Color blindness (part one)

color blindness or color blindness is a reduced ability to perceive differences between certain colors that healthy people can distinguish. Usually the nature of the origin of this disorder is, but the disorder can also occur due to damage to the eyes, nerves, brain, or exposure to certain chemicals.

English chemist John Dalton published the first scientific paper on the subject in 1798 year, she was called "The Extraordinary Facts About Seeing Flowers" after discovering the disease in himself. Today, after the publication of Dalton's work, this disease is known to us as color blindness, although at present this term is used to indicate one type of color blindness, namely.

color blindness are generally classified as a moderate disability, but sometimes affected people have an advantage over those with normal color perception. Several studies indicate that color-blind people are better at distinguishing objects camouflaged in color, and it has been suggested that this may be an evolutionary explanation for the unexpectedly high frequency of congenital.

Prerequisites

The retina of an average person contains two types of light-sensitive cells: sticks (active in low light conditions) and cones (active in normal daylight). Generally, there are three kinds of cones, each containing different pigments that are activated when these pigments absorb light. The spectral sensitivity of the cones is different. Some are maximally sensitive to short waves, the second to medium waves, and the third to long electromagnetic waves of the light range.

Peak Sensitivity these cones located in the blue, yellowish green, and yellow regions of the spectrum, respectively. The absorption spectra of all three cone systems cover the entire visible spectrum. These receptors are often referred to as S cones, M and L cones, respectively for short, medium and long lengths waves, but they are also often referred to as blue, green, and red cones, respectively.

Although the receptors are often referred to as "blue, green and red" receptors, this term is not very accurate, especially considering that the "red" receptor actually has its sensitivity peak in the yellow region. The sensitivity of normal color vision actually depends on overlapping absorption spectra of the three systems: different colors are perceived when stimulated Various types cones. Red light, for example, stimulates the long wavelength cones much more than any other, and also shortens the wavelength, causing more and more stimulation of the other two cone systems, and the hue change gradually wears off.

3. anomalous trichromacy- This is a fairly common type of hereditary color perception disorder in which one of the three types of cones changes spectral sensitivity. That is why, as a result of only some impairment, and not a complete loss, a person with trichromacy has normal, three-dimensional color vision.

Based on the clinical picture, color blindness can be characterized as complete or partial. Complete color blindness is much less common than partial color blindness. There are two main types of disease:

Dichromacy (protanopia and deuteranopia)
- Abnormal trichromia (protanomaly and deuteranomaly)

Blue-yellow:

Dichromia (tritanopia)
- Abnormal trichromia (tritanomaly)

Causes

Diagnostics

To diagnose the perception of red and green colors is most often used Ishihara color test (Ishihara color test), which consists of a series of photographs depicting colored spots that are placed in such a way that they create a specific pattern. The figures (as a rule - Arabic numerals) are depicted as a certain sequence of spots, which are depicted in a color that is somewhat different from the general background. This number can be easily called by a person with normal vision, but not by a person who has certain defects in color perception. The full set of tests consists of various shapes that contain a variety of color combinations, which allows you to diagnose exactly which visual defect was present. As mentioned above, for the diagnosis of anomalous trichromacy is used anomaloscope .

Because colored Ishihara test (Ishihara color test ) contains only numbers, it cannot be used to diagnose young children who do not yet know the numbers. To identify visual problems on early stage use alternative tests with the image of certain symbols (square, circle, car).

In addition to the above test, upon admission to the US armed forces (in particular, to the navy and army), they conduct the so-called FALANT test. When participants are asked to determine from a distance what kind of light the beacon emits. At the same time, two of the possible three colors light up - red, green and white. However, the light is passed through special "filters", due to which the color is "muted". This is done so that colorblind people cannot determine the light according to its brightness. This test can be passed by 30% of people with minor color perception disorders.

Most clinical trials are designed to quickly, easily, and effectively identify broad categories of color blindness. In academic research on color blindness, there is an increased interest in developing flexible tests that would collect detailed data that would identify relevant points and detect subtle deviations.

