Analysis of individuals of the i-th generation by genotype and phenotype. Hardy-Weinberg law in solving genetic problems

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8. ANALYSIS OF SPECIMENS OF THE Ist GENERATION BY GENOTYPE AND PHENOTYPE

The last stage of solving problems is the analysis of individuals obtained by crossing. The purpose of the analysis depends on the question of the task: 1) What is the genotype and phenotype of the expected offspring? 2) What is the probability of the appearance of a carrier of a particular trait in the offspring? To answer the first question, it is necessary to correctly solve the set genetic problem in literal terms and find all possible genotypes and phenotypes of the expected offspring. The answer to the second question is related to the calculation of the probability of the manifestation of a feature. Probability can also be calculated as a percentage. In this case, the total number of offspring is taken as 100% and the percentage of individuals bearing the analyzed trait is calculated.

Analysis by genotype

In most cases, when analyzing the genotypes of offspring, it is necessary to determine the quantitative relationships between different genotypes and record them. First, all genotype variants with various combinations of dominant and recessive genes are written out, and the same variants are combined and summarized. Example 8.1. In monohybrid crossing, the number of homozygous dominant, homozygous recessive and heterozygous organisms is emphasized and their ratio is recorded.

Thus, when two heterozygous organisms with the Aa genotype are crossed, the total number of variants of the offspring genotypes is four (genotypes AA, Aa, Aa, aa). Of these, one part of the individuals will be homozygous for the dominant trait, two - heterozygous, one - homozygous for the recessive gene). Analysis by genotype should be written as

1AA : 2Aa : 1aa

or easier

1 : 2 : 1

Such a record of the ratios of the Aa genotypes among the descendants shows heterogeneity in the genotype. Moreover, for one individual with the AA genotype, there are two individuals with the Aa genotype and one with the aa genotype. The occurrence of such heterogeneity is called splitting.

Phenotype analysis

After recording the ratio of genotypes, the phenotypic manifestation of each of them is determined. When analyzing by phenotype, the manifestation of a dominant trait is taken into account. If a dominant gene is present in the genotype, then an organism with such a genotype will carry a dominant trait. A recessive trait will only appear when the organism is homozygous for the recessive gene. Considering all this, the number of all individuals with one or another sign is counted, and their numerical ratio is recorded. Example 8.2. In a monohybrid crossing of two organisms with the Aa genotype. The record of the genotypes of the descendants of the first generation looked like

1AA : 2Av : 1aa.

The dominant gene A completely suppresses the manifestation of the recessive gene a. Therefore, the phenotype of individuals with the AA and Aa genotypes will be the same: these individuals will carry a dominant trait. The ratio of individuals with different phenotypes will look like 3 : 1. Such a record means that among the offspring there is heterogeneity in phenotype and for three individuals with a dominant trait there is one individual with a recessive trait. Since individuals that were homogeneous in phenotype were crossed, it can be said that in the first generation there was a split in phenotype. When determining the probability of the appearance in the offspring of an individual with a certain phenotype, it is necessary to find the proportion of such individuals among all possible ones. In the example under consideration, the proportion of individuals with a dominant trait is 3 (1AA and 2Aa). The probability calculated as a percentage is 75%. Consider how the ratios of genotypes and phenotypes are recorded and the probability is calculated. Example 8.3. Normal skin pigmentation dominates albinism. What is the probability of an albino child in a family of heterozygous parents? sign : gene norm : And albinism : a

R Aa x Aa norm norm Gametes A; a A; a F 1 AA; Ah; Ah; aa genotype 1AA : 2Aa : 1aa phenotype 3 normal : 1 albinism P albinism = 1 = 1 or 25% 3+1 4

By recording the solution, it can be seen that the offspring will have the following 4 genotypes: AA, Aa, Aa, aa. Analysis by genotype shows that among the children of the first generation one is homozygous dominant (AA), two are heterozygous (Aa) and one is homozygous recessive. The digital ratio of genotypes will look like 1 : 2:1 . Phenotype analysis shows that children with the AA and Aa genotypes will have a dominant trait - normal skin pigmentation. The numerical ratio of children with normal pigmentation and albinism will be 3:1. Children with albinism account for ¼ of all hybrids. This means that the probability (p) of having albino children is ¼, or 25%.

