What eye color will be dominant? Blue eyes are the result of a genetic mutation. Eye color: how it is passed on from parents to child. Calculate child's eye color
My friends know how much I am interested in the question of the color of my son's eyes.
For those who don’t know, I’ll tell you: Our dad has brown eyes. My eyes are green with pronounced heterochromia (there are brown veins in the eyes, the rim of the eyes is gray, the iris is green. That is, the eyes are three-colored).
Eye color: from grandparents to our grandchildren: how it is transmitted genetically.
Tables for calculating the eye color of an unborn child.
During pregnancy, many parents are eager to find out what eye color their unborn child will have. All answers and tables for calculating eye color are in this article.
Good news for those who want to pass on their exact eye color to their descendants: it is possible.
Recent research in the field of genetics has discovered new data on the genes that are responsible for eye color (previously 2 genes were known that were responsible for eye color, now there are 6). At the same time, today genetics does not have answers to all questions regarding eye color. However, there is a general theory that, even with the latest research, provides a genetic basis for eye color. Let's consider it.
So: every person has at least 2 genes that determine eye color: the HERC2 gene, which is located on human chromosome 15, and the gey gene (also called EYCL 1), which is located on chromosome 19.
Let's look at HERC2 first: humans have two copies of this gene, one from their mother and one from their father. HERC2 can be brown and blue, that is, one person has either 2 brown HERC2 or 2 blue HERC2 or one brown HERC2 and one blue HERC2:
HERC2 gene: 2 copies* Human eye color
Brown and Brown brown
Brown and blue brown
Blue and cyan blue or green
(*In all tables in this article, the dominant gene is written with a capital letter, and the recessive gene is written with a small letter, eye color is written with a small letter).
Where does the owner of two blue HERC2 green eyes come from - is explained below. In the meantime, here is some data from the general theory of genetics: brown HERC2 is dominant, and blue is recessive, so a carrier of one brown and one blue HERC2 will have brown eye color. However, a carrier of one brown and one blue HERC2 can pass on both brown and blue HERC2 to their children with a 50x50 probability, that is, the dominance of brown does not in any way affect the transmission of a copy of HERC2 to children.
For example, a wife has brown eyes, even if they are “hopelessly” brown: that is, she has 2 copies of brown HERC2: all children born with such a woman will be brown-eyed, even if the man has blue or green eyes, so how she will pass on one of her two brown genes to her children. But grandchildren can have eyes of any color:
So, for example:
HERC2 from the mother is brown (in the mother, for example, both HERC2 are brown)
HERC2 from father - blue (father, for example, has both HERC2 blue)
The child's HERC2 is one brown and one blue. The eye color of such a child is always brown; at the same time, he can pass on his blue HERC2 to his children (who can also receive blue HERC2 from the second parent and then have eyes either blue or green).
Now let's move on to the gey gene: it can be green and blue (blue, gray), each person also has two copies: a person receives one copy from his mother, the second from his father. Green gey is a dominant gene, blue gey is recessive. A person thus has either 2 blue gey genes or 2 green gey genes, or one blue and one green gey gene. At the same time, this affects the color of his eyes only if he has blue HERC2 from both parents (if he received brown HERC2 from at least one of his parents, his eyes will always be brown).
So, if a person received blue HERC2 from both parents, depending on the gey gene, his eyes may be the following colors:
Gey gene: 2 copies
Human eye color
Green and Green
Green
Green and blue
Green
blue and blue
Blue
General table for calculating the color of a child's eyes, brown eye color is indicated by “K”, green eye color is indicated by “Z” and blue eye color is indicated by “G”:
Eye color
Brown
Brown
Brown
Brown
Brown
Brown
Green
Green
Blue
Using this table, we can say with a high degree of probability that a child will have green eyes if both parents have green eyes or one parent has green eyes and the other has blue eyes. You can also say for sure that the child's eyes will be blue if both parents have blue eyes.
If at least one of the parents has brown eyes, their children may have brown, green or blue eyes.
Statistically:
With two brown-eyed parents, the probability that the child will have brown eyes is 75%, green - 18.75% and blue - 6.25%.
If one of the parents is brown-eyed and the other is green-eyed, the probability that the child will have brown eyes is 50%, green - 37.5%, blue - 12.5%.
If one of the parents is brown-eyed and the other is blue-eyed, the probability that the child will have brown eyes is 50%, blue - 50%, green - 0%.
Thus, if a child’s eyes are not the same color as his parents, there are genetic reasons and justifications for this, because “nothing disappears without a trace and nothing comes out of nowhere.”