Treatment

To date, there are no effective drugs for the complete treatment of color perception disorders. However The use of some types of tinted filters and contact lenses can help a person to distinguish colors better. Doctor optometrist may recommend the use of contact lenses with a special, red tint. This will allow their owners to pass several diagnostic tests, but for practical application they are not very helpful. The effect of wearing such devices is similar to wearing red/blue 3D glasses and often takes some getting used to as certain wavelengths can "jump out" and/or be over-represented. In addition, software and special cybernetic devices have been developed to help sick people, such as i-borg (eyeborg) and " cybernetic eyes, which allow colorblind people to hear specific sounds that match a particular color.

Workspace GNOME provides colorblind people with the ability to use the software GNOME Mag and "l ibcolorblind". Using the GNOME applet, the user can switch the color filter, choosing from a variety of possible color schemes, which will change colors for better recognition. This software allows, for example, a person with color vision impairment to see the numbers in the Ishihara test.

In September 2009 the magazine Nature published a study by scientists at the University of Washington and the University of Florida, who with the help of made a saimiri monkey - a trichromat, because usually, they see only two spectrums of colors.

Example. A woman suffering from color blindness married a man with normal vision. What will be the color perception of the sons and daughters of these parents?

Decision: in humans, color blindness (color blindness) is due to the recessive gene w, and normal color vision is due to its dominant allele W. The gene for color blindness is linked to the X chromosome. The Y chromosome does not have a corresponding locus and does not contain a gene that controls color vision. Therefore, in the diploid set of men, there is only one allele responsible for color perception, while in women there are two, since they have two X chromosomes.

The crossover scheme for this type of task is constructed somewhat differently from the schemes for traits not related to sex. These schemes indicate not only the symbols of genes, but also the symbols of the sex chromosomes of male and female individuals X and Y. This must be done to show the absence of the second allele in male and female individuals.

So, according to the condition of the problem, the crossover scheme will look like this. A woman has two X chromosomes and each has a recessive gene for color blindness (Xw). A man has only one X chromosome, which carries the gene for normal color vision (X W), and a Y chromosome that does not contain the allelic gene. First, we designate the two X chromosomes of a woman and indicate that they contain genes for color blindness (X w X w). Then we denote the male chromosomes (X W Y). Below we write out the gametes that they produce. In a woman, they are the same, they all contain the X chromosome with the gene for color blindness. In a man, half of the gametes carry the X chromosome with the gene for normal color perception, and half the Y chromosome that does not contain the allele. After that, determines the genotypes F 1 .

All girls receive one X chromosome from their father, which contains the gene for normal vision (W), and another X chromosome from their mother, which contains the gene for color blindness (w).

Thus, girls will have two Ww genes, and since the gene for normal vision dominates, they will not have color blindness. All boys receive their single X chromosome from their mother and the gene for color blindness (w) located in it. Since they do not have the second allele, they will suffer from color blindness.

58. The daughter of a color blind person marries the son of a color blind person, and the bride and groom can distinguish colors normally. What will their children's vision be like?

59. Men and women who are healthy in terms of color vision have a son suffering from color blindness, who has a healthy daughter. Determine the genotypes and phenotypes of parents, children and grandchildren.

60. Men and women who are healthy in terms of color vision have a healthy daughter who has one healthy son and one color-blind son. Determine the genotypes and phenotypes of parents, children and grandchildren.

61. Men and women who are healthy in terms of color vision have a healthy daughter who has five healthy sons. Determine the genotypes and phenotypes of parents, children and grandchildren.

62. What phenotypes can be expected for a Y-linked trait in the following descendants of a father with this trait:

a) sons

b) daughters

c) grandchildren from sons,

d) granddaughters from daughters,

e) granddaughters from sons,

f) grandchildren from daughters.