The analysis of hybrids obtained by dihybrid crosses is similar to the description for multihybrid crosses. First, different variants of genotypes are also signed and the same variants are summarized. Variants of genotypes in this case can be very diverse: dihomozygous dominant, dihomozygous recessive, diheterozygous, monoheterozygous. After counting, the results are recorded as a numerical ratio in the same way as in the case of a monohybrid cross. Example 8.4. In humans, non-red hair color dominates over red, and freckles - over the norm. The genes that determine hair color and freckled skin are not linked. What percentage of non-red freckled children can be expected in the marriage of a man and a woman who are heterozygous for both traits? sign : non-red color gene : A red color : rather freckled : In normal skin : in P AaBv x AaBv non-redhead freckled freckled Gametes AB, AB, AB, AB AB, AB, AB, AB lattice cells - genotypes of individuals of the first generation. F 1

Analysis by genotype here is based on the calculation and compilation of numerical ratios of all possible variants of the genotypes of individuals, with the same genotypes summing them up. AT this case splitting by genotype will look like this:

1AABB : 2AABv : 1AAvv : 2AaBB : 4AaBv : 2Aavv : 1aaBB : 2ааВв : 1aavv

When analyzed by phenotype, they write out possible options genotypes of offspring with different combinations of coat color traits with coat length. Then, for each combination of traits, the corresponding genotypes are recorded. Therefore, having determined the phenotypic manifestation of each of the above-recorded genotypes, we will group all of them according to their belonging to some particular combination of traits:

Genotypes

non-red, freckled non-red, normal red, freckled red, normal

1AABB, 2AABB, 2AABB, 4AABB

1AAvv, 2Aavv 1aaBB, 2aaVv1aavv

The proportion of non-red freckled children will be 9 or 56.25%.

9. GENERAL FORMULA FOR SPLITTING

If the genes responsible for the manifestation of two non-alternative traits are located in non-homologous chromosomes, then during meiosis they will enter the gametes independently of each other. Therefore, when crossing two diheterozygotes, splitting according to the genotype:

1AABB:2AABv:1AAvv:2AaBB:4AaVv:2Aavv:1aaVv:2aaVv:1aavv

Is the result of two independent splits

1AA:2Aa:1aa and 1BB:2Vv:1vv

Mathematically, it can be expressed as a product:

(1АА:2Аа:1аа) and (1ВВ:2Вв:1вв)

(1: 2: 1) 2

Such a record shows that among individuals with the AA genotype, one part carries the BB genes, two parts carry the BB genes, and one part carries the cc genes. Similar ratios of individuals according to the set of B genes will be for the Aa and aa genotypes.

If individuals are analyzed for several traits, then the general splitting formula when crossing completely heterozygous individuals will be (1 : 2: 1) n , where n is the number of pairs of alternative features. Maximum value n is equal to the number of pairs of homologous chromosomes.

The situation is similar with splitting according to the phenotype. It is also based on monohybrid cleavage 3 : 1. When crossing two diheterozygous organisms for independent splitting 3 : 1 give splitting 9 : 3: 3: 1. The general formula for splitting by phenotype will be (3 : one). Knowing the splitting formulas, it is possible not to compose the Punnett lattice. Example 9.1. Some forms of cataracts and deafness in humans are transmitted as autosomal recessive traits. What is the probability of having a child with two anomalies in a family where both parents are heterozygous for both pairs of genes.

Solution:

sign: gene without cataract : A R AaBv x AaBv cataract : a healthy healthy without deafness : In deaf-mute : in F 1 aavv - ?

