Morgan's Law - linked inheritance. Linkage of inheritance of genes. Genetics of sex Complete linked inheritance

Chromosomal level of organization of hereditary material. Chromosomes as gene linkage groups.

It follows from the principles of genetic analysis that independent combination of traits can only be carried out under the condition that the genes that determine these traits are located in different pairs of chromosomes. Consequently, in each organism, the number of pairs of characters for which independent inheritance is observed is limited by the number of pairs of chromosomes. On the other hand, it is obvious that the number of characteristics and properties of an organism controlled by genes is extremely large, and the number of pairs of chromosomes in each species is relatively small and constant. It remains to be assumed that each chromosome contains not one gene, but many. If this is so, then it should be recognized that Mendel’s third rule concerns only the distribution of chromosomes, and not genes, i.e. its action is limited. Analysis of the manifestation of the third rule showed that in some cases new combinations of genes were completely absent in hybrids, i.e. complete linkage was observed between the genes of the original forms and a 1:1 split was observed in the phenotype. In other cases, a combination of traits was observed with less frequency than expected from independent inheritance.

In 1906, W. Betson described a violation of the Mendelian law of independent inheritance of two characters. Questions arose: why are not all traits inherited and how are they inherited, how are genes located on chromosomes, what are the patterns of inheritance of genes located on the same chromosome? The chromosomal theory of heredity, created by T. Morgan, in 1911, was able to answer these questions.

T. Morgan, having studied all the deviations, proposed to call the joint inheritance of genes, limiting their free combination, linkage of genes or linked inheritance.

Patterns of complete and incomplete coupling. Clutch groups in humans.

Research by T. Morgan and his school has shown that genes are regularly exchanged in a homologous pair of chromosomes. The process of exchange of identical sections of homologous chromosomes with the genes they contain is called chromosome crossing or crossing over. Crossing over occurs in meiosis. It provides new combinations of genes located on homologous chromosomes. The phenomenon of crossing over, like gene linkage, is characteristic of animals, plants, and microorganisms. The exceptions are male fruit flies and female silkworms. Crossing over ensures the recombination of genes and thereby significantly increases the role of combinative variability in evolution. The presence of crossing over can be judged by taking into account the frequency of occurrence of organisms with a new combination of characteristics. The phenomenon of crossing over was discovered by Morgan in Drosophila.

Recording the genotype of a diheterozygote with independent inheritance:

A IN

Recording the genotype of a diheterozygote with linked inheritance:

Gametes with chromosomes that have undergone crossing over are called crossover, and those that have not undergone are called non-crossover.

AB, AB AB, AB

Non-crossover gametes. Crossover gametes.

Accordingly, organisms that arise from a combination of crossover gametes are called crossovers or recombinants, and those arising from a combination of non-crossover gametes - non-crossovers or non-recombinants .

The phenomenon of crossing over, as well as the linkage of genes, can also be considered in the classic experiment of T. Morgan when crossing Drosophila.

	Sign	P♀ B.V. x♂ bv
	gray body color
	black body color
	normal wings
	vestigial wings
Analysis cross 1. Complete linkage of genes. 2. Incomplete linkage of genes.		1. Full grip P♀ bv x♂ B.V. F 2 bv bv splitting – 1:1
2. Incomplete traction (crossing over) P:♀ B.V. x♂ bv G: BV bv Bv bV bv non-crossover crossover F 2 B.V. bv Bv bV non-crossovers – 83% crossovers – 17%

To measure the distance between genes by test crossing, you can use the formula:

Where:

X– distance between genes in % crossing over or in morganids;

A– number of individuals of the 1st crossover group;

V– number of individuals of the 2nd crossover group;

n– total number of hybrids in the experiment;

100% – coefficient for conversion to percentage.

Based on a study of linked inheritance, Morgan formulated a thesis that was included in genetics under the name Morgan's rule : genes localized on the same chromosome are inherited linked, and the strength of linkage depends on the distance between them.

Linked genes are arranged in a linear order and the frequency of crossing over between them is directly proportional to the distance between them. However, this thesis is typical only for genes that are close to each other. In the case of relatively distant genes, some deviation from this dependence is observed.

Morgan proposed expressing the distance between genes as the percentage of crossing over between them. The distance between genes is also expressed in morganids or centimorganids. Morganidae is the genetic distance between genes where crossing over occurs with a frequency of 1%.

The frequency of crossing over between two genes can be used to infer the relative distance between them. So, if between genes A And IN crossing over is 3%, and between genes IN And WITH– 8% crossing over, then between A And WITH crossing over should occur at a frequency of either 3+8=11% or 8-3=5%, depending on the order in which these genes are located on the chromosome.

A ─ ─ ─ B ─ ─ ─ ─ ─ ─ ─ ─ C B ─ ─ ─ A ─ ─ ─ ─ ─ ─ ─ ─ C

Task 1. Cataracts and polydactyly are inherited as dominant autosomal traits. The woman inherited cataracts from her father and polydactyly from her mother. The genes are linked, the distance between them is 3M. What are the genotypes and phenotypes of the children from the marriage of this woman and a man normal for these characteristics? What is the probability of having healthy children?

cataract		P♀ aB x ♂ aw
polydactyly		X = AB = 3 Morgue.
P♀ aB x ♂ aw		Answer: the probability of having a healthy child is 1.5%, having one characteristic is 48.5%, having both characteristics is 1.5%
G: (аВ) (Ав) (ав)
F1 aB Av aw AB aw aw aw aw 48,5% 48,5% 1,5% 1,5%

Genetic map chromosomes is a diagram showing the order of genes at their relative distance from each other. The distance between linked genes is judged by the frequency of crossing over between them. Genetic maps of all chromosomes have been compiled for the most genetically studied organisms: Drosophila, chickens, mice, corn, tomatoes, Neurospora. Genetic maps of all 23 chromosomes have also been compiled for humans.

After establishing the linear discreteness of chromosomes, the need arose to compile cytological maps in order to compare them with genetic maps compiled on the basis of taking into account recombinations.

