:

The ♀ symbol denotes a female,

symbol ♂ - masculine,

x - crossing,

A, B, C - genes responsible for

dominant feature,

a, b, c - gene responsible for

recessive trait

P - parental generation,

G - gametes,

F 1 - the first generation of descendants,

F 2 - second generation of descendants,

G - genotype

G (F 1) - genotype of the first generation of offspring

XX - sex chromosomes of a woman

XY - male sex chromosomes

X A - dominant gene localized on the X chromosome

X a - recessive gene localized on the X chromosome

Ph - phenotype

Ph (F 1) - phenotype of the first generation of offspring

3.2. Algorithm for solving genetic problems

Read the problem statement carefully.

Make a short note of the problem statement (as given by the problem statement).

Write down the genotypes and phenotypes of the mating species.

Identify and write down the types of gametes that the breeding species form.

Identify and record the genotypes and phenotypes of the offspring obtained from the crossing.

Analyze the crossing results. To do this, determine the number of classes of offspring by phenotype and genotype and write them down as a numerical ratio.

Write down the answer to the problem question.

(When solving problems on certain topics, the sequence of stages can change, and their content can be modified.)

3.3. Registration of tasks

It is customary to write the genotype of the female first, and then the male (the correct entry is ♀ААВВ х ♂аавв; the wrong entry is ♂aavv x ♀ААВВ).

The genes of one allelic pair are always written side by side (the correct entry is ♀AABB; the incorrect entry is ♀ABAB).

When writing a genotype, the letters denoting traits are always written in alphabetical order, regardless of which trait - dominant or recessive - they denote (correct entry - ♀aaBB; incorrect entry -♀ BBaa).

If only the phenotype of an individual is known, then when recording its genotype, only those genes are written, the presence of which is indisputable. A gene that cannot be identified by its phenotype is denoted by the symbol "_" (for example, if the yellow color (A) and smooth form (B) of pea seeds are dominant traits, and the green color (a) and wrinkled form (b) are recessive, then the genotype of an individual with yellow wrinkled seeds is recorded as A_bv).

The phenotype is always written under the genotype.

In individuals, the types of gametes are determined and recorded, not their number.

correct entry incorrect entry

♀ AA ♀ AA

8. Phenotypes and types of gametes are written strictly under the corresponding genotype.

9. The progress of solving the problem is recorded with the justification of each conclusion and the results obtained.

10. When solving problems on di- and polyhybrid crossing, it is recommended to use the Pennet lattice to determine the genotypes of the offspring. The types of gametes from the maternal individual are recorded vertically, and the paternal ones are recorded horizontally. At the intersection of the column and the horizontal line, a combination of gametes corresponding to the genotype of the resulting daughter is recorded.

IV. Examples of solving genetic problems

4.1. Monohybrid crossing

1. Conditions of the problem: In humans, the gene for long eyelashes dominates over the gene for short eyelashes. A woman with long eyelashes, whose father had short eyelashes, married a man with short eyelashes. Answer the questions:

How many types of gametes are formed in a woman, a man?

What is the probability (in%) of giving birth to a child with long eyelashes in this family?

How many different genotypes, phenotypes can there be among the children of this married couple?

Given:An object research - human

Investigated sign- eyelash length:

Gene A - long

Gene a - short

Find : The number of gametes formed in♀, ♂ ; The likelihood of having a baby with long eyelashes; G (F 1), Ph (F 1)

Solution. We determine the genotypes of the parents. The woman has long eyelashes, therefore, her genotype can be AA or Aa. According to the condition of the problem, the woman's father had short eyelashes, which means that his genotype is aa. Each organism from a pair of allelic genes receives one from the father, the other from the mother, which means that the woman's genotype is Aa. The genotype of her husband is aa, since he has short eyelashes.

Let's write down the marriage scheme

R ♀ Aa X ♂ aa

G A a a

F 1 Aa; aa

Phenotype: long short

Let us write down the genotype splitting of the hybrids: 1Aa: 1aa, or 1: 1. The phenotype splitting will also be 1: 1, one half of the children (50%) will be with long
eyelashes, and the other (50%) with short.

Answer:- in a woman of type 2, in a man of type 1; the probability of having a baby with long eyelashes is 50%, with short eyelashes - 50%; genotypes among children - 2 types

4.2. Dihybrid crossing

1. Conditions of the problem: A dominates yellow a, and the disc-shaped form V- over the spherical b .

Answer the questions: how it will look F 1 and F 2

Let's write down the object of research and the designation of genes:

Given:An object research - pumpkin

Investigated signs:

– fruit color: Gene A - white

Gene a - yellow

– fruit shape: Gene B - disc-shaped

Gene b - spherical

Find : G (F 1), Ph (F 1)

Solution. Determine the genotypes of the parent pumpkins. According to the conditions of the problem, pumpkins are homozygous, therefore, they contain two identical alleles for each trait.

Let's write down the crossing scheme

P ♀ AA bb X ♂ aa BB

G A b aB

F 1 ♀АaBb X ♂ АaBb

G AB, A b, aB, ab AB, Ab, aB, ab

5. Find F 2 : building a pinnet lattice and we introduce into it all possible types of gametes: horizontally we bring in the gametes of the male, vertically - the female. At the intersection, we get the possible genotypes of the offspring.

AABb *

Аa ВB *

Aa Bb *

AABb *

AAbb **

AaBb *

Aabb **

AaBB *

AaBb *

Aabb **

Aabb ***

6. Vwe write the splittinghybridsonphenotype: 9 white disc-shaped *, 3 white globular **, 3 yellow disc-shaped, 1 yellow globular ***.