A child with two anomalies has the genotype - aavb. It is known that when two diheterozygotes are crossed among individuals of the first generation, splitting in phenotype is observed. The splitting formula looks like 9:3:3:1. From this formula it can be seen that 1 part falls to the share of an organism dihomozygous for recessive genes. This means that the probability of having a child with two anomalies is 1/16 or 6.25%.

You can approach the solution of problems for di- and polyhybrid crossing even easier. To do this, it is only necessary to know the probabilities of the manifestation of certain genotypes or phenotypes in various variants of monohybrid crossing. So, with monohybrid crossing of heterozygotes (Aa), the probability of the appearance of offspring with the AA genotype is ¼, with the Aa genotype - ½, with the aa I / 4 genotype, and the probability of the appearance of individuals with signs of the dominant gene is ¾, with the sign of the recessive gene - ¼. When crossing diheterozygotes (AaBv), the probability of the appearance of individuals with the aa genotype is - ¼, with the bb genotype - also ¼. To find the probability of a coincidence of two phenomena independent of each other, you need to multiply the probabilities of each of them among themselves. For example, to determine the probability of occurrence of individuals with two recessive traits (aavb), you need to multiply ¼ by ¼, which will give a total of 1/16. Example 9.2. In humans, the gene for brown eyes dominates over blue eyes, and the ability to use predominantly the right hand over left-handedness. Non-allelic genes are located on non-homologous chromosomes. A blue-eyed right-hander married a brown-eyed right-hander. They had two children: a brown-eyed left-hander and a blue-eyed right-hander. Determine the probability that their next child will be blue-eyed and predominantly left-handed. A brief statement of the task condition will look like this: sign: gene Brown eyes : A R A B x aa B blue eyes : and brown-eyed blue-eyed right-handedness : In right-handed right-handed left-handedness : in F 1 A cc aaB brown-eyed blue-eyed left-handed right-handed aavv - ?

Solution:

Before proceeding to calculate the probability of the birth of the next child with the genotype - aavb, it is necessary to determine the genotypes of their children. The presence in the genotype of children of two pairs of recessive genes (aa and cc) indicates that both parents carry genes a and c. Thus, the genotypes of the mother will be AaBv, the father - aaBv. The probability of the birth of the next child with a certain genotype does not depend on what characteristics already born children have. Therefore, the solution of the problem is reduced to determining the probability of birth for this pair of blue-eyed left-handers. P AaBv x aaBv p (aa) \u003d 1/2 p (vv) \u003d 1/4 p (aavv) \u003d 1/2 x ¼ \u003d 1/8

The probability of the appearance of the genotype aa when crossing individuals with the genotypes Aa and aa is ½, the probability of the appearance of the genotype bb when crossing BB and BB is ¼. Therefore, the probability of meeting these two genotypes in one organism will be ½ x ¼ = 1/8.

Answer: The probability of having a child with blue eyes in the family, owning mainly the left hand, is 1/8. This technique should be used only when the skill of ordinary problem solving is fully mastered.

10. ANALYSIS CROSS

Individuals showing dominant traits can be both homozygous and heterozygous, since they do not differ in phenotype. It is impossible to determine the genotype of such organisms by the phenotype. In these cases, an analyzing cross is performed and the genotype of the parent individuals is recognized by the phenotype of the offspring. In an analysis cross, an individual whose genotype is to be determined is crossed with a homozygous recessive individual. If from such a crossing all the offspring will be uniform, then the analyzed individual is homozygous. If ½ of the offspring shows a recessive trait, then the analyzed individual is heterozygous for one pair of allelic genes. Example 10.1 . In dogs, hard coat is dominant, soft coat is recessive. With whom should a wirehaired dog be crossed to find out its genotype? sign: gene rough wool : And soft wool : a If the dog has a coarse coat, then it is impossible to determine the genotype of the dog from the phenotype. It can be either homozygous (AA) or heterozygous (Aa). To determine the genotype of a wire-haired dog, it must be crossed with a soft-haired individual (aa), i.e., an analyzing cross should be made. 2 cases are possible. 1. R AA x aa 2. R Aa x aa hard soft hard soft wool wool wool wool G A; a G A; a a F1 according to the Aa genotype F1 according to the genotype 1Aa: 1aa hard hard soft wool wool wool according to the phenotype 100% according to the phenotype 50%: 50% In the first case, all offspring are homogeneous; in the second case, ½ of the offspring shows a dominant trait, and ½ - a recessive one. Thus, according to the result of crossing, one can judge the genotype of a wire-haired dog. If in her offspring all individuals will have hard hair, then she will be homozygous (AA), if in her offspring half of the individuals will have soft hair, and half will have hard hair, then she will be heterozygous (Aa).