Cytological card is a map of a chromosome that determines the location and relative distance between genes on the chromosome itself. They are constructed based on the analysis of chromosomal rearrangements, differential coloring of polytene chromosomes, radioactive labels, etc.

To date, genetic and cytological maps have been constructed and compared for a number of plants and animals. The reality of this comparison confirms the correctness of the principle of the linear arrangement of genes on a chromosome.

In humans, some cases of linked inheritance can be named.

The genes that control the inheritance of ABO blood groups and nail and patella defect syndrome are inherited linked.

The genes for the Rh factor and the oval shape of red blood cells are linked.

The third autosome contains the genes for the Lutheran blood group and the secretion of antigens A and B with saliva.

The genes for polydactyly and cataracts are inherited linked.

The X chromosome contains the genes for hemophilia and color blindness, as well as the genes for color blindness and Duchenne muscular dystrophy.

Autosome 6 contains subloci A, B, C, D/DR of the HLA system, which control the synthesis of histocompatibility antigens.

Inheritance of X-linked and holandric traits.

Traits controlled by genes located on the sex chromosomes are called adhered to the floor. More than 60 sex-linked diseases have been described in humans, most of which are inherited recessively. Genes on sex chromosomes can be divided into 3 groups:

Genes partially linked to sex. They are located in paired segments X And Y chromosomes . Partially sex-linked diseases include: hemorrhagic diathesis, convulsive disorders, retinitis pigmentosa, xeroderma pigmentosa, and general color blindness.

Genes are completely sex-linked. They are located in the area X chromosome , for which there is no homologous region in Y chromosome (heterological). These genes control diseases: optic atrophy, Duchenne muscular dystrophy, color blindness, hemophilia, and the ability to smell hydrocyanic acid.

Genes located in the region Y chromosomes , for which there is no homologous locus in X chromosome are called holandric . They control symptoms: syndactyly, hypertrichosis of the auricle.

The color blindness gene occurs in 7% of men and 0.5% of women, but 13% of women are carriers of this gene.

Sex-linked inheritance was described by T. Morgan using the example of inheritance of the eye color trait in Drosophila.

Several patterns of inheritance of sex-linked traits have been noted:

passed cross to cross (from father to daughter, from mother to son);

the results of direct and back crossings do not coincide;

in the heterogametic sex, the trait manifests itself in any state (dominant or recessive).

Basic provisions of the chromosomal theory of heredity.

The main provisions of the chromosomal theory of heredity can be formulated as follows:

Genes are located on chromosomes. Each gene on a chromosome occupies a specific locus. Genes are arranged linearly on chromosomes.

Each chromosome represents a group of linked genes. The number of linkage groups in each species is equal to the number of pairs of chromosomes.

Allelic genes are exchanged between homologous chromosomes—crossing over.

The distance between genes on a chromosome is proportional to the percentage of crossing over between them. Knowing the distance between genes, you can calculate the percentage of genotypes in the offspring.

Chained inheritance. Independent distribution of genes (Mendel's second law) is based on the fact that genes belonging to different alleles are located in different pairs of homologous chromosomes. The question naturally arises: how will the distribution of different (non-allelic) genes occur in a number of generations if they lie in the same pair of chromosomes? This phenomenon must take place, because the number of genes is many times greater than the number of chromosomes. Obviously, the law of independent distribution (Mendel's second law) does not apply to genes located on the same chromosome. It is limited only to those cases where the genes of different alleles are on different chromosomes.

Pattern of inheritance when genes are found on the same chromosome, it was carefully studied by T. Morgan and his school. The main object of research was the small fruit fly Drosophila

This insect is extremely convenient for genetic work. The fly is easily bred in the laboratory, produces a new generation every 10–15 days at its optimal temperature of 25–26° C, has numerous and varied hereditary characteristics, and has a small number of chromosomes (8 in the diploid set).

Experiments have shown that genes localized on one chromosome are linked, i.e., they are inherited predominantly together, without showing independent distribution. Let's look at a specific example. If you cross a Drosophila with a gray body and normal wings with a fly that has a dark body color and rudimentary wings, then in the first generation all the flies will be gray with normal wings. This is a heterozygote for two pairs of alleles (gray body - dark body and normal wings - rudimentary wings). Let's crossbreed. Let us cross the females of these diheterozygous flies (gray body and normal wings) with males possessing recessive traits - a dark body and rudimentary wings. Based on the second, one would expect to obtain four flies in the offspring: 25% gray, with normal wings; 25% gray, with rudimentary wings; 25% dark, with normal wings; 25% dark, with rudimentary wings.

In fact, in the experiment there are significantly more flies with the original combination of characteristics (gray body - normal wings, dark body - rudimentary wings) (in this experiment, 41.5%) than flies with recombined characters (gray body - rudimentary wings and dark body - normal wings).

There will be only 8.5% of each type. This example shows that the genes that determine the characteristics of a gray body - normal wings and a dark body - rudimentary wings are inherited predominantly together, or, in other words, turn out to be linked with each other. This linkage is a consequence of the localization of genes on the same chromosome. Therefore, during meiosis, these genes do not separate, but are inherited together. The phenomenon of linkage of genes localized on the same chromosome is known as Morgan's law.

Why, after all, among the second generation hybrids do a small number of individuals appear with recombination of parental characteristics? Why is gene linkage not absolute? Research has shown that this recombination of genes is due to the fact that during the process of meiosis, during the conjugation of homologous chromosomes, they sometimes exchange their sections, or, in other words, crossover occurs between them.

It is clear that in this case, genes that were originally located in one of two homologous chromosomes will end up in different homologous chromosomes. There will be a recombination between them. The crossover frequency is different for different genes. It depends on the distance between them. The closer the genes are located on the chromosome, the less often they are separated during crossover. This happens because chromosomes exchange different regions, and genes that are closely located are more likely to end up together. Based on this pattern, it was possible to construct genetic maps of chromosomes for well-studied organisms, on which the relative distance between genes is plotted.

The biological significance of chromosome crossing is very great. Thanks to it, new hereditary combinations of genes are created, hereditary variability increases, which supplies material for.