7. Answer: F 1 - all white disc-shaped, F 2 - 9 white disc-shaped, 3 white spherical, 3 yellow disc-shaped, 1 yellow spherical.

4.3. Sex-linked inheritance

1. Conditions of the problem: The recessive gene for color blindness (color blindness) is located on the X chromosome. The girl's father suffers from color blindness, and the mother, like all her ancestors, distinguishes colors normally. The girl marries a healthy young man.

Answer the questions:

What can be said about their future sons, daughters?

Let's write down the object of research and the designation of genes:

Given:An object research - human

Investigated sign- color perception (the gene is localized on the X chromosome):

Gene A - normal color perception

Gene a - color blindness

Find : G (F 1), Ph (F 1)

Solution. We determine the genotypes of the parents. Sex chromosomes of women XX, men - XY. The girl receives one X chromosome from her mother and one from her father. According to the condition of the problem, the gene is localized on the X chromosome. The girl's father suffers from color blindness, which means he has the X and Y genotype, the mother and all her ancestors are healthy, which means her genotype is X A X A. Each organism from a pair of allelic genes receives one from the father, the other from the mother, which means that the girl's genotype is X A X a. The genotype of her spouse is X A Y, since he is healthy according to the condition of the problem.

Let's write down the marriage scheme

P ♀ X A X a X ♂ X A Y

G X A X a X A Y

F 1 X A X A X A Y X A X a X a Y

Phenotype: healthy healthy healthy sick

Answer: Daughter can be healthy ( X A X A) or be healthy, but be a carrier of the hemophilia gene ( X A X), and the son can be healthy ( X A Y) and sick ( X a Y).

V. Tasks for determining the number and types of formed gametes

Vi. Genetic tasks

In mice, long ears are inherited as a dominant trait, while short ears are inherited as recessive. A male with long ears was crossed with a female with short ears. In F 1, all the offspring came out with long ears. Determine the genotype of the male.

(3 points)

How many and what types of gametes does an organism of the AabbCcDd genotype form?

(3 points)

3. Husband and wife have curly (A) and dark (B) hair. They had a baby with curly (A) and light (b) hair. What are the possible genotypes of the parents?

(3 points)

When crossing a shaggy (A) white rabbit (c) with a shaggy (A) black rabbit, several white smooth, non-shaggy rabbits were born. What are the genotypes of the parents?

(5 points)

A blue-eyed man whose both parents had Brown eyes, married a brown-eyed woman, whose father has brown eyes, and whose mother's are blue. From this marriage, a blue-eyed son was born. Determine the genotype of all individuals and chart the pedigree.

(3 points)

The hemophilia gene is recessive and localized on the X chromosome. A healthy woman, whose mother was healthy and whose father was hemophiliac, married a hemophiliac man. What kind of children can you expect from this marriage?

(5 points)

2 varieties of strawberries are crossed: beardless red and beardless white. In F 1, all mustachioed red, in F 2 splitting: 331 mustachioed red, 98 mustachioed white, 235 mustachioed red, 88 mustache white. How are traits inherited?

(b points)

Beige mink crossed with gray. In F 1 all minks are brown, in F 2 there are 14 gray, 46 brown, 5 cream, 16 beige. How is the trait inherited? ( 6 points)

1. In dogs, black coat color dominates over brown. From crossing a black female with a brown male, 4 black and 3 brown puppies were obtained. Determine the genotypes of the parents and offspring.

(3 points)

2. In humans, phenylketonuria is inherited as a recessive trait. The disease is associated with a lack of an enzyme that breaks down phenylalanine. An excess of this amino acid in the blood leads to damage to the central nervous system and the development of dementia. Determine the likelihood of developing the disease in children in a family where both parents are heterozygous for this trait.

(3 points)

Sickle cell anemia in humans is inherited as an incompletely dominant autosomal trait. Homozygotes die in early childhood, heterozygotes are viable and resistant to malaria. What is the likelihood of having children resistant to malaria in a family where one parent is heterozygous for the trait of sickle cell disease and the other is normal for that trait?

(3 points)

There are two types of blindness in humans, and each is defined by its own recessive autosomal gene. The genes for both traits are on different pairs of chromosomes. What is the probability of having a blind child if the father and mother suffer from the same type of blindness, but are normal on the other pair of genes?

(4 points)

In fruit flies, genes for body color and wing shape are linked. A female with normal wings and a gray body was crossed with a male with a black body and reduced wings. In the first generation, all offspring had a gray body and normal wings. Determine the genotypes of the parents and offspring.

(6 points)

The mother has a blood group I, and the father has an IV. Can children inherit a parent's blood type?

Reference.

(3 points)

When two silkworm lines were crossed, the caterpillars of which form white cocoons, in the first generation, all cocoons were yellow. During the subsequent crossing of the hybrids in the second generation, splitting occurred: 9 yellow cocoons to 7 white ones. Determine the genotypes of all individuals.

(6 points)

In cats, the black gene and the red gene are sex-linked, located on the X chromosome and give incomplete dominance. When combined, a tortoiseshell color is obtained. 5 kittens were born from a tortoiseshell cat, one of which turned out to be red, 2 had a tortoiseshell color and 2 were black. The ginger kitten turned out to be a female. Determine the genotype of the cat, as well as the genotypes of the cat and offspring.

(5 points)

A blue-eyed man whose parents had brown eyes married a brown-eyed woman whose father had blue eyes and whose mother had brown. What offspring can be expected from this marriage if it is known that the brown color is the dominant trait? Determine the genotypes of the male and female parents.