3) One of the forms of glaucoma is determined by a dominant autosomal gene, and the second has an autosomal recessive type of inheritance. These genes are located on different pairs of chromosomes.
What is the probability of having a sick child if both parents are diheterozygous?
4) The gene for black color in cats is sex-linked. The other allele of this gene corresponds to the red color; heterozygous animals have a spotted ("tortoiseshell") coloration.
What will be the offspring from crossing a “tortoiseshell” cat with a ginger cat?
5) The forensic medical examination is tasked with finding out whether the boy in the family is his own or adopted. A blood test of the husband, wife and child showed: the wife is Rh-AB-IV blood type, the husband is Rh+O (I) blood type, the child is Rh-B (III) blood type.
What conclusion should the expert give and on what is it based?
6) When crossing a brown Great Dane with a white one, all the offspring are white. When breeding the resulting offspring "in themselves" we got 40 white, 11 black and 3 brown.
What type of gene interaction determines the inheritance of coat color in Great Danes? What is the genotype of the parents and offspring?
7) In chickens, striped color is due to the sex-linked dominant (B), and the absence of such striping is due to its recessive alleles (c). The presence of a crest on the head is a dominant autosomal gene (C), and its absence is a recessive allele (c). Two striped, combed birds were crossed and produced two chicks: a striped cockerel with a comb and a non-striped hen without a comb. Indicate the genotypes of the parental forms and offspring.
heard; but one has smooth hair and the other has curly hair, a deaf child was born with smooth hair. Their second child heard well and had curly hair.
What is the probability of having a deaf child with curly hair in such a family?
2) The gene for brown eyes is dominant over the gene for blue eyes. A brown-eyed man whose parent had blue eyes married a brown-eyed woman whose father had blue eyes and whose mother had brown eyes.
What kind of offspring can be expected from this marriage?
Right-handedness predominates over the left-handedness gene. What is the probability of having blue-eyed, left-handed children from the marriage of two diheterozygous parents?
in humans, the gene for brown eyes dominates the gene causing blue eyes, a blue-eyed man, one of whose parents has brown eyes, marriesa brown-eyed woman whose father has brown eyes and whose mother has blue eyes. What kind of offspring can be expected from this marriage.
1. A blue-eyed man, whose parents had brown eyes, married a brown-eyed woman, whose father had blue eyes and whose mother had blue eyes.brown. What kind of offspring can be expected from this marriage if it is known that the gene for brown eyes dominates the gene for blue eyes?
2. There were two brothers in the family. One of them, a patient with hemorrhagic diathesis, married a woman also suffering from this disease. All three of their children (2 girls and 1 boy) were also sick. The second brother was healthy and married a healthy woman. Of their four children, only one was sick with hemorrhagic diathesis. Determine which gene determines hemorrhagic diathesis.
3. In a family where both parents had normal hearing, a deaf child was born. Which trait is dominant? What are the genotypes of all members of this family?
4. A man suffering from albinism marries a healthy woman whose father suffered from albinism. What kind of children can be expected from this marriage, given that albinism is inherited in humans as an autosomal recessive trait?
Incredible facts
Researchers have proven that Blue eye color is the result of a genetic mutation that probably occurred between 6,000 and 10,000 years ago. Scientists say they have discovered the reason why some of us have blue pigment in the iris.
Professor Hans Eiberg, leader of the research team at the University of Copenhagen, claims that all humans originally had brown eyes. As a result of a genetic mutation, the color of the eyes has changed, and both of these pigments are present in the iris of the eyes of modern people.
According to experts, most likely blue eye color comes from the countries of the Middle East or the northern part of the Black Sea coast. It was in this area that the largest migration took place during the Neolithic period (about 6,000 – 10,000 years ago). People moved in huge groups to the northern part of Europe.
"These are just our guesses," says Professor Eyberg. According to him, this could also be the territory of the northern part of Afghanistan.
Genetic mutation
This mutation, which occurred thousands of years ago, affected the so-called OCA2 gene and literally, “turned off” the ability of brown eyes to produce dark pigment.
For those less educated on this issue, it is worth explaining that the OCA2 gene is involved in the production of melanin, the pigment that gives color to hair, eyes and skin. A mutation in neighboring genes does not completely immobilize the OCA2 gene, but it certainly limits its action, thereby reducing the production of melanin in the iris. Thus, brown eyes are “diluted” with blue pigment.
If the OCA2 gene were completely turned off, those who inherited this mutation would lose melanin for their skin, hair and iris. Sometimes this happens. We call people with a complete lack of melanin albinos.