63. Hypertrichosis (growth of hair on the edge of the auricle) is inherited as a trait linked to the Y chromosome, which manifests itself only by the age of 17. Common ichthyosis (scaly and patchy thickening of the skin) is inherited as a recessive, X-linked trait. In a family where the woman is healthy in relation to both signs, and the husband suffers only from hypertrichosis, a boy was born with signs of ichthyosis. Determine: a) the probability of manifestation of hypertrichosis in this child; b) the probability of the birth of children in this family without both anomalies. What gender will they be?

64. In humans, the dominant gene P determines persistent rickets, which is inherited sex-linked. What is the probability of having sick children if the mother is heterozygous for the rickets gene, and the father is healthy?

65. The gene for eye color in Drosophila flies is located on the X chromosome. Red (normal) eyes (W) are dominant over white eyes (w). Determine the phenotype and genotype of the offspring F 1 and F 2 if you cross a white-eyed female with a red-eyed male.

66. In Drosophila flies, the gene for normal eye color (red) W dominates over white-eyed w, the gene for abnormal abdominal structure - over the gene for its normal structure. These pairs of genes are located on the X chromosome at a distance of three morganids. Determine the probability of different genotypes and phenotypes in the offspring from crossing a heterozygous female for both traits with a male with a normal eye color and a normal abdominal structure.

67. The H gene determines normal blood clotting in humans, and the h gene determines hemophilia. A woman heterozygous for the hemophilia gene married a man with normal blood clotting. Determine the phenotype and genotype of children that can be born from such a marriage. What is the type of trait inheritance?

68. In Plymouth Rock chickens, the dominant variegated gene is linked to the X chromosome. The recessive allele, the black gene, is observed in Austrolon chickens. Which cross will allow early sex marking of chicks?

Example. A woman suffering from color blindness married a man with normal vision. What will be the color perception of the sons and daughters of these parents?

Decision: in humans, color blindness (color blindness) is due to the recessive gene w, and normal color vision is due to its dominant allele W. The gene for color blindness is linked to the X chromosome. The Y chromosome does not have a corresponding locus and does not contain a gene that controls color vision. Therefore, in the diploid set of men, there is only one allele responsible for color perception, while in women there are two, since they have two X chromosomes.

The crossover scheme for this type of task is constructed somewhat differently from the schemes for traits not related to sex. These schemes indicate not only the symbols of genes, but also the symbols of the sex chromosomes of male and female individuals X and Y. This must be done to show the absence of the second allele in male and female individuals.

So, according to the condition of the problem, the crossover scheme will look like this. A woman has two X chromosomes and each has a recessive gene for color blindness (Xw). A man has only one X chromosome, which carries the gene for normal color vision (X W), and a Y chromosome that does not contain the allelic gene. First, we designate the two X chromosomes of a woman and indicate that they contain genes for color blindness (X w X w). Then we denote the male chromosomes (X W Y). Below we write out the gametes that they produce. In a woman, they are the same, they all contain the X chromosome with the gene for color blindness. In a man, half of the gametes carry the X chromosome with the gene for normal color perception, and half the Y chromosome that does not contain the allele. After that, determines the genotypes F 1 .

All girls receive one X chromosome from their father, which contains the gene for normal vision (W), and another X chromosome from their mother, which contains the gene for color blindness (w).

Thus, girls will have two Ww genes, and since the gene for normal vision dominates, they will not have color blindness. All boys receive their single X chromosome from their mother and the gene for color blindness (w) located in it. Since they do not have the second allele, they will suffer from color blindness.

58. The daughter of a color blind person marries the son of a color blind person, and the bride and groom can distinguish colors normally. What will their children's vision be like?

59. Men and women who are healthy in terms of color vision have a son suffering from color blindness, who has a healthy daughter. Determine the genotypes and phenotypes of parents, children and grandchildren.

60. Men and women who are healthy in terms of color vision have a healthy daughter who has one healthy son and one color-blind son. Determine the genotypes and phenotypes of parents, children and grandchildren.

61. Men and women who are healthy in terms of color vision have a healthy daughter who has five healthy sons. Determine the genotypes and phenotypes of parents, children and grandchildren.