(3 score)

In humans, the polydactyly allele (6 fingers) dominates the normal five-fingered hand. In a family where one parent has a six-fingered hand and the other has a normal hand structure, a child was born with a normal hand. Determine the likelihood of having a second child without an anomaly,

(3 points)

In dogs, short hair dominates over long. The hunter has bought a short-haired dog and wants to be sure it does not carry the long-haired allele. What phenotype and genotype partner should be selected for crossing in order to check the genotype of the purchased dog? Make a crossing scheme. What should be the result if the dog is purebred?

(3 points)

In cattle, hornlessness (hornlessness) and black coat color dominate over horniness and red color. The genes for both traits are on different chromosomes. When crossing a hornless black bull with three red hornless cows, the calves were all black, but one of them was horned. Determine the likely genotypes of the parents and offspring.

(4 points)

In tomatoes, the genes that determine the height of the stem and the shape of the fruit are linked, with the tall stem dominating over dwarfism, and the spherical shape of the fruit over the pear-shaped. What offspring should be expected from crossing a plant heterozygous for both traits with a dwarf one that has spherical fruits,

(b points)

Parents have II and III blood groups, and their son is I. Determine the genotypes of the parents.

Reference. The blood group depends on the action of not two, but three allelic genes, denoted by the symbols A, B, 0. They, when combined in diploid cells by two, can form 6 renootypes (00 - 1 blood group; AA, AO -IIblood type; VO, BB -IIIblood type; AB -IV blood type). It is assumed that both allelic gene A and B dominate over recessive gene 0, but A and B do not suppress each other.

(3 points)

When crossing two varieties of rye with white and yellow grain in the first generation, all plants had green grains. When these green hybrids were crossed, 450 green, 150 yellow and 200 white were obtained. Determine the genotypes of the parents and offspring. How is the trait inherited?

( b points)

In the fruit fly Drosophila, white-eyed is inherited as a recessive trait linked to the X chromosome. What offspring will turn out if you cross a white-eyed female with a red-eyed male?

(5 points)

The black color of the coat is dominant in dogs over brown. The black female crossed several times with the brown male. A total of 15 black and 13 brown puppies were born. Determine the genotypes of the parents and offspring.

(3 points)

One of the forms of cystinuria (violation of the exchange of four amino acids) is inherited as an autosomal recessive trait. However, in heterozygotes, only an increased content of cystine in the urine is observed, and in homozygotes, the formation of cyetine stones in the kidneys. Define possible forms manifestations of cystinuria in children in a family where one of the spouses suffered from this disease, and the other had only an increased content of cystine in the urine.

(3 points)

V maternity hospital confused two boys. The parents of one of them have G and II blood groups, the parents of the other - II and IV. Studies have shown that Children have blood groups I and II. Determine who is whose son. Is it possible to do this for sure with other combinations of blood groups? Give examples.

Reference. The blood group depends on the action of not two, but three allelic genes, denoted by the symbols A, B, 0. They, when combined in diploid cells by two, can form 6 genotypes (00 - 1 blood group; AA, AO - II blood group; BO , BB - III blood group; AB - IV blood group). It is assumed that both allelic gene A and B dominate over recessive gene 0, but A and B do not suppress each other.

(6 points)

V A deaf child with smooth hair was born to a family where the parents could hear well and had one smooth hair and the other curly hair. Their second child could hear well and had curly hair. What is the probability of having a deaf child with curly hair in this family if the allele for curly hair is known to dominate over the allele for smooth hair; is deafness a recessive trait, and both genes are on different chromosomes?

(5 points)

The sex-linked gene B in canaries determines the green color of the plumage, in - brown. The green male was crossed with the brown female. The offspring were obtained: 2 brown males and 2 green females. What are the genotypes of the parents?

(5 points)

When crossing two varieties of Levkoy, one of which has double red flowers, and the second - double white, all the hybrids of the first generation had simple red flowers, and in the second generation splitting was observed: 68 - with double white flowers, 275 - with simple red flowers, 86 - with simple white and 213 - with double red flowers. How is the color and shape of a flower inherited?

(9 points)

Science knows dominant genes, which have never been obtained in homozygotes, since the offspring homozygous for this gene dies at the embryo stage. Such "killing" genes are called lethal. The dominant gene of platinum coloration in fur-bearing foxes is lethal, the allele of which is a recessive gene that determines the silvery coloration of the animals. Determine which offspring and in what respect will be born from the crossing of two platinum parents.

(3 points)

The curly pumpkin has a white fruit color A dominates yellow a, and the disc-shaped form V- over the spherical b... How it will look F 1 and F 2 from crossing a homozygous white globular pumpkin with a homozygous yellow discoid?

(4 points)

An early maturing oat variety of normal growth was crossed with a late maturing giant growth variety. Determine what the first generation hybrids will be. What will be the offspring from crossing hybrids among themselves by genotype and phenotype, as well as their quantitative ratio? (The early maturity gene dominates the late maturity gene, the normal growth gene dominates the giant growth gene.)

(4 points)

What will the kittens be from crossing a tortoiseshell cat with a black cat, a tortoiseshell cat with a ginger cat? The gene for black and red color is located on the X chromosome (a sign of color is linked to the floor); none of them dominates the other; in the presence of both genes in the X chromosome, the color turns out to be spotty: "tortoiseshell" /

(6 points)

The recessive gene for color blindness (color blindness) is located on the X chromosome. The girl's father suffers from color blindness, and the mother, like all her ancestors, distinguishes colors normally. The girl marries a healthy young man. What can be said about their future sons, daughters, as well as grandchildren of both sexes (provided that the sons and daughters will not marry carriers of the color blindness gene)?