Genetics of sex. It is well known that in dioecious organisms (including humans) the sex ratio is usually 1:1. What reasons determine the sex of a developing organism? This question has long been of interest to humanity due to its great theoretical and practical significance. The chromosome set of males and females in most dioecious organisms is not the same. Let's get acquainted with these differences using the example of the set of chromosomes in Drosophila.

Males and females do not differ from each other in three pairs of chromosomes. But for one couple there are significant differences. The female has two identical (paired) rod-shaped chromosomes; The male has only one such chromosome, the pair of which is a special, double-armed chromosome. Those chromosomes in which there are no differences between males and females are called autosomes. The chromosomes on which males and females differ from each other are called sex chromosomes. Thus, the chromosome set of Drosophila consists of six autosomes and two sex chromosomes. The sex, rod-shaped chromosome, present in a double number in a female, and in a single number in a male, is called the X chromosome; the second, sexual (two-armed chromosome of the male, absent in the female) - the Y chromosome.

How are the considered sex differences in the chromosome sets of males and females maintained in the process? To answer this question, it is necessary to clarify the behavior of chromosomes in meiosis and during fertilization. The essence of this process is presented in the figure.

During the maturation of germ cells in a female, each egg cell, as a result of meiosis, receives a set of four chromosomes: three autosomes and one X chromosome. Males produce two types of sperm in equal quantities. Some carry three autosomes and an X chromosome, others carry three autosomes and a Y chromosome. During fertilization, two combinations are possible. An egg can be equally likely to be fertilized by a sperm with an X or Y chromosome. In the first case, a female will develop from a fertilized egg, and in the second, a male. The sex of an organism is determined at the time of fertilization and depends on the chromosome complement of the zygote.

In humans, the chromosomal mechanism for determining sex is the same as in Drosophila. The diploid number of human chromosomes is 46. This number includes 22 pairs of autosomes and 2 sex chromosomes. In women there are two X chromosomes, in men there is one X and one Y chromosome.

Accordingly, men produce sperm of two types - with X- and Y-chromosomes.

In some dioecious organisms (for example, some insects), the Y chromosome is completely absent. In these cases, the male has one less chromosome: instead of the X and Y chromosomes, he has one X chromosome. Then, during the formation of male gametes during meiosis, the X chromosome does not have a partner for conjugation and goes into one of the cells. As a result, half of all sperm have an X chromosome, while the other half lack it. When an egg is fertilized by sperm with an X chromosome, a complex with two X chromosomes is obtained, and a female develops from such an egg. If an egg is fertilized by a sperm without an X chromosome, an organism with one X chromosome (received through the egg from the female) will develop, which will be a male.

In all the examples discussed above, sperm of two categories develop: either with the X and Y chromosomes (Drosophila, humans), or half of the sperm carry the X chromosome, and the other is completely devoid of it. Eggs are all the same in terms of sex chromosomes. In all these cases we have male heterogamety (different gamety). The female sex is homogametic (equal gametic). Along with this, another type of sex determination occurs in nature, characterized by female heterogamety. Here the opposite relationships to those just discussed take place. Different sex chromosomes or only one X chromosome are characteristic of the female sex. The male sex has a pair of identical X chromosomes. Obviously, in these cases, female heterogamety will occur. After meiosis, two types of egg cells are formed, while with regard to the chromosomal complex, all sperm are the same (all carry one X chromosome). Consequently, the sex of the embryo will be determined by which egg - with an X or Y chromosome - will be fertilized.

1. There are two types of blindness in humans, and each is determined by its own recessive autosomal gene, which are not linked. What is the probability of having a blind child if the father and mother suffer from the same type of blindness and are both dihomozygous? What is the probability of having a blind child if both parents are dihomozygous and suffer from different types of hereditary blindness?

Explanation:

First crossing:

R: AAbb x AAbb

G: Av x Av

F1: AAbb is a blind child.

The law of uniformity appears. The probability of having a blind child is 100%.

Second crossing:

R: ААвв x ааВВ

G: Av x aV

F1: AaBv is a healthy child.

The law of uniformity appears. Both types of blindness are absent. The probability of having a blind child is 0%.

2. In humans, color blindness is caused by a recessive gene linked to the X chromosome. Thalassemia is inherited as an autosomal dominant trait and is observed in two forms: in homozygotes it is severe, often fatal, in heterozygotes it is mild.

A woman with a mild form of thalassemia and normal vision, married to a man who is colorblind but healthy for the thalassemia gene, has a colorblind son with a mild form of thalassemia. What is the probability of this couple having children with both anomalies? Determine the genotypes and phenotypes of possible offspring.

Explanation:

R: AaХDХd x aaХdУ

D: АХD, аХd, AXd, aXD x аХd, аУ

F1: АаХдУ - colorblind boy with a mild form of thalassemia

AaXDXd - girl with normal vision and mild thalassemia

aaXdXd - colorblind girl without thalassemia

AaXdXd - colorblind girl with mild thalassemia

aaXDХd - girl with normal vision without thalassemia

AaXDY - boy with normal vision and mild thalassemia

aaXdY - colorblind boy without thalassemia

aaXDY - boy with normal vision and without thalassemia

That is, there are eight variants of the genotype with an equal probability of occurrence. The chance of having a child with mild thalassemia and color blindness is 2/8 or 25% (12.5% ​​chance of having a boy and 12.5% ​​chance of having a girl). The probability of having a colorblind child with severe thalassemia is 0%.

3. A blue-eyed, fair-haired man and a diheterozygous brown-eyed, dark-haired woman entered into marriage. Determine the genotypes of the married couple, as well as the possible genotypes and phenotypes of the children. Establish the probability of having a child with a dihomozygous genotype.

Explanation: A - brown eyes

a - blue eyes

B - dark hair

c - blond hair

R: aavv x AaVv

G: av x AB, av, Av, aV

F1: AaBv - brown eyes, dark hair

aavv - blue eyes, blond hair

Aaav - brown eyes, blond hair

aaVv - blue eyes, dark hair

The probability of having a child with each genotype is 25%. (and the probability of having a child with a dihomozygous genotype (aabv) is 25%)

Signs are not gender-linked. This is where the law of independent inheritance comes into play.