62. What phenotypes can be expected for a Y-linked trait in the following descendants of a father with this trait:

a) sons

b) daughters

c) grandchildren from sons,

d) granddaughters from daughters,

e) granddaughters from sons,

f) grandchildren from daughters.

63. Hypertrichosis (growth of hair on the edge of the auricle) is inherited as a trait linked to the Y chromosome, which manifests itself only by the age of 17. Common ichthyosis (scaly and patchy thickening of the skin) is inherited as a recessive, X-linked trait. In a family where the woman is healthy in relation to both signs, and the husband suffers only from hypertrichosis, a boy was born with signs of ichthyosis. Determine: a) the probability of manifestation of hypertrichosis in this child; b) the probability of the birth of children in this family without both anomalies. What gender will they be?

64. In humans, the dominant gene P determines persistent rickets, which is inherited sex-linked. What is the probability of having sick children if the mother is heterozygous for the rickets gene, and the father is healthy?

65. The gene for eye color in Drosophila flies is located on the X chromosome. Red (normal) eyes (W) are dominant over white eyes (w). Determine the phenotype and genotype of the offspring F 1 and F 2 if you cross a white-eyed female with a red-eyed male.

66. In Drosophila flies, the gene for normal eye color (red) W dominates over white-eyed w, the gene for abnormal abdominal structure - over the gene for its normal structure. These pairs of genes are located on the X chromosome at a distance of three morganids. Determine the probability of different genotypes and phenotypes in the offspring from crossing a heterozygous female for both traits with a male with a normal eye color and a normal abdominal structure.

67. The H gene determines normal blood clotting in humans, and the h gene determines hemophilia. A woman heterozygous for the hemophilia gene married a man with normal blood clotting. Determine the phenotype and genotype of children that can be born from such a marriage. What is the type of trait inheritance?

68. In Plymouth Rock chickens, the dominant variegated gene is linked to the X chromosome. The recessive allele, the black gene, is observed in Austrolon chickens. Which cross will allow early sex marking of chicks?

If their children will marry healthy people?

the girl suffers from color blindness, and the father, like all his ancestors, distinguishes colors normally, the girl marries a healthy young man. person. The probability of being born healthy and suffering from color blindness

2. The recessive gene is on the X chromosome, the girl's father is healthy, her mother is healthy, but her father was ill with hemophilia (mother's father), the girl marries a healthy young man. What can you say about their future children and grandchildren?

is a carrier of the gene for color blindness, and her husband suffers from color blindness. What is the probability that this family will have children with normal color discrimination?

1.

to be born a colorblind daughter? What is the probability of having the first two colorblind sons?

In chickens, silver plumage is a dominant trait. Silver plumage gene S (Silver) is localized in Z-chromosome. A silver rooster was crossed with a black hen. What are possible consequences this crossing?

Enamel hypoplasia is inherited as linked to X-chromosome dominant trait. In a family where both parents suffered from the marked anomaly, a son was born with normal teeth. Determine the probability that the next child will also have normal teeth.

One form of hemophilia is inherited as a recessive trait linked to X-chromosome. A man with hemophilia marries a normal woman (whose father and mother are healthy). Will children suffer from this disease?

* A tortoiseshell (spotted) cat was crossed with a red cat. How will the splitting of hybrids by genotype and phenotype (by coat color and sex) go?

*Hypertension in humans is determined by a dominant autosomal gene, optic atrophy is caused by a sex-linked recessive gene. A woman with optic atrophy marries a man with hypertension whose father also had hypertension and whose mother was healthy. What is the probability that a child in this family will suffer from both anomalies (in %)? What is the probability of having a healthy baby (in %)?

* What are the genotypes of the parents if a blue-eyed hemophilic boy was born from the marriage of a brown-eyed healthy (according to blood clotting) woman with a blue-eyed healthy man. Hemophilia and Blue eyes- recessive traits, but the blue eye gene is located on the autosome, and the hemophilia gene is linked to X-chromosome.

colorblind with good hearing. determine the genotypes of parents and children, if both signs (deafness and color blindness) are recessive, deafness is an autosomal trait, and color blindness is linked to the x chromosome

1. A person has two types of blindness, and each is determined by its 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 both are dihomozygous? What is the probability of having a blind child if both parents are dihomozygous and suffer from different types hereditary blindness?