(6 points)

A blue-eyed man whose parents had brown eyes married a brown-eyed woman whose father had blue eyes and whose mother had brown. From this marriage, one child was born, whose eyes turned out to be brown. What are the genotypes of all the individuals mentioned here?

(3 points)

Human blood is divided into four groups. Blood type is a hereditary trait that depends on one gene. This gene has not two, but three alleles, denoted by the symbols A, B, 0. Individuals with genotype 00 have the first blood group, those with the AA or AO genotype - the second, with the BB and BO genotypes - the third, and with the AB genotype - the fourth ( alleles A and B dominate allele 0, but do not suppress each other). What blood groups are possible in children if their mother has a second group, and a father has a fourth?

(4 points)

I.'s mother found on her child a tag with the surname of a roommate in N.'s ward. Blood tests were taken from the children's parents. The blood groups were distributed as follows: I. - I, her husband - IV, child - I; N. - I, her husband - I, child - III .

What is the conclusion from this?

(4 points)

Drosophila grey colour body (B) dominates over black (c). When gray parents were crossed, the offspring turned out to be gray as well. Identify possible parental genotypes.

(3 points)

Write down the possible genotypes of a person, if his phenotype is:

a). big brown eyes - ...

b) big blue eyes - ...

c) thin lips and a Roman nose - ...

d) thin lips and straight nose - ...

Reference. Dominant features: big eyes, brown eyes, Roman nose. Recessive signs: blue eyes, thin lips, straight nose.

(4 score)

What blood types are possible in children if their mother has a blood type II and a father has a blood type VI?

(3 points)

When two different varieties of white-flowered sweet pea are crossed, all P1 hybrid plants are red-flowered. How can this be explained?

(4 score)

Reference. The synthesis of red pigment in a pea flower occurs only in the presence of two non-allelic dominant genes A and B; in the absence of at least one of them, the flower loses its red pigment. These genes are localized in non-homologous chromosomes and are inherited independently, as in a dihybrid crossing.

You have purchased a male rabbit with black fur (dominant trait), but the exact genotype of this animal is unknown. How can you find out its genotype?

(3 points)

Can white rabbits be unclean (heterozygous) for coat color?

(3 points)

In the Datura plant, the purple color of the flowers (A) dominates over the white (a), the prickly seed pods (B) - over the smooth ones (c). A plant with purple flowers and thornless bolls, crossed with a plant with white flowers and thorny bolls, produced 320 offspring with purple flowers and thorny bolls and 312 purple flowers and smooth bolls. What are the genotypes of the parent plants?

(5 points)

From gray rabbits and gray rabbits, offspring were obtained: 503 gray and 137 white rabbits. Which coat color is dominant and which is recessive?

(3 points)

There were black and red cows in the herd. The bull was black. All calves that appeared in this herd were black. Determine the recessive suit. What offspring will these calves have when they grow up?

(4 score)

What can be said about the nature of the inheritance of the color of apple fruits when crossing Antonovka (green fruits) with Wesley (red fruits), if all the fruits of the hybrids obtained from this crossing were red? Record the genotypes of the parents and hybrids. Make a scheme of fruit color inheritance in F 1 and F 2.

(4 points)

As a result of hybridization of plants with red and white flowers, all hybrid plants had pink flowers. Record the genotype of the parent plant. What is the nature of inheritance?

(4 points)

A normal plant is crossed with a dwarf plant, the first generation is normal. Determine what the offspring will be from the self-pollination of the first generation hybrids.

(4 points)

Two gray rabbits (female and male) were brought to the school corner of wildlife, considering them purebred. But in F 2, black rabbits appeared among their grandchildren. Why?

(4 points)

Chickens with white plumage, when crossed with each other, always give white offspring, and chickens with black plumage - black. Offspring from crossing white and black individuals turns out to be gray. What part of the offspring from crossing a gray rooster and a hen will have gray plumage?

(4 points)

The father and son are color blind, and the mother can distinguish colors normally. From whom did the son inherit the color blindness gene?

(4 points)

When a clean line of mice with brown hair is crossed with a clean line of mice with a clean line of mice with gray hair, descendants with brown hair are obtained. BF 2 from crossing between these F 1 mice are brown and gray mice in a ratio of 3: 1. Please give a full explanation of these results.

(3 points)

Two black female mice mated with a brown male. One female gave 20 black and 17 brown offspring, and the other - 33 black. What are the genotypes of the parents and offspring?

(4 points)

When plants of red-fruited strawberries are crossed among themselves, plants with red fruits are always obtained, and white-fruited plants with white ones. As a result of crossing both varieties, pink berries are obtained. What offspring will be obtained when red strawberries are pollinated with the pollen of a plant with pink berries?

(5 points)

Phenylketonuria (metabolic disorder of the amino acid - phenylalanine) is inherited as a recessive trait. The husband is heterozygous for the phenylketonuria gene, and the wife is homozygous for the dominant allele of this gene. What is the likelihood of having a sick child?

(4 points)

The color of flowers in a night beauty is inherited by an intermediate type, and the height of the plant dominates over dwarfism. A crossing of a homozygous plant of a night beauty with red flowers, normal growth and a plant with white flowers, dwarf growth was made. What will be the first and second generation hybrids? What kind of splitting will be observed in the second generation for each trait separately?

(5 points)

The father and mother are healthy, and the child has hemophilia. What is the child's gender?