4. When crossing a gray (a) shaggy rabbit with a black shaggy rabbit, a split was observed in the offspring: black shaggy rabbits and gray shaggy rabbits. In the second crossing of phenotypically the same rabbits, the following offspring were produced: black shaggy rabbits, black smooth-haired rabbits, gray shaggy-haired rabbits, gray smooth-haired rabbits. What law of heredity is manifested in these crosses?

Explanation:

A - black color

a - gray color

B - furry rabbit

c - smooth-haired rabbit

First crossing:

P: aaBB x AaBB

F1: AaBB - black furry rabbits

aaBB - gray furry rabbits

Second crossing:

P: aaBv x AaBv

G: aV, av x AB, av, Av, aV

F1: 8 genotypes and 4 phenotypes are obtained

AaBB, 2AaBB - gray furry rabbits

Aavv - black smooth-haired rabbits

aaBB, aaBB - gray furry rabbits

aavv - gray smooth-haired rabbits

In this case, the law of independent inheritance applies, since the presented characteristics are inherited independently.

5. An analytical cross was carried out for the crested (A) green (B) female, and four phenotypic classes were obtained in the offspring. The resulting crested offspring were crossed with each other. Can this cross produce offspring without a tuft? If so, what gender and phenotype will it be? In canaries, the presence of a crest depends on an autosomal gene, and the color of the plumage (green or brown) depends on a gene linked to the X chromosome. The heterogametic sex in birds is the female.

Explanation:

First crossing:

R: AaHVU x aaHvHv

G: AHV, aHV, AU, aU x aHv

F1: AaХВХв - tufted green male

aaХВХв - green male without crest

AaHvU - tufted brown female

We cross a male and a female with a crest:

R: AaХВХв x АаХвУ

G: AXB, AXB, AXB, AXB x AXB, AU, AXB, AU

F2: we get 16 genotypes, among which only 4 phenotypes can be distinguished.

Phenotypes of individuals without a crest:

Females: aaХВУ - green female without crest

aaHvU - brown female without crest

Males: aaХВХв - green male without crest

aaХвХв - brown male without a crest.

6. In crossing female fruit flies with normal wings and normal eyes and males with reduced wings and small eyes, all offspring had normal wings and normal eyes. The resulting females in the first generation were backcrossed with the original parent. The shape of the wings in Drosophila is determined by an autosomal gene, the gene for eye size is located on the X chromosome. Draw up crossbreeding schemes, determine the genotypes and phenotypes of parental individuals and offspring in crosses. What laws apply to crossbreeding?

Explanation:

A - normal wings

a - reduced wings

HB - normal eyes

First crossing:

R: ААХВХВ x ввХвУ

G: АХВ x аХВ, аУ

AaХВХв - normal wings, normal eyes

AaHVU - normal wings, normal eyes

Second crossing:

R: AaХВХв x ааХвН

G: AHB, aHv, AHv, aHv x aHv, aU

AaХВХв, АаХВУ - normal wings, normal eyes

aaХвХв, ааХвУ - reduced wings, small eyes

AaHvHv, AaHvU - normal wings, small eyes

aaХВХв, ааХВУ - reduced wings, normal eyes

The law of sex-linked inheritance applies here (the eye shape gene is inherited with the X chromosome), and the wing gene is inherited independently.

7. When crossing a Drosophila fly with a gray body (A) and normal wings (B) with a fly with a black body and curled wings, 58 flies with a gray body and normal wings were obtained, 52 flies with a black body and curled wings, 15 - with a gray body and curled wings, 14 - with a black body and normal wings. Make a diagram for solving the problem. Determine the genotypes of the parents and offspring. Explain the formation of four phenotypic classes. What law applies in this case?

Explanation: A - gray body

a - black body

B - normal wings

c - curled wings

Crossing: P: AaBv x aavv

G: AB, AB, AB, AB x AB

F1: AaBv - gray body, normal wings - 58

aavv - black body, curled wings - 52

Aavv - gray body, curled wings - 15

aaВв - black body, normal wings - 14

Genes A and B and a and b are linked, so they form groups of 58 and 52 individuals, and in the case of the other two groups, crossing over occurred and these genes ceased to be linked, which is why they formed 14 and 15 individuals.

8. When analyzing the crossing of a diheterozygous tall tomato plant with round fruits, a splitting of the offspring according to phenotype was obtained: 38 tall plants with rounded fruits, 10 tall plants with pear-shaped fruits, 10 dwarf plants with rounded fruits, 42 dwarf plants with pear-shaped fruits. Draw up a crossbreeding scheme, determine the genotypes and phenotypes of the original individuals and offspring. Explain the formation of four phenotypic classes.

Explanation:

A - tall plant

a - dwarf plant

B - round fruits

c - pear-shaped fruits

R: AaVv x aavv

G: AB, AB, AB, AB x AB

F1: AaBv - tall plants with round fruits - 38

aavv - dwarf plants with pear-shaped fruits - 42

aaBv - dwarf plants with round fruits - 10

Aavv - tall plants with pear-shaped fruits - 10

Here two groups of signs can be distinguished:

1. AaBb and aabv - in the first case, A and B are inherited concatenated, and in the second - a and b.

2. aaBv and Aavv - crossing over occurred here.

9. In humans, non-red hair dominates red hair. Father and mother are heterozygous redheads. They have eight children. How many of them might be redheads? Is there a clear answer to this question?

Explanation: A - non-red hair

a - red hair

R: Aa x Aa

G: A, a x A, a

F1: AA: 2Aa: aa

Genotype splitting is 1:2:1.

Phenotype splitting is 3:1. Therefore, the probability of having a non-red-haired child is 75%. The probability of having a red-haired child is 25%.

There is no definite answer to the question, since it is impossible to guess the genotype of the unborn child, since germ cells with different genotypes may occur.