Explanation:

First cross:

R: AAvv x AAvv

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 cross:

R: AAvv x aaBB

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 an X-linked recessive gene. Thalassemia is inherited as an autosomal dominant trait and occurs in two forms: in homozygotes it is severe, often fatal, in heterozygotes it is mild.

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

Explanation:

R: AaXDXd x aaXdY

G: AXD, aXd, AXd, aXD x aXd, aY

F1: AaXdY - colorblind boy with mild thalassemia

AaXDXd - a girl with normal vision and mild thalassemia

aaXdXd - colorblind girl without thalassemia

AaXdXd - colorblind girl with mild thalassemia

aaXDХd - a 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 no thalassemia

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

Explanation: BUT - Brown eyes

a blue eyes

B - dark hair

c - blond hair

R: aavv x AaVv

G: av x AB, av, av, aB

F1: AaBv - brown eyes, dark hair

aavv - blue eyes, blonde hair

Aaww - brown eyes, blond hair

aaBv - blue eyes, dark hair

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

The signs are not sex-linked. This is where the law of independent inheritance comes into play.

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

Explanation:

A - black color

a - gray color

B - furry rabbit

c - smooth-haired rabbit

First cross:

R: aaBB x AaBB

F1: AaBB - black fluffy rabbits

aaBB - gray furry rabbits

Second cross:

R: aaBv x AaBv

G: аВ, АВ x АВ, av, АВ, аВ

F1: 8 genotypes and 4 phenotypes

АаВВ, 2АаВв - gray furry rabbits

Aavv - black smooth-haired rabbits

ааВВ, ааВв - gray furry rabbits

aavv - gray smooth-haired rabbits

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

5. For a crested (A) green (B) female, an analyzing cross was carried out, four phenotypic classes were obtained in the offspring. The resulting crested offspring were crossed among themselves. Is it possible to get offspring without a crest in this crossing? If so, what gender will it be, what phenotype? In canaries, the presence of a crest depends on an autosomal gene, the color of plumage (green or brown) depends on a gene linked to the X chromosome. The heterogametic sex in birds is female.

Explanation:

First cross:

R: AaHVU x aaHvHv

G: AHV, AHV, AU, AU X AHV

F1: АаХВХв - crested green male

ааХВХв - green male without crest

AaHvU - crested brown female

We cross a male and a female with a crest:

R: AaHVHv x AaHvU

G: AHV, AHV, AHV, AHV x AHV, AU, AHV, AU

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

Phenotypes of individuals without crest:

Females: ааХВУ - green female without crest

ааХвУ - brown female without crest

Males: ааХВХв - green male without crest

ааХвХв - brown male without crest.

6. In crossing female Drosophila with normal wings and normal eyes and males with reduced wings and small eyes, all offspring had normal wings and normal eyes. The females obtained 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. Make crossbreeding schemes, determine the genotypes and phenotypes of parental individuals and offspring in crosses. What are the laws of crossbreeding?

Explanation:

A - normal wings

a - reduced wings

XB - normal eyes

First cross:

R: AAHVHV x vvHvU

G: AHV x AHV, AU

AaHVHv - normal wings, normal eyes

AaHVU - normal wings, normal eyes

Second cross:

R: AaHVHv x aaHvN

G: AHV, AHV, AHV, AHV x AHV, AU

АаХВХв, АаХВУ - normal wings, normal eyes

ааХвХв, ааХвУ - reduced wings, small eyes

AaHvHv, AaHvU - normal wings, small eyes

ааХВХв, 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 twisted wings, 58 flies were obtained with a gray body and normal wings, 52 with a black body and twisted wings, 15 - with a gray body and swirling wings, 14 - with a black body and normal wings. Make a scheme for solving the problem. Determine the genotypes of the parent individuals, 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 - swirling wings

Crossing: R: AaVv x aavv

G: AB, av, av, aB x av

F1: AaBv - gray body, normal wings - 58

aavv - black body, swirling wings - 52

Aavv - gray body, swirling wings - 15

aaBv - 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 remaining two groups, crossing over occurred and these genes ceased to be linked, and therefore formed 14 and 15 individuals.