(4 points)

In cats, short hair dominates over long hair. The long-haired cat, when crossed with the short-haired cat, yielded three short-haired and two long-haired kittens. Determine the genotypes of the parental and hybrid forms.

(4 points)

What offspring should be expected from the marriage of a color blind man and a healthy woman whose father suffered from color blindness?

(5 points).

Creative level

All tasks in genetics can be classified according to two main criteria: A) by the type of inheritance; B) on the issue of the task (that is, what needs to be found or determined). Based on the tasks of options 1-4, as well as the literature recommended by the teacher, make a classification of genetic tasks for each of the specified criteria. (20 points). For each class of problems, create and solve an example problem (40 points) ... To avoid biological errors, it is best to take fictitious organisms and characters for tasks. (For examples of problems with fictitious organisms and traits, see below.) Suggest your criterion for the classification of genetic problems.

(5 points).

In the Aldebaran nostril, the allele defining 3 nostrils is incompletely dominant over the allele defining one nostril. How many tail nostrils can the cubs have if both parents have 2 nostrils.

(3 points)

A brainless woman, whose father and mother were also brainless, married a brainless man. They had a child with a brain. Suggest at least 2 options for inheriting this trait.

(6 points)

In the family of marsupial microcephals, a heterozygous square-headed saber-toothed female and a tri-headed saber-toothed male were born 83 square-headed saber-toothed, 79 tri-headed saber-toothed, 18 triple-headed saber-toothed and 17 square-headed saber-toothed microcephalates. Determine how the traits are inherited.

(9 points)

A female pari-headed swelling with a mouth on the abdomen and a long ovipositor crossed with a male with a mouth on the back and a short ovipositor. The female laid her eggs on the asteroid, ate the male and flew away. The eggs hatched into cubs with mouths on their belly and a long ovipositor. They interbred randomly, resulting in the birth of 58 females with a mouth on the belly and a long ovipositor, 21 females with a mouth on the belly and a short ovipositor, 29 males with a mouth on the belly and a long ovipositor, 11 males with a mouth on the belly and a short ovipositor, 9 males with dorsal mouth and short ovipositor, and 32 males with dorsal mouth and long ovipositor. Determine how the traits are inherited.

(12 points)

Vii. List of sources used

Anastasova L.P. Independent work Students in General Biology: A Guide for Teachers. M .: Education, 1989 - 175 p.

Biology: 1600 problems, tests and verification work for schoolchildren and those entering universities / Dmitrieva T.A., Gulenkov S.I., Sumatikhin S.V. and others - M .: Bustard, 1999.-432 p.

Borisova, L.V. Thematic and lesson planning in biology: grade 9: kuchebnik, Mamontova S.G.,., Zakharova VB, Sonina N.I. "Biology. General patterns. Grade 9 ": Method. allowance / Borisova L.V. - M.: Publishing house "Examination", 2006. - 159 p.

Kozlova T.A. Thematic and lesson planning in biology for the textbook Kamensky A.A., Kriksunov E.A., Pasechnik V.V " General biology: 10-11 grades ". - M .: Publishing house "Exam", 2006. - 286 p.

Krasnovidova S.S. Didactic materials in general biology: 10-1 grades: A manual for general education students. Institutions / Krasnovidova S.S., Pavlov S.A., Khvatov A.B. - M .: Education, 2000 .-- 159 p.

Lovkova T.A. Biology. General patterns. Grade 9: Methodological manual for the textbook Mamontov S.G., Zakharova VB, Sonina N.I. "Biology. General patterns. Grade 9 "/ Lovkova T.A., Sonin N.I. - M.; Bustard, 2003 .-- 128 p.

Pepelyaeva O..A., Suntsova I.V. Lesson development in general biology: grade 9. - M .: BAKO, 2006 .-- 464 p.

Sukhova T.S. Shining biology. 10-11 grades: workbook to the textbooks “General biology. Grade 10 "and" General Biology. Grade 11 "/ Sukhova TS, Kozlova TA, Sonin NI; edited by V.B. Zakharov - M .: Bustard, 2006.-171 p.

Instructions

To solve genetic problems, certain types of research are used. The method of hybridological analysis was developed by G. Mendel. It allows you to identify the patterns of inheritance of individual traits during sexual reproduction. The essence this method is simple: when analyzing certain alternative characters, they are traced in the offspring. Also, an accurate record of the manifestation of each alternative trait and each individual individual of the offspring is carried out.

The main patterns of inheritance were also developed by Mendel. The scientist deduced three laws. Subsequently, they are so - Mendel's laws. The first is the law of uniformity of the first hybrids. Take two heterozygous individuals. When crossed, they will give two types of gametes. The offspring of such parents will appear in a 1: 2: 1 ratio.

Mendel's second law is the law of splitting. it is based on the statement that a dominant gene does not always suppress a recessive one. In this case, not all individuals among the first generation reproduce the traits of their parents - the so-called intermediate nature of inheritance appears. For example, when crossing homozygous plants with red flowers (AA) and white flowers (aa), offspring with pink ones is obtained. Incomplete dominance is fairly common. It is also found in some biochemical characteristics of a person.

The third and last law is the law of independent combination of features. For the manifestation of this law, several conditions must be met: there must not be lethal genes, dominance must be complete, genes must be on different chromosomes.

