The dominant eye color is blue or brown. How eye color is inherited. Formation of iris color
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.
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%).
Answer: 18.75%.
TASKS FOR INDEPENDENT SOLUTION
1. In oats, normal growth dominates over gigantism. A heterozygous plant with normal growth was crossed with a giant one. Determine the genotypes and phenotypes of the offspring.
2. The presence of a white strand of hair above the forehead is determined by a dominant autosomal gene. In a family where the father had a gray strand of hair and the mother did not, a son was born with a white strand of hair above his forehead. Determine the probability of the next child being born without a gray strand of hair.
3. In humans, long eyelashes and cataracts are determined by dominant autosomal unlinked genes. A man with short eyelashes and cataracts and a woman with long eyelashes and normal vision entered into marriage. It is known that only the man’s father suffered from cataracts, while his mother had normal vision and had short eyelashes. The first child in the family was born with short eyelashes and normal vision. Determine the probability of the next child being born with cataracts.
4. The ability to taste phenylthiourea is determined in humans by a dominant autosomal gene. Polydactyly is another dominant gene. Both genes are located on different chromosomes.
A) Determine the probability of the appearance of children with polydactyly in a family of heterozygous parents.
B) In the family, the father has polydactyly, is able to taste phenylthiourea and is heterozygous for both pairs of genes, and the mother is healthy and does not taste phenylthiourea. Determine possible genotypes and phenotypes of children from this marriage.
5. Humans have two forms of hereditary deafness, determined by recessive genes. Determine the probability of sick children appearing in a family where:
A) both parents suffer from the same forms of deafness, and they are heterozygous for another pair of genes;
B) the parents suffer from different forms of deafness, and are they also heterozygous for another form of the disease?
6. In Drosophila, the rugged wing margin and forked setae are determined by recessive unlinked autosomal genes. The laboratory acquired flies with normal wings and straight bristles.
A) how can you be sure that the acquired individuals do not carry the genes for rugged wing margins and forked setae?
B) determine the possible genotypes and phenotypes of the offspring from crossing heterozygous flies with a cut wing and straight bristles and flies with a normal wing and forked bristles.
C) A line with a jagged wing and forked setae and a line with a normal wing and straight setae are crossed. 78 F 1 fruit flies were obtained. 96 descendants were obtained from crossing them with each other. How many types of gametes do flies form from F 1? How many F 1 flies are heterozygous? How many F 2 flies are not capable of producing offspring with jagged wings and forked setae when crossed with completely recessive individuals?
7. In chickens, the genes for black plumage and the presence of a crest dominate over the genes for brown color and the absence of a crest. A) Black crested hen and rooster are crossed. The offspring produced 16 chickens - 8 black crested, 3 black without crest, 4 brown crested and 1 brown without crest. Determine the genotypes of the parental forms. B) Black crested hen and rooster are crossed. All offspring (17 chicks) are similar to their parents. Determine the genotypes of the parental forms. C) A black crested hen and a brown rooster without a crest are crossed. 16 chickens were obtained: 4 black crested, 3 black without crest, 4 brown crested and 5 brown without crest. Determine the genotypes of the parental forms. D) A black crested hen and a brown rooster without a crest are crossed. 17 chickens were obtained: 9 black crested and 8 brown crested. Determine the genotypes of the parental forms.
8. In dogs, the genes for black coat color, hard coat and floppy ear are dominant over the genes for brown coat, soft coat and erect ear. At the kennel, a brown soft-haired puppy with erect ears appeared from crossing a black soft-haired dog with a floppy ear and a brown wire-haired male with a floppy ear. Determine the genotypes of the parent dogs and the likelihood of them having offspring with brown hair.
9. Gray hair is a dominant autosomal trait. In the family, the mother has a strand of gray hair, and the father is healthy. Determine the probability of having children with gray hair in this family if it has been established that the mother is heterozygous for this trait.
10. The ability to taste phenylthiourea is a dominant trait, the gene for which is located on the autosome. In a family of parents who sense the taste of phenylthiourea, a daughter was born who does not sense it. To determine the likelihood of children in the family who can taste phenylthiourea. What is the likelihood of them having grandchildren who are unable to sense this taste?