11. HERITAGE OF CHARACTERISTICS WITH INCOMPLETE DOMINATION

In the cases discussed above, there is a complete suppression of one gene by another (complete dominance). In nature, incomplete dominance is often observed when the dominant gene does not completely suppress the action of the allelic gene in heterozygotes. Both genes function, and as a result, they have an intermediate trait that is different from the traits of individuals homozygous for either of these allelic genes. A gene with incomplete dominance is indicated by a dash above its symbolic designation - Ā. In the table for recording the conditions of the problem, the column "genotype" is entered. And throughout the entire time while this task is being solved, each time writing A, one should not forget to put this line, thereby demonstrating the relative (incomplete) dominance of this gene. No lines are placed above the recessive gene (a). Example 11.1. In humans, curly hair is determined by the A gene. Smooth hair is determined by the a gene. The A gene is not completely dominant over the a gene. The interaction of genes A and a gives wavy hair. Determine the phenotypes of F 1 children in the marriage of a man with curly hair and a smooth-haired woman. sign: Gene: Genotype curly hair : BUT : AA smooth hair : a : ah wavy hair : : Ah

Solution:

R ĀĀ x aa curly smooth Gametes Ā a F 1 genotype Āa phenotype wavy 100% In the first generation of offspring, all children are uniform, have wavy hair. Example 11.2. In humans, normal hemoglobin is determined by the A gene. Sickle cell anemia is determined by the a gene. The A gene is not completely dominant over the a gene. The interaction of genes A and a gives a sickle-shaped part of the erythrocytes. In the analyzed marriage, half of the children had partially sickle erythrocytes, and half had sickle cell anemia. Determine the genotype and phenotype of the parents. sign : Gene : Genotype normal hemoglobin : BUT : AA sickle cell anemia : a : aa Part of crescent-shaped erythrocytes : : Ah

Solution:

P Āa x aa part of erythrocytes sickle-shaped anemia Gametes Ā; a a F 1 genotype Āa; aa phenotype part anemia of erythrocytes sickle-shaped 50% 50% Since 50% of children have genes A and a, they should be present in parents as well. then the genotypes of the parents will look like: A and a. If we take into account that the other half of the children have only recessive genes, then we can fill the points in the genotypes of the parents with the genome a, then the parental genotypes will look like: Aa and aa.

12. INHERITANCE OF SEX-LINKED TRAITS

Unlike autosomes, sex chromosomes are not homologous, and they are often called heterosomes (from the Greek "heteros" - different, different). On the X chromosome it is known big number genes that control the development of somatic traits that are absent on the Y chromosome, and vice versa. Therefore, the female organism, on the basis of the similarity of their sex chromosomes, is homogametic, and according to the gene composition it can be homo- or heterozygous. The male (44+XY) organism is heterogametic, because forms two types of gametes on the basis of the content of sex chromosomes and hemizygous (Greek "hemi" - half), because the Y chromosome lacks many of the somatic trait genes found on the X chromosome. Therefore, recessive alleles of X-linked traits in men, as a rule, appear. If there is a gene on the Y chromosome that does not have an allelic counterpart on the X chromosome, the trait will only appear in males.