The tasks of gender genetics stand apart. There are two types of sex chromosomes: the X chromosome (female) and the Y chromosome (male). Sex with two identical sex chromosomes is called homogametic. Gender determined different chromosomes, is called heterogametic. The sex of the future individual is determined at the time of fertilization. In the sex chromosomes, in addition to genes that carry information about the sex, there are others that have nothing to do with this. For example, the gene responsible for blood clotting is carried by the female X chromosome. Sex-linked traits are transmitted from the mother to the sons and daughters, and from the father - only to the daughters.

Basic terms of genetics

Gene is a section of a DNA molecule that carries information about the primary structure of one protein. A gene is a structural and functional unit of inheritance.

Allelic genes (alleles) - different variants one gene encoding an alternative manifestation of the same trait. Alternative signs are signs that cannot be in the body at the same time.

Homozygous organism- an organism that does not split in one way or another. Its allelic genes equally affect the development of this trait.

Heterozygous organism- an organism that splits according to one or another characteristic. 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 is manifested in a homozygous organism containing two recessive genes.

Genotype- a set of genes in a diploid set of an organism. The set of genes in a haploid set of chromosomes is called genome.

Phenotype- the totality of all the signs of the body.

G. Mendel's laws

Mendel's first law - the law of uniformity of hybrids F 1

This law is derived from the results monohybrid crossing... For the experiments, two varieties of peas were taken, differing from each other by one pair of signs - the color of the seeds: one variety had a yellow color, the second - green. The crossed plants were homozygous.

To record the results of crossing, Mendel proposed the following scheme:

A - yellow color of seeds
a - green color of seeds

Formulation of the law: when crossing organisms that differ in one pair of alternative traits, the first generation is uniform in phenotype and genotype.

Mendel's second law - the law of splitting

From seeds obtained by crossing a homozygous plant with a yellow seed color with a plant with a green seed color, plants were grown, and F 2 was obtained by self-pollination.

P (F 1)	Aa	Aa
G	A; a	A; a
F 2	AA; Aa; Aa; aa (75% of plants are dominant, 25% are recessive)

The wording of the law: in the offspring obtained from crossing the first generation hybrids, there is a splitting according to the phenotype in a ratio of 3: 1, and according to the genotype - 1: 2: 1.

Mendel's third law - the law of independent inheritance

This law was deduced from the data obtained from the dihybrid crossing. Mendel considered the inheritance of two pairs of traits in peas: color and shape of seeds.

Mendel used plants homozygous for both pairs of traits as parental forms: one variety had yellow seeds with smooth skin, the other had green and wrinkled seeds.

A - yellow color of seeds, a - green color of seeds,
B - smooth shape, b - wrinkled shape.

Then Mendel grew plants from F1 seeds and got second-generation hybrids by self-pollination.

AaBv

AB, AB, AB, AB

F 2

Punnett grid is used to record and define genotypes.

Gametes	AB	Av	aB	aw
AB	AABB	AABv	AaBB	AaBv
Av	AABv	Aavb	AaBv	Aavb
aB	AaBB	AaBv	aaBB	aaBv
aw	AaBv	Aavb	aaBv	aavv

F 2 was split into 4 phenotypic classes in a ratio of 9: 3: 3: 1. 9/16 of all seeds had both dominant traits (yellow and smooth), 3/16 - the first dominant and the second recessive (yellow and wrinkled), 3/16 - the first recessive and the second dominant (green and smooth), 1/16 - both recessive traits (green and wrinkled).

When analyzing the inheritance of each pair of traits, the following results are obtained. F 2 contains 12 parts of yellow seeds and 4 parts of green seeds, i.e. ratio 3: 1. Exactly the same ratio will be for the second pair of traits (seed shape).

Formulation of the law: when organisms are crossed that differ from each other by two or more pairs of alternative traits, genes and their corresponding traits are inherited independently of each other and are combined in all possible combinations.

Mendel's third law is fulfilled only if the genes are in different pairs of homologous chromosomes.

The law (hypothesis) of "purity" of gametes

When analyzing the traits of hybrids of the first and second generations, Mendel found that the recessive gene does not disappear and does not mix with the dominant one. Both genes are manifested in F 2, which is possible only if the F 1 hybrids form two types of gametes: some carry a dominant gene, others a recessive one. This phenomenon is called the gamete purity hypothesis: each gamete carries only one gene from each allelic pair. The hypothesis of the purity of gametes was proved after studying the processes occurring in meiosis.

The hypothesis of "purity" of gametes is the cytological basis of Mendel's first and second laws. With its help, it is possible to explain the segregation by phenotype and genotype.

Analyzing cross

This method was proposed by Mendel to elucidate the genotypes of organisms with a dominant trait that have the same phenotype. For this, they were crossed with homozygous recessive forms.

If, as a result of crossing, the entire generation turned out to be the same and similar to the analyzed organism, then it could be concluded that the original organism is homozygous for the studied trait.

If, as a result of crossing in a generation, splitting was observed in a 1: 1 ratio, then the original organism contains genes in a heterozygous state.

Inheritance of blood groups (AB0 system)

The inheritance of blood groups in this system is an example of multiple allelism (this is the existence of more than two alleles of the same gene in a species). In the human population, there are three genes (i 0, I A, I B), encoding proteins-antigens of erythrocytes, which determine the blood groups of people. The genotype of each person contains only two genes that determine his blood group: the first group i 0 i 0; the second I A i 0 and I A I A; the third I B I B and I B i 0 and the fourth I A I B.

Inheritance of sex-linked traits

In most organisms, sex is determined during fertilization and depends on the set of chromosomes. This method is called chromosomal sex determination. Organisms with this type of sex determination have autosomes and sex chromosomes - Y and X.