10. Determine the genotypes of the parents in a family where all the sons are color blind and the daughters are healthy.

Explanation: XDXd - healthy girl

XdY - boy - colorblind

This situation will be more possible if the mother is colorblind (since the female sex is homogametic) and the father is healthy (heterogametic sex).

Let's write a crossing diagram.

P: XdXd x XDY

G: Xd x XD, Y

F1: XDXd - a healthy girl, but a carrier of the color blindness gene.

XdY - colorblind boy

11. In humans, glaucoma is inherited as an autosomal recessive trait (a), and Marfan syndrome, accompanied by an anomaly in the development of connective tissue, is inherited as an autosomal dominant trait (B). The genes are located in different pairs of autosomes. One of the spouses suffers from glaucoma and had no ancestors with Marfan syndrome, and the second is diheterozygous for these characteristics. Determine the genotypes of the parents, possible genotypes and phenotypes of the children, and the likelihood of having a healthy child. Make a diagram for solving the problem. What law of heredity is manifested in this case?

Explanation: glaucoma is a recessive trait and manifests itself only in homozygotes, and Marfan syndrome manifests itself in both hetero- and homozygotes, but is a dominant trait; accordingly, we will determine the genotypes of the parents: one parent suffers from glaucoma - aa, but does not suffer from Marfan syndrome - bb, and the second parent is heterozygous for both characteristics - AaBb.

R: aavv x AaVv

G: av x AB, aB, Aw, aB

F1: AaBv - normal vision + Marfan syndrome

aavv - glaucoma

AAVV - normal vision, no Marfan syndrome - healthy child

aaBv - glaucoma + Marfan syndrome

By drawing a Punnett grid, you can see that the probability of giving birth to each child is the same - 25%, which means the probability of having a healthy child will be the same.

The genes for these traits are not linked, which means the law of independent inheritance is manifested.

12. We crossed short (dwarf) tomato plants with ribbed fruits and plants of normal height with smooth fruits. In the offspring, two phenotypic groups of plants were obtained: low-growing and smooth fruits and normal height with smooth fruits. When low-growing tomato plants with ribbed fruits were crossed with plants that had normal stem height and ribbed fruits, all offspring had normal stem height and ribbed fruits. Make crossbreeding schemes. Determine the genotypes of the parents and offspring of tomato plants in two crosses. What law of heredity is manifested in this case?

Explanation: in the first crossing, a digomozygote is crossed with a plant that is homozygous for one trait and heterozygous for another (to understand this, you need to write several options; this offspring is obtained only with such parents). in the second crossing everything is simpler - two digomozygotes are crossed (only in the second parent one trait will be dominant).

a - short individuals

A - normal height

c - ribbed fruits

B - smooth fruits

P: aavv x AaBB

F1: aaBv - short individuals with smooth fruits

AaBv - normal height, smooth fruits

P: aavv x AAvv

F1: Aavv - normal height, smooth fruits.

In both cases, the law of independent inheritance manifests itself, since these two characteristics are inherited independently.

13. Based on the pedigree shown in the figure, determine and explain the nature of inheritance of the trait highlighted in black. Determine the genotypes of parents and offspring, indicated in the diagram by numbers 2, 3, 8, and explain their formation.

Explanation: since in the first generation we see uniformity, and in the second generation - 1:1 splitting, we conclude that both parents were homozygous, but one for a recessive trait, and the other for a dominant one. That is, in the first generation all children are heterozygous. 2 - Aa, 3 - Aa, 8 - aa.

14. When a pied crested (B) hen was crossed with the same rooster, eight chickens were obtained: four pied crested chickens, two white (a) crested chickens and two black crested chickens. Make a diagram for solving the problem. Determine the genotypes of parents and offspring, explain the nature of inheritance of traits and the appearance of individuals with variegated colors. What laws of heredity are manifested in this case?

Explanation: such splitting is possible only if the parents are heterozygous in color, that is, the variegated color has the genotype - Aa

AA - black color

aa - white color

Aa - variegated color

P: AaBB x AaBB

G: AB, aB

F1: AaBB - mottled crested (4 chicks)

aaBB - white crested (two chickens)

AABB - black crested

In terms of color, the split between genotype and phenotype is the same: 1:2:1, since there is a phenomenon of incomplete dominance (an intermediate variant appears between black and white coloration), the characters are inherited independently.

15. In humans, the gene for normal hearing (B) dominates over the gene for deafness and is located in the autosome; The gene for color blindness (color blindness - d) is recessive and linked to the X chromosome. In a family where the mother suffered from deafness but had normal color vision, and the father had normal hearing (homozygous) and was color blind, a color blind girl with normal hearing was born. Make a diagram for solving the problem. Determine the genotypes of the parents, daughters, possible genotypes of the children and the likelihood of future births in this family of color-blind children with normal hearing and deaf children.

Explanation: from the conditions of the problem it is clear that the mother is heterozygous for the deafness gene and homozygous for the blindness gene, and the father has the blindness gene and is heterozygous for the deafness gene. Then the daughter will be homozygous for the blindness gene and heterozygous for the deafness gene.

P: (mother)XDXd x (father)XdYBB

daughter - XdXdBb - colorblind, normal hearing

Gametes - XDb, Xdb, XdB, YB

Children: XDXdBb - normal vision, normal hearing

XDYBb - normal vision, normal hearing

XdXdBb - colorblind, normal hearing

XdYBb - colorblind, normal hearing

Splitting: 1:1:1:1, that is, the probability of being born colorblind with normal hearing is 50%, and the probability of being born deaf colorblind is 0%.

16. The husband and wife have normal vision, despite the fact that the fathers of both spouses suffer from color blindness (color blindness). The color blindness gene is recessive and linked to the X chromosome. Determine the genotypes of the husband and wife. Make a diagram for solving the problem. What is the probability of having a son with normal vision, a daughter with normal vision, a color-blind son, and a color-blind daughter?

Explanation: Let's say the mothers of the husband and wife were healthy.

Let us also describe the possible genotypes of the parents of the husband and wife.