Professor Eyberg and his colleagues examined the DNA of blue-eyed people from countries where the majority of the population has brown eyes. Residents of Jordan, India, Denmark and Turkey took part in a number of experimental observations.
The results of Professor Eyberg's research are very important for genetics in general. For the first time in 1996, a scientist suggested that the OCA2 gene is responsible for eye and hair color. From this moment, a very important stage began in the study of the OCA2 gene, as well as all processes in the body associated with this gene.
The results of this study were published in the journal Human Genetics, which clearly indicate that all blue-eyed inhabitants of our planet were once the owners of brown eyes, and only as a result of the mutation that occurred, the pigment of the eyes changed.
Albinism in humans
It is known that the cause of albinism is the absence of the enzyme tyrosinase, which is involved in the normal synthesis of melanin.
There are several main types of this genetic disorder:
1. Oculocutaneous albinism.
2. Temperature-sensitive albinism.
3. Ocular albinism.
Treatment of any of these types is unsuccessful. It is impossible to compensate for the lack of melanin or prevent various visual disorders that are an integral part of the disease.
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 the future birth of color-blind children with normal hearing and deaf children in this school.
Answer
B – normal hearing, b – deafness.
The mother is deaf, but has normal color vision bbX D X _ .
Father with normal hearing (homozygous), color blind BBX d Y.
The colorblind girl X d X d received one X d from her father and the second from her mother, therefore the mother is bbX D X d .
| P | bbX D X d | x | BBX d Y | ||
| G | bX D | BX d | |||
| bXd | BY | ||||
| F1 | BbX D X d | BbX D Y | BbX d X d | BbX d Y | |
| girls from normal hearing and vision |
boys from normal hearing and vision |
girls from normal hearing, colorblind |
boys from normal hearing, colorblind |
Daughter BbX d X d . Probability of having colorblind children = 2/4 (50%). All of them will have normal hearing, the probability of being born deaf = 0%.
In humans, the gene for brown eyes dominates over blue eyes (A), and the gene for color blindness is recessive (color blindness - d) and linked to the X chromosome. A brown-eyed woman with normal vision, whose father had blue eyes and suffered from color blindness, marries a blue-eyed man with normal vision. Make a diagram for solving the problem. Determine the genotypes of the parents and possible offspring, the likelihood of having color-blind children with brown eyes and their gender in this family.
Answer
A – brown eyes, and – blue eyes.
X D – normal vision, X d – color blindness.
Brown-eyed woman with normal vision A_X D X _ .
The woman's father is aaX d Y, he could only give his daughter aX d, therefore, the brown-eyed woman is AaX D X d.
The woman's husband is aaX D Y.
P AaX D X d x aaX D Y
The probability of having a colorblind child with brown eyes is 1/8 (12.5%) and it is a boy.
One form of anemia (blood disease) is inherited as an autosomal dominant trait. In homozygotes this disease leads to death, in heterozygotes it manifests itself in a mild form. A woman with normal vision, but a mild form of anemia, gave birth to two sons from a healthy (by blood) color-blind man - the first, suffering from a mild form of anemia and color blindness, and the second, completely healthy. Determine the genotypes of the parents, sick and healthy sons. What is the probability of having the next son without anomalies?
Answer
AA – death, Aa – anemia, aa – normal.
X D – normal vision, X d – color blindness.
A woman with normal vision but mild anemia AaX D X _ .
A healthy blood color-blind man aaX d Y.
The first child is AaX d Y, the second child is aaX D Y.
The first child received Y from his father, therefore, he received X d from his mother, therefore, his mother AaX D X d.
P AaX D X d x aaX d Y
The probability of having the next son without anomalies is 1/8 (12.5%).
Deafness is an autosomal trait; color blindness is a gender-linked trait. In the marriage of healthy parents, a child was born who was deaf and color blind. Make a diagram for solving the problem. Determine the genotypes of the parents and the child, its gender, genotypes and phenotypes of possible offspring, the likelihood of having children with both anomalies. What laws of heredity are manifested in this case? Justify your answer.
Answer
Healthy parents gave birth to a sick child, therefore, deafness and color blindness are recessive traits.
A - normal. hearing, a - deafness
X D - normal. vision, X d - color blindness.
The child has aa, the parents are healthy, therefore they are Aa.
The father is healthy, therefore he is X D Y. If the child were a girl, then she would have received X D from her father and would not be color blind. Consequently, the child is a boy and received the color blindness gene from his mother. The mother is healthy, therefore fire X D X d .