In mammals (including humans), the female sex has a set of sex chromosomes XX, the male sex - XY. The female sex is called homogametic (forms one type of gametes); and the male is heterogametic (forms two types of gametes). In birds and butterflies, males (XX) are homogametic sex, and females are heterogametic (XY).

The USE includes tasks only for characters linked to the X chromosome. Basically, they relate to two signs of a person: blood clotting (X H is the norm; X h is hemophilia), color vision (X D is the norm, X d is color blindness). Much less common is the problem of inheriting sex-linked traits in birds.

In humans, the female sex can be homozygous or heterozygous for these genes. Let's consider the possible genetic sets in a woman using the example of hemophilia (a similar picture is observed with color blindness): X H X H - healthy; X H X h - healthy, but is a carrier; X h X h - sick. The male sex for these genes is homozygous, because The Y chromosome does not have alleles of these genes: X H Y - healthy; X h Y - sick. Therefore, men most often suffer from these diseases, and women are their carriers.

Typical USE tasks in genetics

Determination of the number of gamete types

Determination of the number of gamete types is carried out according to the formula: 2 n, where n is the number of pairs of genes in a heterozygous state. For example, an organism with the AAvCC genotype has no genes in a heterozygous state, i.e. n = 0, therefore, 2 0 = 1, and it forms one type of gametes (ABC). An organism with the AaBBcc genotype has one pair of genes in a heterozygous state (Aa), i.e. n = 1, therefore 2 1 = 2, and it forms two types of gametes. An organism with the AaBbCc genotype has three pairs of genes in a heterozygous state, i.e. n = 3, therefore, 2 3 = 8, and it forms eight types of gametes.

Problems for mono- and dihybrid crossing

For monohybrid crossing

Task: White rabbits were crossed with black rabbits (black is dominant). F 1 has 50% white and 50% black. Determine the genotypes of the parents and offspring.

Solution: Since in the offspring there is a splitting according to the studied trait, therefore, the parent with the dominant trait is heterozygous.

For dihybrid crossing

Dominant genes are known

Task: Tomatoes of normal growth with red fruits were crossed with dwarf tomatoes with red fruits. In F1, all plants were of normal growth; 75% with red fruits and 25% with yellow ones. Determine the genotypes of parents and offspring if it is known that in tomatoes, red fruit color dominates over yellow, and normal growth over dwarfism.

Solution: Let's designate dominant and recessive genes: A - normal growth, a - dwarfism; B - red fruits, c - yellow fruits.

Let's analyze the inheritance of each trait separately. In F1, all offspring are of normal height, i.e. splitting for this trait is not observed, therefore the original forms are homozygous. According to the color of the fruit, a splitting of 3: 1 is observed, therefore the original forms are heterozygous.

Dominant genes unknown

Task: Two varieties of phlox were crossed: one has red saucer-shaped flowers, the second has red funnel-shaped flowers. In the offspring, 3/8 red saucer-shaped, 3/8 red funnel-shaped, 1/8 white saucer-shaped and 1/8 white funnel-shaped were obtained. Identify the dominant genes and genotypes of the parental forms, as well as their offspring.

Solution: Let's analyze the splitting for each attribute separately. Among the descendants, plants with red flowers make up 6/8, with white flowers - 2/8, i.e. 3: 1. Therefore, A is red, and - White color, and the parental forms are heterozygous for this trait (since there is a splitting in the offspring).

Splitting is also observed in the shape of the flower: half of the offspring have saucer-shaped flowers, half are funnel-shaped. Based on these data, it is not possible to unambiguously determine the dominant feature. Therefore, we will assume that B - saucer-shaped flowers, in - funnel-shaped flowers.

AaBv
(red flowers, saucer shape)

Aavb
(red flowers, funnel-shaped)

AB, AB, AB, AB

Av, av

F 1

Gametes	AB	Av	aB	aw
Av	AABv	AAvv	AaBv	Aavb
aw	AaBv	Aavb	aaBv	aavv

3/8 A_B_ - red saucer-shaped flowers,
3/8 A_vv - red funnel-shaped flowers,
1/8 aaBv - white saucer-shaped flowers,
1/8 aavv - white funnel-shaped flowers.

Solving problems for blood groups (AB0 system)

Task: the mother has the second blood group (she is heterozygous), the father has the fourth. What blood types are possible in children?

Solution:

Solving problems on the inheritance of sex-linked traits

Such tasks may well be encountered both in part A and in part C of the exam.

Task: carrier of hemophilia married a healthy man. What kind of children can be born?

Solution:

Mixed problem solving

Task: A man with brown eyes and blood group 3 married a woman with brown eyes and blood group 1. They had a blue-eyed baby with 1 blood group. Determine the genotypes of all persons indicated in the task.

Solution: Brown eyes dominate over blue, so A - brown eyes, a - blue eyes. The child has blue eyes, so his father and mother are heterozygous for this trait. The third blood group can have the genotype I B I B or I B i 0, the first - only i 0 i 0. Since the child has the first blood group, therefore, he received the gene i 0 from both his father and mother, therefore his father has the genotype I B i 0.

Task: The man is color blind, right-handed (his mother was left-handed), married to a woman with normal vision (her father and mother were completely healthy), left-handed. What kind of children can this couple have?

Solution: A person has better right-handed control over left-handedness, so A is right-handed and A is left-handed. The male genotype is Aa (because he received the a gene from his left-handed mother), and the female - aa.

A color-blind man has the genotype X d Y, and his wife - X D X D, because her parents were completely healthy.