8. When analyzing the crossing of a diheterozygous tall tomato plant with round fruits, a splitting of offspring according to the 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. Make a crossbreeding scheme, determine the genotypes and phenotypes of the original individuals, offspring. Explain the formation of four phenotypic classes.

Explanation:

A is a tall plant

a - dwarf plant

B - round fruits

c - pear-shaped fruits

R: AaBv x aavb

G: AB, av, aB, Av x av

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 we can distinguish two groups of signs:

1. AaBv and aavb - in the first case, A and B are inherited linked, and in the second - a and c.

2. aaBv and Aavb - there was a crossing over.

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

Explanation: A - red hair

a red hair

R: Aa x Aa

G: A, a x A, a

F1: AA: 2Aa: aa

Splitting by genotype - 1:2:1.

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

There is no unequivocal answer to the question, since it is impossible to assume the genotype of the unborn child, since sex cells with different genotypes can be found.

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

Explanation: XDXd - healthy girl

XdY - the boy is colorblind

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

Let's write a crossover scheme.

P: XdXd x XDY

G: Xd x XD, Y

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

XdY - colorblind boy

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

Explanation: glaucoma is a recessive trait and manifests itself only with homozygote, and Marfan syndrome manifests itself both with hetero- and homozygote, but is a dominant trait, respectively, we will determine the genotypes of the parents: one parent suffers from glaucoma - aa, but does not suffer from Marfan syndrome - cc, and the second parent is heterozygous for both traits - AaBv.

R: aavv x AaVv

G: av x AB, av, av, aB

F1: AaBv - normal vision + Marfan's syndrome

aavv - glaucoma

Aavv - normal vision, no Marfan syndrome - healthy child

aaBv - glaucoma + Marfan's syndrome

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

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

12. Undersized (dwarf) tomato plants with ribbed fruits and plants of normal height with smooth fruits were crossed. In the offspring, two phenotypic groups of plants were obtained: undersized and smooth fruits and normal height with smooth fruits. When crossing undersized tomato plants with ribbed fruits with plants having a normal stem height and ribbed fruits, all offspring had a normal stem height and ribbed fruits. Create crossover patterns. Determine the genotypes of parents and offspring of tomato plants in two crosses. What law of heredity is manifested in this case?

Explanation: in the first crossing, a dihomozygote is crossed with a homozygous plant for one trait and a heterozygous plant 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 dihomozygotes are crossed (only the second parent will have one dominant trait).

a - undersized individuals

A - normal height

c - ribbed fruits

B - smooth fruits

P: aavb x aabb

F1: aaBb - undersized individuals with smooth fruits

AaBv - normal height, smooth fruits

P: aavv x AAiv

F1: Aavv - normal height, smooth fruits.

In both cases, the law of independent inheritance is manifested, since these two traits are inherited independently.

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

Explanation: since in the first generation we see uniformity, and in the second generation - splitting 1: 1, 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 crossing a motley crested (B) chicken with the same rooster, eight chickens were obtained: four motley crested chickens, two white (a) crested and two black crested. Make a scheme for solving the problem. Determine the genotypes of parents and offspring, explain the nature of the inheritance of traits and the appearance of individuals with variegated coloration. 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 coloring

P: AaBB x AaBB

G: AB, aB

F1: AaBB - pied crested (4 chicks)

aaBB - white crested (two chickens)

AABB - black crested

By color, the splitting by genotype and phenotype is the same: 1:2:1, since there is a phenomenon of incomplete dominance (an intermediate variant appears between both black and white), the signs are inherited independently.

15. In humans, the normal hearing gene (B) dominates the deafness gene 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), colorblind, a colorblind girl with normal hearing was born. Make a scheme for solving the problem. Determine the genotypes of parents, daughters, possible genotypes of children and the probability of future birth in this family of color-blind children with normal hearing and deaf children.