P: XDXD x XdY XDXD x XdY

↓ ↓

XDXd x XDY

↓

Possible genotypes of children:

XDXD - healthy girl

XDY - healthy boy

XDXd - healthy girl

XdY - colorblind boy

The probability of having a child with each genotype is 25%. The probability of having a healthy girl is 50% (in one case the child is heterozygous, in the other - homozygous). The probability of having a colorblind girl is 0%. The probability of having a colorblind boy is 25%.

17. In common peas, the yellow color of the seeds dominates over the green, the convex shape of the fruits dominates over the constricted fruits. When crossing a plant with yellow convex fruits with a plant with yellow seeds and constricted fruits, 63 plants with yellow seeds and convex fruits were obtained. 58 - with yellow seeds and constricted fruits, 18 - with green seeds and convex fruits, and 20 - with green seeds and constricted fruits. Make a diagram for solving the problem. Determine the genotypes of the original plants and descendants. Explain the emergence of different phenotypic groups.

Explanation:

A - yellow color

a - green color

B - convex shape

c - fruits with constriction

Having carefully read the conditions of the problem, you can understand that one parent plant is diheterozygous, and the second is homozygous for the shape of the fruit, and heterozygous for the color of the seed.

Let's write a diagram for solving the problem:

P: AaBv x Aavv

G: AB, AB, AB, AB x Av, AB

F1: This results in a 3:1 split and the following first generation offspring:

63 - А_Вв - yellow seeds, convex fruits

58 - А_вв - yellow seeds, fruits with constriction

18 - aaВв - green seeds, convex shape of the fruit

20 - aavv - green seeds, fruits with constriction

Here we observe the law of independent inheritance, since each trait is inherited independently.

18. In snapdragon, the red color of the flowers is incompletely dominant over the white, and the narrow leaves are over the wide ones. Genes are located on different chromosomes. Plants with pink flowers and leaves of intermediate width are crossed with plants with white flowers and narrow leaves. Make a diagram for solving the problem. What offspring and in what ratio can be expected from this cross? Determine the type of crossing, genotypes of parents and offspring. What law applies in this case?

Explanation: AA - red color

Aa - pink color

aa - white color

BB - narrow leaves

BB - leaves of intermediate width

bb - wide leaves

Crossing:

R: AaBB x aaBB

G: AB, AB, AB, AB x AB

F1: AaBB - pink flowers, narrow leaves

aaВв - white flowers, leaves of intermediate width

AaBv - pink flowers, leaves of intermediate width

aaBB - white flowers, narrow leaves

The probability of flowers appearing with each genotype is 25%.

The crossing is dihybrid (since the analysis is based on two characteristics).

In this case, the laws of incomplete dominance and independent inheritance of traits apply.

Tasks for independent solution

1. In dogs, black hair dominates over brown, and long hair dominates over short hair (genes are not linked). The following offspring were obtained from a black long-haired female during testing crossing: 3 black long-haired puppies, 3 brown long-haired puppies. Determine the genotypes of parents and offspring that correspond to their phenotypes. Make a diagram for solving the problem. Explain your results.

2. In sheep, gray wool color (A) dominates over black, and hornedness (B) dominates over polled (hornless). The genes are not linked. In the homozygous state, the gray color gene causes the death of embryos. What viable offspring (by phenotype and genotype) and in what ratio can be expected from crossing a diheterozygous sheep with a heterozygous gray polled male? Draw up a diagram for solving hadachi. Explain your results. What law of heredity is manifested in this case?

3. In corn, the recessive gene “shortened internodes” (b) is located on the same chromosome with the recessive gene “incipient panicle” (v). When conducting an analytical crossing of a diheterozygous plant with normal internodes and a normal panicle, the following offspring were obtained: 48% with normal internodes and a normal panicle, 48% with shortened internodes and a rudimentary panicle, 2% with normal internodes and a rudimentary panicle, 2% with shortened internodes and a rudimentary panicle. a normal panicle. Determine the genotypes of parents and offspring. Make a diagram for solving the problem. Explain your results. What law of heredity is manifested in this case?

4. When a corn plant with smooth, colored seeds was crossed with a plant that produces wrinkled, uncolored seeds (the genes are linked), the offspring ended up with smooth, colored seeds. When analyzing the crossing of hybrids from F1, plants with smooth colored seeds, with wrinkled uncolored seeds, with wrinkled colored seeds, and with smooth uncolored seeds were obtained. Make a diagram for solving the problem. Determine the genotypes of parents, offspring F1 and F2. What laws of heredity are manifested in these crosses? Explain the appearance of four phenotypic groups of individuals in F2?

5. When a corn plant with smooth, colored seeds was crossed with a plant with wrinkled, uncolored seeds (the genes are linked), the offspring ended up with smooth, colored seeds. With further analytical crossing of the hybrid from F1, plants with seeds were obtained: 7115 with smooth colored seeds, 7327 with wrinkled uncolored seeds, 218 with wrinkled colored seeds, 289 with smooth uncolored ones. Make a diagram for solving the problem. Determine the genotypes of parents, offspring F1, F2. What law of heredity is manifested in F2? Explain the basis for your answer.

6. In humans, cataracts (an eye disease) depend on a dominant autosomal gene, and ichthyosis (a skin disease) depends on a recessive gene linked to the X chromosome. A woman with healthy eyes and normal skin, whose father suffered from ichthyosis, marries a man with cataracts and healthy skin, whose father did not have these diseases. Make a diagram for solving the problem. Determine the genotypes of the parents, possible genotypes and phenotypes of the children. What laws of heredity are manifested in this case?

7. When crossing white rabbits with shaggy hair and black rabbits with smooth hair, the following offspring were obtained: 50% black shaggy and 50% black smooth. When another pair of shaggy-haired white rabbits was crossed with smooth-haired black rabbits, 50% of the offspring were black shaggy and 50% were white shaggy. Make a diagram of each cross. Determine the genotypes of parents and offspring. Explain what law is manifested in this case?

8. When a watermelon plant with long striped fruits was crossed with a plant with round green fruits, the offspring were plants with long green and round green fruits. When the same watermelon with long striped fruits was crossed with a plant that had round striped fruits, all the offspring had round striped fruits. Make a diagram of each cross. Determine the genotypes of parents and offspring. What is this crossing called and why is it carried out?