Tasks for independent solution

Determine the number of gamete types in an organism with the AaBBCc genotype.
Determine the number of gamete types in an organism with genotype AaBvX d Y.
Determine the number of gamete types in an organism with genotype aaBBI B i 0.
Crossed tall plants with low plants. In F 1, all plants are medium-sized. What is F 2?
We crossed a white rabbit with a black rabbit. In F1, all rabbits are black. What is F 2?
We crossed two rabbits with gray hair. In F 1 - 25% with black wool, 50% with gray and 25% with white. Identify genotypes and explain this splitting.
They crossed a black hornless bull with a white horned cow. In F1 received 25% black hornless, 25% black horned, 25% white horned and 25% white hornless. Explain this cleavage if black color and lack of horns are dominant.
Drosophila with red eyes and normal wings were crossed with fruit flies with white eyes and defective wings. In the offspring, all flies with red eyes and defective wings. What will be the offspring from crossing these flies with both parents?
A blue-eyed brunette married a brown-eyed blonde. What kind of children can be born if both parents are heterozygous?
A right-handed man with a positive Rh factor married a left-handed woman with a negative rhesus factor. What children can be born if a man is heterozygous only for the second trait?
Both mother and father have blood group 3 (both parents are heterozygous). What blood group is possible in children?
The mother has 1 blood group, the child has 3 group. What blood type is impossible for a father?
The father has the first blood group, the mother has the second. What is the probability of having a baby with the first blood group?
A blue-eyed woman with a 3 blood group (her parents had a third blood group) married a brown-eyed man with a 2 blood group (his father had blue eyes and a first blood group). What kind of children can be born?
A right-handed hemophilic man (his mother was left-handed) married a left-handed woman with normal blood (her father and mother were healthy). What kind of children can be born from this marriage?
We crossed strawberry plants with red fruits and long-petiolate leaves with strawberry plants with white fruits and short-peeled leaves. What kind of offspring can there be if the red color and short-petiolate leaves are dominant, while both parent plants are heterozygous?
A man with brown eyes and a 3 blood group married a woman with brown eyes and a 3 blood group. They had a blue-eyed baby with 1 blood group. Determine the genotypes of all persons indicated in the task.
Melons with white oval fruits were crossed with plants that had white globular fruits. The offspring produced the following plants: 3/8 with white oval, 3/8 with white globular, 1/8 with yellow oval and 1/8 with yellow globular fruits. Determine the genotypes of the original plants and offspring, if the white color of the melon dominates over the yellow, the oval shape of the fruit - over the spherical.

Answers

4 types of gametes.
8 types of gametes.
2 types of gametes.
1/4 high, 2/4 medium and 1/4 low (incomplete dominance).
3/4 black and 1/4 white.
AA - black, aa - white, Aa - gray. Incomplete dominance.
Bull: AaBb, cow - aavb. Offspring: AaBv (black hornless), Aavv (black horned), aaBv (white hornless), aavv (white hornless).
A - red eyes, a - white eyes; B - defective wings, b - normal. The original forms are ААвв and ааВВ, the offspring of АаВв.
Crossing results:
a) AaBv x AAbv
- F 2
- AaBB red eyes, defective wings
- AAb red eyes, normal wings
- Aavb red eyes, normal wings
b) AaBv x aaBB
- F 2 AaBB red eyes, defective wings
- AaBB red eyes, defective wings
- aaBb white eyes, defective wings
- aaBB white eyes, defective wings
A - brown eyes, a - blue; B - dark hair, b - blonde. Father aaBv, mother - Aavv.
A - right-handed, a - left-handed; B - Rh positive, B - negative. Father AABv, mother - Aavv. Children: 50% AaBb (right-handed, Rh positive) and 50% Aavb (right-handed, Rh negative).
Father and mother - I В i 0. Children may have a third blood group (probability of birth - 75%) or first blood group (probability of birth - 25%).
Mother i 0 i 0, child I B i 0; from his mother he received the gene i 0, and from the father - I B. The following blood groups are impossible for the father: the second I A I A, the third I B I B, the first i 0 i 0, the fourth I A I B.
A child with the first blood group can be born only if his mother is heterozygous. In this case, the probability of birth is 50%.
A - brown eyes, a - blue. Woman aaI B I B, man AaI A i 0. Children: AaI A I B (brown eyes, fourth group), AaI B i 0 (brown eyes, third group), aaI A I B (blue eyes, fourth group), aaI B i 0 (blue eyes, third group).
A is right-handed, a is left-handed. Male AaX h Y, female aaX H X H. Children AaX H Y (healthy boy, right-handed), AaX H X h (healthy girl, carrier, right-handed), aaX H Y (healthy boy, left-handed), aaX H X h (healthy girl, carrier, left-handed).
A - red fruits, a - white; B - short petiolate, c - long petiolate.
Parents: Aavb and aaBv. Offspring: AaBb (red fruits, short petiolate), Aavv (red fruits, long petiolate), aaBb (white fruits, short petiolate), aavv (white fruits, long petiolate).
We crossed strawberry plants with red fruits and long-petiolate leaves with strawberry plants with white fruits and short-peeled leaves. What offspring can there be if the red color and short-petiolate leaves are dominant, while both parent plants are heterozygous?
A - brown eyes, a - blue. Woman AaI B I 0, man AaI B i 0. Child: aaI 0 I 0
A - white color, and - yellow; B - oval fruits, c - round. Initial plants: AaBv and Aavv. Offspring:
А_Вв - 3/8 with white oval fruits,
А_вв - 3/8 with white globular fruits,
aaBv - 1/8 with yellow oval fruits,
aavv - 1/8 with yellow spherical fruits.