Educational and methodological complex of the discipline "Fundamentals of Genetics" Specialties

Training and metodology complex

Basic concepts and provisions of modern genetics. The human genome. Interaction of genes. reaction rate. The laws of heredity. Chromosomal theory of heredity.

Bulletin of new arrivals 2008 (7)

Bulletin

This Bulletin includes books received by all departments scientific library. The "Bulletin" is compiled on the basis of electronic catalog entries. Recordings were made in the RUSMARC format using the Ruslan program.

The gene pool of a population can be described either by gene frequencies or by genotype frequencies. Imagine that there are N diploid individuals in a population that differ in one pair of alleles (A and a); D - means the number of homozygotes for the dominant allele (AA); P is the number of homozygotes for the recessive allele (aa); H is the number of heterozygotes (Aa). Thus, in the population there will be three types of individuals with genotypes AA, Aa, aa, respectively. Since each individual with the AA genotype has two A alleles, and each Aa individual has one A allele, the total number of A alleles is 2D+H. Then p - the frequency of occurrence of the dominant allele A is equal to:

The frequency of the recessive allele (a) is usually denoted by q. The sum of the frequencies of genes A and a is equal to one, p+q=1, hence q=1-p. If a gene is represented by only two alleles (A and a) with a frequency of p and q, then what will be the frequencies of the three possible genotypes?

This question is answered by the Hardy-Weinberg law. At first glance, it may seem that individuals with a dominant phenotype will occur more often than with a recessive one. However, the 3:1 ratio is observed only in the offspring of two individuals heterozygous for the same alleles. Mendel's laws say nothing about the frequencies of genotypes and phenotypes in populations. They are discussed in the said law. It was formulated independently by the mathematician J. Hardy in England and the physician Wilhelm Weinberg in Germany. To understand the meaning of this law, suppose that males and females in a population interbreed randomly, or, equivalently, male and female gametes will combine randomly to form zygotes. In the zygote, the maternal and paternal chromosomes are combined, each of the homologous chromosomes carries one allele from this pair. The formation of individuals with the AA genotype is due to the probability of obtaining the A allele from the mother and the A allele from the father, i.e. pxp = p2.

Similarly, the emergence of the aa genotype, the frequency of occurrence of which is equal to q2. The Aa genotype can arise in two ways: the body receives the A allele from the mother, the a allele from the father, or, conversely, the A allele from the father, the a allele from the mother. The probability of both events is equal to pq, and the total probability of occurrence of the Aa genotype is equal to 2pq. Thus, the frequency of the three possible genotypes can be expressed by the equation: (p+q)2=p2+2pq+q2=1

It follows from the equation that if the crossing is random, then the genotype frequencies are related to the allele frequencies by simple ratios according to the Newton binomial formula.

Let's analyze an example when the allele frequencies of a given gene in a population are 0.1A; 0.3a (the geometric expression of the Hardy-Weinberg law for this case is shown in Fig. 21). In the offspring for 100 zygotes there will be 49 AA homozygotes, 9 aa homozygotes and 42 Aa heterozygotes, i.e. this corresponds to the ratio of genotypes already known to us - p2(AA) : 2pq(Ad): q2(aa).

Interestingly, in the next generation, gametes with the A allele will occur with a frequency of 0.7 (0.49 from AA homozygotes + 0.21 from Aa heterozygotes). This ratio will continue in the future. The frequencies of genes, and, accordingly, genotypes remain unchanged from generation to generation - this is one of the main provisions of the Hardy-Weinberg law. However, this law is of a probabilistic nature and therefore is realized in an infinitely large population. At the same time, gene frequencies remain unchanged if: there is unlimited panmixia; there is no natural selection; no new mutations of the same genes occur; there is no migration of individuals with other genotypes from neighboring populations.