P AaX D X d x AaX D Y
| AX D | AY | aX D | aY | |
| AX D | AAX D X D normal hearing normal vision girl |
AAX D Y normal hearing normal vision boy |
AaX D X D normal hearing normal vision girl |
AaX D Y normal hearing normal vision boy |
| AX d | AAX d X D normal hearing normal vision girl |
AAX d Y normal hearing colorblind boy |
AaX d X D normal hearing normal vision girl |
AaX d Y normal hearing colorblind boy |
| aX D | AaX D X D normal hearing normal vision girl |
AaX D Y normal hearing normal vision boy |
aaX D X D deafness normal vision girl |
aaX D Y deafness normal vision boy |
| aX d | AaX d X D normal hearing normal vision girl |
AaX d Y normal hearing colorblind boy |
aaX d X D deafness normal vision girl |
aaX d Y deafness color blindness boy |
The probability of having a child with two anomalies is 1/16 (6.25%).
In this case, Medel's third law (the law of independent inheritance) appeared.
The shape of the wings in Drosophila is an autosomal gene, the gene for eye color is located on the X chromosome. The male sex is heterogametic in Drosophila. When female fruit flies with normal wings and red eyes were crossed and males with reduced wings and white eyes, all offspring had normal wings and red eyes. The resulting F1 males were crossed with the original parent female. Make a diagram for solving the problem. Determine the genotypes and phenotypes of parents and offspring in two crosses. What laws of heredity appear in two crosses?
Answer
In the first generation, uniform offspring were obtained (Mendel's first law), therefore the parents were homozygotes, F1 were heterozygotes, and heterozygotes showed dominant genes.
A - normal wings, a - reduced wings
B - red eyes, b - white eyes
P AAX B X B x aaX b Y
F1 AaX B X b , AaX B Y
AaX B Y x AAX B X B
| AX B | aX B | AY | aY | |
| AX B | AAX B X B |
AaX B X B |
AAX B Y |
AaX B Y |
All offspring turned out to have normal wings and red eyes. In the second crossing, Mendel's third law (the law of independent inheritance) appeared.
In Drosophila, the heterogametic sex is male. Drosophila females with a gray body, red eyes and males with a black body, white eyes were crossed, all the offspring were uniform in body and eye color. In the second crossing of Drosophila females with a black body, white eyes and males with a gray body, red eyes, the offspring were females with a gray body, red eyes and males with a gray body, white eyes. Draw up crossing schemes, determine the genotypes and phenotypes of the parents, offspring in two crosses and the sex of the offspring in the first cross. Explain why the characteristics split in the second crossing.
Answer
A - gray body, a - black body
X E - red eyes, X E - white eyes
Since in the first crossing all the offspring were uniform, therefore, homozygotes were crossed:
P AA X E X E x aaX e Y
F1 AaX E X e, AaX E Y (all with gray body and red eyes)
Second crossing:
P aa X e X e x AAX E Y
F1 AaX e X E, AaX e Y (females with a gray body and red eyes, males with a gray body and white eyes)
The splitting of characters in the second generation occurred because the eye color trait is linked to the X chromosome, and males receive the X chromosome only from the mother.
In humans, the inheritance of albinism is not sex-linked (A - the presence of melanin in skin cells, and - the absence of melanin in skin cells - albinism), and hemophilia is sex-linked (X H - normal blood clotting, X h - hemophilia). Determine the genotypes of the parents, as well as the possible genotypes, sex and phenotypes of children from the marriage of a dihomozygous woman, normal for both alleles, and an albino man with hemophilia. Make a diagram for solving the problem.
Answer
A is normal, a is albinism.
X H - normal, X h - hemophilia.
Woman ААХ Н Х Н, man ааХ Н Х h.
Wing shape in Drosophila is an autosomal gene; the gene for eye size is located on the X chromosome. The male sex is heterogametic in Drosophila. When two fruit flies with normal wings and normal eyes were crossed, the offspring produced a male with curled wings and small eyes. This male was crossed with the parent. Make a diagram for solving the problem. Determine the genotypes of the parents and the resulting F1 male, and the genotypes and phenotypes of the F2 offspring. What proportion of females from the total number of offspring in the second cross is phenotypically similar to the parent female? Determine their genotypes.
Answer
Since crossing two fruit flies with normal wings resulted in a child with curled wings, therefore A - normal wings, a - curled wings, parents Aa x Aa, child aa.