9. A dark-haired, blue-eyed woman, dihomozygous, married a dark-haired, blue-eyed man, heterozygous for the first allele. Dark hair color and brown eyes are dominant signs. Determine the genotypes of parents and offspring, types of gametes and probable genotypes of children.

10. A dark-haired woman with curly hair, heterozygous for the first allele, married a man with dark, smooth hair, heterozygous for the first allele. Dark and curly hair are dominant characteristics. Determine the genotypes of the parents, the types of gametes they produce, and the likely genotypes and phenotypes of the offspring.

11. A dark-haired, brown-eyed woman, heterozygous for the first allele, married a fair-haired, brown-eyed man, heterozygous for the second allele. Dark hair and brown eyes are dominant traits, while blond hair and blue eyes are recessive traits. Determine the genotypes of the parents and the gametes they produce, the likely genotypes and phenotypes of the offspring.

12. We crossed a red-eyed gray (A) Drosophila, heterozygous for two alleles, with a red-eyed black (XB) Drosophila, heterozygous for the first allele. Determine the genotypes of the parents, the gametes they produce, and the numerical ratio of the splitting of the offspring by genotype and phenotype.

13. A black hairy rabbit, heterozygous for two alleles, was crossed with a white hairy rabbit, heterozygous for the second allele. Black shaggy fur is dominant traits, white smooth fur is recessive traits. Determine the genotypes of the parents and the gametes they produce, the numerical ratio of the cleavage of the offspring by phenotype.

14. The mother has the 3rd blood group and a positive Rh factor, and the father has the 4th blood group and a negative Rh factor. Determine the genotypes of the parents, the gametes they produce, and the possible genotypes of the children.

15. One tortoiseshell and several black kittens were born from a black cat. These characteristics are sex-linked, that is, color genes are found only on the sex X chromosomes. The black color gene and the red color gene give incomplete dominance; when these two genes are combined, a tortoiseshell color is obtained. Determine the genotype and phenotype of the father, the gametes produced by the parents, and the sex of the kittens.

16. A heterozygous gray female Drosophila was crossed with a gray male. These traits are sex-linked, that is, the genes are found only on the sex X chromosomes. Gray body color dominates yellow. Determine the genotypes of the parents and gametes. which they produce, and the numerical splitting of offspring by sex and body color.

17. In tomato, the genes that cause tall plant growth (A) and the round shape of the fruit (B) are linked and localized on one chromosome, and the genes that cause short stature and a pear-shaped shape are in the autosome. A heterozygous tomato plant, which has a tall stature and a round fruit shape, was crossed with a short pear-bearing plant. Determine the genotypes and phenotypes of the offspring of the parents, gametes formed in meiosis, if there was no crossing of chromosomes.

18. In Drosophila, the dominant genes for a normal wing and gray body color are linked and localized on one chromosome, and the recessive genes for the rudimentary wing and black body color are in another homologous chromosome. Two diheterozygous fruit flies with normal wings and gray body color were crossed. Determine the genotype of the parents and gametes formed without crossing chromosomes, as well as the numerical ratio of the splitting of the offspring by genotype and phenotype.

19. What are the genotypes of the parents and children if a fair-haired mother and a dark-haired father had five children, all dark-haired? What law of heredity is manifested?

20. What are the genotypes of the parents and offspring if, from crossing a cow with a red coat color with a black bull, all the offspring are black? Identify dominant and recessive genes and patterns of dominance.

21. What phenotypes and genotypes are possible in children if the mother has the first blood group and a homozygous Rh-positive factor, and the father has the fourth blood group and the Rh-negative factor (recessive trait)? Determine the probability of having children with each of these characteristics.

22. A blue-eyed child was born into the family, similar in this feature to his father. The child's mother is brown-eyed, his maternal grandmother is blue-eyed, and his grandfather is brown-eyed. On the paternal side, grandparents are brown-eyed. Determine the genotypes of your parents and paternal grandparents. What is the probability of having a brown-eyed child in this family?

23. A woman with blond hair and a straight nose married a man with dark hair and a Roman nose, heterozygous for the first characteristic and homozygous for the second. Dark hair and a Roman nose are dominant characteristics. What are the genotypes and gametes of the parents? What are the likely genotypes and phenotypes of the children?

24. Several kittens were born from a tortoiseshell cat, one of which turned out to be a ginger cat. In cats, coat color genes are sex-linked and are found only on the X chromosomes. Tortoiseshell coat color is possible with a combination of the black and red color genes. Determine the genotypes of the parents and the phenotype of the father, as well as the genotypes of the offspring.

25. A heterozygous gray female Drosophila was crossed with a gray male. Determine the gametes produced by the parents, as well as the numerical ratio of the splitting of hybrids by phenotype (by sex and body color) and genotype. These characteristics are sex-linked and are found only on the X chromosomes. Gray body color is a dominant feature.

26. In corn, the dominant genes for brown color and smooth seed shape are linked and localized on one chromosome, and the recessive genes for white color and wrinkled shape are in another homologous chromosome. What genotype and phenotype of offspring should be expected when crossing a diheterozygous plant with white smooth seeds with a plant with white wrinkled seeds. Crossing over did not occur in meiosis. Identify the gametes produced by the parents.

27. In corn, the dominant genes for brown color and smooth seed shape are linked and localized on one chromosome, and the recessive genes for white color and wrinkled shape are in another homologous chromosome. What genotype and phenotype of offspring should be expected when crossing a diheterozygous plant with white smooth seeds with a homozygous plant with dark smooth seeds. Crossing over occurs in meiosis. Identify the gametes produced by the parents without crossing over and after crossing over.

28. When crossing a shaggy white rabbit with a shaggy black rabbit, one smooth white rabbit appeared in the offspring. Determine the genotypes of the parents. In what numerical ratio can we expect the offspring to be split into genotype and phenotype?