Explanation: It can be seen from the conditions of the problem 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 a color-blind child with normal hearing is 50%, and the probability of a color-blind child being born 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 gene for color blindness is recessive and linked to the X chromosome. Determine the genotypes of husband and wife. Make a scheme for solving the problem. What is the probability of having a son with normal vision, a daughter with normal vision, a color blind son, a color blind daughter?

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

We will also write down 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 of the genotypes 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 color-blind boy is 25%.

17. In sowing peas, the yellow color of the seeds dominates over the green, the convex shape of the fruit - over the fruits with a constriction. When a plant with yellow convex fruits was crossed with a plant with yellow seeds and fruits with constriction, 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 scheme 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

After carefully reading the condition of the problem, one 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 scheme for solving the problem:

P: AaBv x Aavb

G: AB, av, av, aB x av, av

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

63 - A_Vv - yellow seeds, convex fruits

58 - A_vv - yellow seeds, fruits with constriction

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

Explanation: AA - red color

Aa - pink coloring

aa - white color

BB - narrow leaves

Вв - leaves of intermediate width

cc - wide leaves

Crossbreeding:

R: AaVv x aaVV

G: AB, av, AV, aB x aB

F1: AaBB - pink flowers, narrow leaves

aaBv - 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 of the genotypes is 25%.

Crossing is dihybrid (since the analysis is based on two traits).

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

Tasks for independent solution

1. In dogs, black coat dominates over brown, and long coat over short (genes are not linked). Offspring were obtained from a black long-haired female during analyzing crosses: 3 black long-haired puppies, 3 brown long-haired puppies. Determine the genotypes of parents and offspring corresponding to their phenotypes. Make a scheme for solving the problem. Explain your results.

2. In sheep, gray color (A) of wool dominates over black, and horn (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? Make a scheme for solving the hadachi. Explain your results. What law of heredity is manifested in this case?

3. In maize, the recessive gene "shortened internodes" (b) is on the same chromosome as the recessive gene "rudimentary panicle" (v). When conducting an analyzing cross of a diheterozygous plant with normal internodes and a normal panicle, 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 normal panicle. Determine the genotypes of parents and offspring. Make a scheme for solving the problem. Explain your results. What law of heredity is manifested in this case?

4. When crossing a corn plant with smooth colored seeds with a plant that produces wrinkled uncolored seeds (genes are linked), the offspring turned out to be with smooth colored seeds. When analyzing the crossing of hybrids from F1, plants were obtained with smooth colored seeds, wrinkled uncolored seeds, wrinkled colored seeds, and smooth uncolored seeds. Make a scheme for solving the problem. Determine the genotypes of the 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 (genes linked), the offspring turned out to have smooth colored seeds. With further analyzing crosses 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 seeds. Make a scheme for solving the problem. Determine the genotypes of parents, offspring F1, F2. What law of heredity is manifested in F2? Explain what your answer is based on.

6. In humans, cataracts (eye disease) depend on a dominant autosomal gene, and ichthyosis (skin disease) depends on an X-linked recessive gene. 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 scheme for solving the problem. Determine the genotypes of the parents, the 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, offspring were obtained: 50% black shaggy and 50% black smooth. When crossing another pair of white rabbits with shaggy hair and black rabbits with smooth hair, 50% of the offspring turned out to be black shaggy and 50% 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, plants with long green and round green fruits were obtained in the offspring. When the same watermelon with long striped fruits was crossed with a plant with round striped fruits, all offspring had round striped fruits. Make a diagram of each cross. Determine the genotypes of parents and offspring. What is the name of this crossing 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 color hair and brown eyes are dominant traits. 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 trait, married a man with dark smooth hair, heterozygous for the first allele. Dark and curly hair are dominant traits. Determine the genotypes of the parents, the types of gametes they produce, the probable 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 trait. Dark hair and brown eyes are dominant traits, blond hair and blue eyes are recessive traits. Determine the genotypes of the parents and the gametes they produce, the probable genotypes and phenotypes of the offspring.