1) homozygous for a recessive trait

2) homozygous for a dominant trait

3) heterozygous

4) forms two types of gametes

5) forms one type of gametes

6) clean line

6. An individual with the AA genotype:

1) homozygous for a recessive trait;

2) homozygous for a dominant trait;

3) heterozygous;

4) forms two types of gametes;

5) forms one type of gametes;

6) clean line;

7. Mendelian signs in humans include

2) blood pressure

3) a white strand of hair above the forehead

4) adherent earlobe

6) the ability to predominantly own the right hand

eight . Varieties of interallelic interaction of genes:

1) coding

2) epistasis

3) complementarity

4) complete dominance.

5) polymer

6) incomplete dominance

9. Match the genotypes of people with their blood types:

Genotypes: blood types:

1) I A I O A. first blood group

2) I O I O B. second blood type

3) I A I A B. third blood group

4) I B I O G. fourth blood group

Part 3:

SITUATIONAL TASKS

1. Determine the penetrance of the allele responsible for the manifestation of the trait, if 80 children were born carrying this gene, but phenotypically manifested in 30 descendants. a) 20% b) 75% c) 12% d) 10%

2. Can parents with achondroplasia (short bones, autosomal dominant trait) have a healthy child? If yes, with what probability?

a) yes, 25% b) yes, 50% c) yes, 75% d) no

3. A father with blood type MM has a child with blood types MN. What genotype can NOT be in the mother of the child?

a) NN b) MN c) MM

4. Below are the various combinations of blood group phenotypes of the parents and the child. Which of them are actually impossible?

FATHER MOTHER CHILD

a) AB A0 B

STANDARDS OF ANSWERS:

Part 1

Part 2

Part 3 1 - b 2 - c 3 - c 4 - c

The date ____________________

LAB #5

Topic: Patterns of inheritance of traits in di - and polyhybrid crossing. Independent inheritance of traits. Interaction of non-allelic genes

Purpose of the lesson :

on the basis of knowledge of the basic laws of Mendel and the forms of interaction of non-allelic genes, be able to predict the manifestation of traits in offspring.

Lesson objectives :

be able to solve problems for di- and polyhybrid crossing and for the interaction of non-allelic genes.

Tests to control the final level of knowledge (answer the questions proposed by the teacher).

OPTION #

1______ 6______

2______ 7______

3______ 8______

4______ 9______

5______ 10______

Number of points: _______

G. Mendel's laws apply to traits that are inherited monogenously with complete dominance. The genotype is a system of interacting genes. Interaction occurs between allelic and non-allelic genes located on the same and different chromosomes. The gene system forms a balanced genotypic environment that influences the function and expression of each gene. As a result, a certain phenotype of the organism is formed, all the signs of which are strictly coordinated in time, place and strength of manifestation. Physicians should draw up genetic schemes of inheritance of Mendelian and non-Mendelian traits and calculate the probability of their occurrence in offspring.

Questions for self-preparation:

1. The law of independent inheritance of traits.

2. Hybridological analysis for di- and polyhybrid crosses.

3. Conditions under which the third law of G. Mendel is observed and signs are inherited independently.

4. Non-allelic genes: definition, designation, location

5. Types of interaction of non-allelic genes. Hybridological analysis of the interaction of non-allelic genes.

6. Define complementarity. What traits are inherited in humans complementary.

7. Justify the phenomenon of epistasis.

8. What are the types of epistasis?

9. Epistatic (suppressors, inhibitors) and hypostatic (suppressed) genes. What signs in humans are inherited by the type of epistasis?

10. Explain the phenomenon of polymerization. What traits are polymerically inherited in humans?

11. Does the interaction between non-allelic genes violate the law of their independent inheritance?

12. Explain the mechanism of the "effect" of the position of genes, give examples of inheritance

signs in humans.

Basic terms

The last stage of solving problems is the analysis of individuals obtained by crossing. The purpose of the analysis depends on the question of the task:

1) What is the genotype and phenotype of the expected offspring?