The gene for eye size is linked to the X chromosome, therefore, a male with small eyes received Y from his father, and the gene for small eyes from his mother, but the mother herself had normal eyes, therefore, she was a heterozygote. X B - normal eyes, X b - small eyes, mother X B X b, father X B Y, child X b Y.
F1 AaX B X b x aaX b Y
Phenotypically similar to the parent female will be F2 AaX B X b, their 1/8 (12.5%) of the total number of offspring.
Problem 1. In humans, the gene for brown eyes is dominant over the gene for blue eyes. What is the probability of having blue-eyed children in a family where the mother had blue eyes and the father had brown eyes, and it is known that he is heterozygous for this trait?
Solution: Let's write down the crossing scheme.
R: ♀ aa x ♂ Aa
blue brown
G: (a) (A) , (a)
brown blue
Answer: the probability of having a blue-eyed child is 50%.
Task 2. Phenylketonuria is inherited as an autosomal recessive trait. In a family where both parents were healthy, a child with phenylketonuria was born. What is the probability that the second child in this family will also be sick?
Solution. R: ♀ A- x ♂ A-
Reasoning. Since both parents are healthy, they can have both the AA and Aa genotypes. Since the first child in this family was sick, his genotype is aa. According to the gamete purity hypothesis, the body receives one allele of a gene from the father, and the other from the mother. Consequently, both parents are heterozygous for the analyzed trait - Aa.
Now you can determine the probability of having a second child with a patient:
R: ♀ Aa x ♂ Aa
Norm norm
G: (A), (a) (A), (a)
F 1: AA, 2 Aa, aa
Normal normal phenylketonuria
Thus, 75% of children will be healthy, and 25% will be sick.
Answer: 25%.
Task 3. In some breeds of cattle, polledness is dominant over hornedness.
A) When crossing polled and horned animals, 14 horned and 15 polled offspring were born. Determine the genotypes of the parental forms.
C) As a result of crossing horned and polled animals, all 30 offspring were polled. Determine the genotypes of the parental forms.
C) Crossing polled animals with each other produced 12 polled and 3 horned calves. Determine the genotypes of the parental forms.
Solution. The genotypes of the parents can be determined by segregation in the offspring. In the first case, the splitting was 1:1, therefore there was an analyzing cross:
R: ♀ Aa x ♂ aa
polled horniness
G: (A), (a) (a)
polled hornedness
In the second case, there was uniformity of the offspring, therefore homozygous horned and polled animals were crossed:
R: ♀ AA x ♂ aa
Norm norm
polled
In the third case, a 3:1 split occurred, which is only possible when crossing two heterozygotes:
R: ♀ Aa x ♂ Aa
polled polled
G: (A), (a) (A), (a)
F 1: AA, 2 Aa, aa
polled hornedness
75% - polled
25% - horned
Answer: A) Aa and aa
B) AA and aa
C) Aa and Aa
Task 4. In humans, brown eyes dominate over blue ones, and right-handedness dominates over left-handedness.
1. What is the probability of having a left-handed, blue-eyed child in a family where the mother is blue-eyed and right-handed (although her father was left-handed), and the father has brown eyes and is predominantly left-handed, although his mother was blue-eyed and right-handed?
2. In a family of brown-eyed, right-handed people, a left-handed child with blue eyes was born. What is the probability that your next child will be right-handed and have blue eyes?
♂--вв ♀ааВ-
R: ♀ aaB- x ♂A-bb
Reasoning. First you need to determine the genotypes of the parents. Since the woman’s father was left-handed, she is therefore heterozygous for the B gene; the man is heterozygous for gene A, since his mother had blue eyes.
Examination. Let's write down the crossing scheme:
R: ♀ aaVv x ♂Aavv
Blue, right Car., lion.
G: (аВ), (ав) (Ав), (ав)
F 1: AaBv, aaBv, Aavv, aavv
Kar., right. Goal, right Kar, lion. Blue, lion
Answer: the probability of having a blue-eyed, left-handed child is 25%.
R: ♀ A-B- x ♂ A-B-
Brown-eyed Brown-eyed
Right-handed right-handed
Blue-eyed lefty
Reasoning. Since, according to the gamete purity hypothesis, the organism receives one gene allele from one parent, and the other from the other, then both parents are heterozygous for both pairs of analyzed genes; their genotype is AaBb.
P: ♀ AaBv x ♂ AaBv
Kar. Right Kar. Right
Thus, the probability of having the next right-handed child with blue eyes (genotype aaBb or aaBB) is 3/16 (or 18.75%).