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

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

15. From a black cat, one tortoiseshell and several black kittens were born. These traits are sex-linked, that is, color genes are found only on the sex X chromosomes. The gene for black color and the gene for red color gives incomplete dominance, when these two genes are combined, a tortie color is obtained. Determine the genotype and phenotype of the father, the gametes that the parents produce, 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. The gray color of the body dominates over the yellow. Determine the genotypes of the parents, gametes. which they produce, and the numerical splitting of offspring by sex and body color.

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

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

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

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

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

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

23. Woman with blonde hair and with a straight nose, married a man with dark hair and a Roman nose, heterozygous for the first trait and homozygous for the second. Dark hair and a Roman nose are dominant traits. 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 red cat. In cats, coat color genes are sex-linked and are found only on the X chromosome. Tortoiseshell coat color is possible with a combination of the gene for black and red color. 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 traits are sex-linked and are found only on the X chromosomes. Gray body color is a dominant trait.

26. In maize, 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 on the other homologous chromosome. What offspring according to the genotype and phenotype should be expected when a diheterozygous plant with white smooth seeds is crossed with a plant with white wrinkled seeds. Crossing over did not occur during meiosis. Determine 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 on the other homologous chromosome. What offspring according to the genotype and phenotype should be expected when crossing a diheterozygous plant with white smooth seeds with a homozygous plant with dark smooth seeds. In meiosis, crossing over occurs. Determine the gametes produced by the parents without crossing over and after crossing over.

28. When a shaggy white rabbit was crossed 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 splitting of the offspring according to the genotype and phenotype?

29. A hunter bought a dog that has short hair. It is important for him to know that she is purebred. What actions will help the hunter to determine that his dog does not carry recessive genes - long hair? Make a scheme for solving the problem and determine the ratio of the 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 on this basis. The hemophilia gene (h) is located on the sex X chromosome. Make a scheme for solving the problem. Determine the genotypes of a married couple, possible offspring, the probability of having daughters who are carriers of this disease.

31. Hypertrichosis is transmitted in a person with a Y chromosome, and polydactyly (multi-fingeredness) is an autosomal dominant trait. In a family where the father had hypertrichosis and the mother had polydactyly, a normal daughter was born. Make 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 features.

32. Diheterozygous tomato plants with rounded fruits (A) and with pubescent leaves (B) were crossed with plants having oval fruits and non-pubescent leaf epidermis. The genes responsible for the structure of the leaf epidermis and the shape of the fruit are inherited linked. Make a scheme for solving the problem. Determine the genotypes of the parents, the genotypes and phenotypes of the offspring, the probability of the appearance of plants with recessive traits 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. Dominant genes the purple color of the stem and the red color of the fruit are inherited independently. Make a scheme for solving the problem. Determine the genotypes of the 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 boxes (c) was crossed with a plant with purple flowers and spiny boxes. The following phenotypes were obtained in the offspring: with purple flowers and spiny boxes, with purple flowers and smooth boxes, with white flowers and spiny boxes, with white flowers and smooth boxes. Make a scheme for solving the problem. Determine the genotypes of parents, offspring and the possible ratio of phenotypes. Establish the nature of inheritance of traits.

35. Two snapdragon plants with red and white flowers were crossed. Their offspring turned out to be with pink flowers. Determine the genotypes of parents, hybrids of the first generation 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 (genes are not linked). Draw up a scheme for solving the problem and determine the genotypes and phenotype ratio of the offspring of their first generation. What is the ratio of genotypes and phenotypes of the second generation from crossing diheterozygotes. What genetic patterns are manifested in this crossing?

37. In pigs, black bristle color (A) dominates over red, long bristle (B) - over short (genes are not linked). A diheterozygous black with long stubble was crossed with a homozygous black with short stubble female. Make a scheme for solving the problem. Determine the genotypes of parents, offspring, offspring phenotypes and their ratio.

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. Make a scheme for solving the problem and determine the probability of having children who do not have small molars.