2) What is the probability of the appearance of a carrier of a particular trait in the offspring?

To answer the first question, it is necessary to correctly solve the set genetic problem in literal terms and find all possible genotypes and phenotypes of the expected offspring.

The answer to the second question is related to the calculation of the probability of the manifestation of a feature. Probability can also be calculated as a percentage. In this case, the total number of offspring is taken as 100% and the percentage of individuals bearing the analyzed trait is calculated.

Analysis by genotype

Example 8.1. In monohybrid crossing, the number of homozygous dominant, homozygous recessive and heterozygous organisms is emphasized and their ratio is recorded.

1AA : 2Aa : 1aa

or easier

1 : 2 : 1

Phenotype analysis

Example 8.2. In a monohybrid crossing of two organisms with the Aa genotype. The record of the genotypes of the descendants of the first generation looked like

1AA : 2Av : 1aa.

The ratio of individuals with different phenotypes will look like 3 : 1. Such a record means that among the offspring there is heterogeneity in phenotype and for three individuals with a dominant trait there is one individual with a recessive trait. Since individuals that were homogeneous in phenotype were crossed, it can be said that in the first generation there was a split in phenotype.

When determining the probability of the appearance in the offspring of an individual with a certain phenotype, it is necessary to find the proportion of such individuals among all possible ones. In the example under consideration, the proportion of individuals with a dominant trait is 3 (1AA and 2Aa). The probability calculated as a percentage is 75%.

Consider how the ratios of genotypes and phenotypes are recorded and the probability is calculated.

Example 8.3. Normal skin pigmentation dominates albinism. What is the probability of an albino child in a family of heterozygous parents?

sign : gene

norm : BUT

albinism : a

norm norm

Gametes A; a A; a F 1 AA; Ah; Ah; aa

by genotype 1AA : 2Aa : 1aa

phenotype 3 normal : 1 albinism

P albinism = 1 = 1 or 25%

Example 8.4. In humans, non-red hair color dominates over red, and freckles - over the norm. The genes that determine hair color and freckled skin are not linked. What percentage of non-red freckled children can be expected in the marriage of a man and a woman who are heterozygous for both traits?

sign : gene

non-red color : BUT

ginger colour : a

freckling : AT

normal skin : in

P AaVv x AaVv

non-redhead

freckled freckled

Gametes AB, AB, AB, AB AB, AB, AB, AB

To determine the genotypes of individuals of the first generation, we use the Punnett lattice, in which we write down the female gametes horizontally, the male gametes vertically, and the genotypes of the first generation individuals in the lattice cells.

	AB	Av	aB	aw
AB	AABB	AAVv	AaBB	AaVv
Av	AAB in	AAvv	AaVv	aww
aB	AaBB	AaVv	aaBB	aawww
aw	AaVv	aww	aaBB	awav

Analysis by genotype here is based on the calculation and compilation of numerical ratios of all possible variants of the genotypes of individuals, with the same genotypes summing them up. In this case, the splitting by genotype will look like this:

1AABB : 2AABv : 1AAvv : 2AaBB : 4AaBv : 2Aavv : 1aaBB : 2ааВв : 1aavv

When analyzing by phenotype, possible variants of the genotypes of offspring are written out with various combinations of coat color traits with coat length. Then, for each combination of traits, the corresponding genotypes are recorded. Therefore, having determined the phenotypic manifestation of each of the above-recorded genotypes, we will group all of them according to their belonging to some particular combination of traits:

Phenotype	Genotypes	Total
non-red, freckled non-red, normal red, freckled red, normal	1AABB, 2AABB, 2AABB, 4AABB 1AABB, 2AABB 1AABB, 2AABB 1AABB

The proportion of non-red freckled children will be 9 or 56.25%.