29. The hunter bought a dog that has short hair. It is important for him to know that she is purebred. What actions will help a hunter determine that his dog does not carry recessive genes - long hair? Draw up a scheme for solving the problem and determine the ratio of genotypes of the offspring obtained from crossing a purebred dog with a heterozygous one.

30. A man suffers from hemophilia. His wife's parents are healthy according to this criterion. The hemophilia gene (h) is located on the sex X chromosome. Make a diagram for solving the problem. Determine the genotypes of the married couple, possible offspring, and the likelihood of having daughters who are carriers of this disease.

31. Hypertrichosis is transmitted in a person with a Y chromosome, and polydactyly (polydactyly) is an autosomal dominant trait. In a family where the father had hypertrichosis and the mother had polydactyly, a normal daughter was born. Draw up a scheme for solving the problem and determine the genotype of the born daughter and the probability that the next child will have two abnormal characteristics.

32. Diheterozygous tomato plants with rounded fruits (A) and pubescent leaves (B) were crossed with plants with oval fruits and non-hairy leaf epidermis. The genes responsible for the structure of the leaf epidermis and the shape of the fruit are inherited linked. Make a diagram for solving the problem. Determine the genotypes of the parents, the genotypes and phenotypes of the offspring, and the likelihood of plants with recessive traits appearing in the offspring.

33. When crossing a tomato with a purple stem (A) and red fruits (B) and a tomato with a green stem and red fruits, 750 plants with a purple stem and red fruits and 250 plants with a purple stem and yellow fruits were obtained. The dominant genes for purple stem color and red fruit color are inherited independently. Make a diagram for solving the problem. Determine the genotypes of parents, offspring in the first generation and the ratio of genotypes and phenotypes in the offspring.

34. A Datura plant with purple flowers (A) and smooth capsules (c) was crossed with a plant with purple flowers and spiny capsules. The following phenotypes were obtained in the offspring: with purple flowers and spiny capsules, with purple flowers and smooth capsules, with white flowers and spiny capsules, with white flowers and smooth capsules. Make a diagram for solving the problem. Determine the genotypes of parents, offspring and possible relationships between phenotypes. Establish the nature of inheritance of traits.

35. Crossed two snapdragon plants with red and white flowers. Their offspring turned out to have pink flowers. Determine the genotypes of parents, first-generation hybrids and the type of inheritance of traits.

36. A brown (a) long-haired (c) female is crossed with a homozygous black (A) short-haired (B) male (the genes are not linked). Draw up a scheme for solving the problem and determine the genotypes and the ratio according to the phenotype of the descendants of their first generation. What is the ratio of genotypes and phenotypes of the second generation from crossing diheterozygotes. What genetic patterns are evident in this crossbreeding?

37. In pigs, black bristles (A) dominate over red ones, long bristles (B) dominate over short ones (genes are not linked). A black diheterozygous male with long bristles was crossed with a homozygous black female with short bristles. Make a diagram for solving the problem. Determine the genotypes of parents, offspring, phenotypes of offspring and their relationship.

38. The absence of small molars in humans is inherited as a dominant autosomal trait. Determine the genotypes and phenotypes of parents and offspring if one of the spouses has small molars and the other is heterozygous for this gene. Draw up a scheme for solving the problem and determine the probability of having children who are missing small molars.

The concept of inheritance of traits is widely studied in genetics. It is they who explain the similarity between offspring and parents. It is curious that some manifestations of traits are inherited together. This phenomenon, first described in detail by the scientist T. Morgan, came to be called “linked inheritance.” Let's talk about it in more detail.

As you know, each organism has a certain number of genes. At the same time, chromosomes are also a strictly limited number. For comparison: a healthy human body has 46 chromosomes. There are thousands of times more genes in it. Judge for yourself: each gene is responsible for one or another trait that manifests itself in a person’s appearance. Naturally, there are a lot of them. Therefore, they began to talk about the fact that several genes are localized on one chromosome. These genes are called a linkage group and determine linked inheritance. A similar theory has been floating around in the scientific community for quite a long time, but only T. Morgan gave it a definition.

Unlike the inheritance of genes that are localized in different pairs of identical chromosomes, linked inheritance causes a diheterozygous individual to form only two types of gametes, repeating the combination of parental genes.

Along with this, gametes arise, the combination of genes in which differs from the chromosomal set of the parents. This result is a consequence of crossing over, a process whose importance in genetics is difficult to overestimate, since it allows the offspring to receive different traits from both parents.

In nature, there are three types of gene inheritance. In order to determine which type is inherent in a particular pair of them, they use. The result will necessarily result in one of the three options given below:

1. Independent inheritance. In such a case, hybrids differ from each other and from their parents in appearance, in other words, as a result we have 4 variants of phenotypes.

2. Complete linkage of genes. First generation hybrids, resulting from crossing parental individuals, completely repeat the phenotype of the parents and are indistinguishable from each other.

3. Incomplete linkage of genes. Just as in the first case, when crossed, 4 classes of different phenotypes are obtained. In this case, however, new genotypes are formed that are completely different from the parent stock. It is in this case that crossing over, mentioned above, interferes with the process of gamete formation.

It has also been established that the smaller the distance between inherited genes on the parent chromosome, the higher the likelihood of their complete linked inheritance. Accordingly, the farther they are located from each other, the less often crossover occurs during meiosis. The distance between genes is the factor that primarily determines the probability of linked inheritance.

Separately, it is necessary to consider linked inheritance associated with gender. Its essence is the same as with the option discussed above, however, the inherited genes in this case are located on the sex chromosomes. Therefore, we can talk about this type of inheritance only in the case of mammals (including humans), some reptiles and insects.

Taking into account the fact that XY is a set of chromosomes corresponding to the male sex, and XX to the female sex, we note that all the main characteristics responsible for the viability of the organism are located in the chromosome present in the genotype of each organism. Of course, we are talking about the X chromosome. In females, both recessive and chromosomal ones may be present. Males can inherit only one of the variants - that is, either the gene manifests itself in the phenotype or not.

Sex-linked inheritance is often heard in the context of diseases that are specific to men, while women are only their carriers:

• hemophilia,
• color blindness;
• Lesch-Nyhan syndrome.