End of work -

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All topics in this section:

General recommendations for solving genetic problems
When solving problems, the student must be free to navigate the basic genetic provisions, special terminology and alphabetic symbols. Allelic gametes are designated by the same b

Recording the genotypes of the parents
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Recording the gametes of the parents
Monohybrid cross Allelic genes are located on homologous chromosomes. During the formation of gametes, homologous chromosomes diverge into different gametes. Allelic genes found in

Formation of gametes in dihybrid crosses
In this case, it must be remembered that the alleles of two different genes can be located either in different or in the same pair of homologous chromosomes. For example, the chromosomal record of gamete formation in an organism

Pennet lattice
In order to make it easier to solve almost any problem in genetics, it is recommended to build the so-called Punnett lattice, which got its name from the English geneticist R.Pennett, for the first time

Recording the genotypes of individuals of the i-th generation
Individuals in the F1 generation are the result of crossing two parental organisms: male and female. Each of them can form a certain number of types of gametes. Each gamete is one

General splitting formulas
If the genes responsible for the manifestation of two non-alternative traits are located in non-homologous chromosomes, then during meiosis they will enter the gametes independently of each other. Therefore, when crossing two

Analyzing cross
Individuals showing dominant traits can be both homozygous and heterozygous, since they do not differ in phenotype. It is impossible to determine the genotype of such organisms by the phenotype. In these words

Inheritance of traits with incomplete dominance
In the cases discussed above, there is a complete suppression of one gene by another (complete dominance). In nature, incomplete dominance is often observed, when the dominant gene does not completely suppress

Inheritance of sex-linked traits
Unlike autosomes, sex chromosomes are not homologous, and they are often called heterosomes (from the Greek "heteros" - different, different). The X chromosome contains a large number of genes that control

The phenomenon of multiple allelism
Quite often, the same gene has not two (dominant and recessive) allelic varieties, but much more: A, a1, a2, a3, etc. They arise as a result of several mutations of one and

Population genetics
Population genetics considers patterns of distribution of dominant and recessive alleles and genotypes of diploid organisms within populations. Calculations are carried out in accordance with the law X

Recording a task response
Finally, the final stage of the process of solving the problem is the formulation of the answer, which should be extremely concise, absolutely accurate, not allowing for discrepancies. At the same time, it is necessary to remember

Encoding of hereditary information
A gene is a specific section of DNA, in the sequence nitrogenous bases which encoded the structure of some cellular products - proteins, t-RNA and r-RNA. DNA itself has the ability to

Genetic code (triplet and RNA)
First letter Second letter Third letter U C A G U Fen

Task number 27.
Human hair is dark and light, curly and smooth. The gene for dark hair is dominant, the gene for smooth hair is recessive. Both pairs of genes are on different chromosomes. 1. What kind of children can

Task number 36.
In humans, myopia (B) dominates over normal vision, and brown (A) eyes over blue. a) The only child of myopic brown-eyed parents has blue eyes and normal vision.

Syndromes with autosomal dominant inheritance
Aarsky syndrome. Brief description: ocular anomalies include ophthalmoplegia, strabismus, astigmatism, enlarged cornea, auricular anomalies 76%. Type of inheritance - autos

Syndromes with autosomal recessive inheritance
Enteropathic acrodermatitis. Brief description: severe stomatitis, photophobia, mental changes, schizoid type. The type of inheritance is autosomal recessive.

X-linked recessive inheritance syndromes
Hydrocephalus. Brief description: the face is small, the forehead hangs down, lack of coordination, mental retardation. Type of inheritance - X - linked recessive with stenosis with

X-linked dominant inheritance syndromes
Focal dermal hypolasia. Brief description: malocclusion, mixed deafness, mental retardation is characteristic. Type of inheritance - X - linked dominant.

Glossary of genetic terms
X-CHROMOSOMA - sex chromosome; in the karyotype of a woman it is presented twice, in the karyotype of a man there is only one X chromosome. Y-CHROMOSOME - a sex chromosome devoid of a homologous partner