11. In guinea pigs, the gene for shaggy hair is dominant over the gene for smooth hair. By crossing two breeds of pigs, one with shaggy fur and the other with smooth hair, 18 furry babies were born. Later they were crossed with each other and got 120 descendants. How many F 2 offspring are heterozygotes?
12. In oats, early ripeness dominates over late ripeness. The gene that determines this trait is located on the autosome. Two heterozygous early ripening plants are crossed. There are 900 plants in the offspring. How many F 1 plants have the trait of late ripening? How many early maturing plants will be homozygous?
13. Hereditary blindness in some dog breeds is a recessive autosomal trait. A pair of dogs with normal vision gave birth to 3 puppies, one of which turned out to be blind. 1) One of the normal puppies from this litter was allowed to continue breeding. What is the probability that his descendants will be blind again? 2) Determine the genotypes of all individuals participating in the crossing. 3) How many types of gametes does a blind dog produce?
14. When two fruit flies were crossed, out of 98 offspring, 23 turned out to be black, the rest were gray in color. Which trait is dominant? Determine the genotypes of the parental forms.
15. When gray chickens were crossed with black ones, all the offspring were gray in color. In F 2, 96 chickens were obtained. How many F 2 hens, when crossed with roosters of the same genotype, can produce black chickens?
16. Hypophosphatemia (a disease of the skeletal system) is inherited in humans as an autosomal recessive trait. Myopia is determined by a dominant autosomal gene. Both traits are inherited independently. 1) In a family, both parents are heterozygous for the analyzed genes. Determine the probability of having a child suffering from both diseases. How many children (%) in this family are completely homozygous? 2) In the family, the mother and all her relatives are healthy. The father suffers from both diseases. Determine the probability of having healthy children if it is known that only the father suffered from myopia.
17. Aniridia is an autosomal dominant disease characterized by the absence of the iris. 1) What is the probability of having sick children in a family of healthy parents if the wife’s parents and all relatives are healthy, and the husband’s mother suffered from the specified disease? 2) What is the probability of having sick children in a family of healthy parents if both the husband and wife have one of the parents suffering from aniridia? 3) A child with aniridia was born into a family of healthy parents. What is the probability of having a second child healthy?
18. In humans, two forms of hereditary deafness are known, both are inherited as recessive autosomal unlinked traits. 1) determine the probability of having sick children in a family where both parents have the same form of deafness, for which they are heterozygous; 2) determine the probability of having sick children in a family where both parents have various forms of deafness, and for other forms they are homozygous; 3) determine the probability of having sick children in a family where both parents have both forms of deafness, if it is known that the mothers of both spouses suffered from one form of deafness, and the fathers from another.
19. In humans, two forms of fructosuria are known and both are inherited as autosomal recessive unlinked traits. One of the forms is characterized by a rather severe course, and the second by a mild course. A myopic man with a mild form of fructosuria is married to a woman with signs of a severe form. The first child in this family had normal vision, but was sick with a severe form of fructosuria. Determine the probability of having healthy children in this family if the maternal grandmother suffered from a mild form of fructosuria, and the grandfather and all his relatives are healthy.
20. How many types of gametes will be produced by an organism having the following genotype: a) AaBbCs; b) АаВВСс; c) Aavvss; d) aaВвСС.
Related information.
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?
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%).
Blue and green, their inheritance results in two pairs of genes. The shades of these colors are determined by the individual characteristics of the body to distribute melanin in chromatophores, which are located in the iris. Other genes that are responsible for hair color and skin tone also affect the shade of eye color. For blond people with light skin are typical, and representatives of the Negroid race have dark brown eyes.
The gene that is responsible only for eye color is located on chromosome 15 and is called HERC2, the second gene - EYCL 1 is located on chromosome 19. The first gene carries information regarding brown and blue colors, the second - about green and blue.
The dominant color in the HERC2 allele is brown, the dominant color in the EYCL 1 allele is green, and blue eyes are inherited in the presence of a recessive trait in two genes. In genetics, it is customary to denote a dominant trait with a capital letter of the Latin alphabet, and a recessive trait is a lowercase letter. If there are uppercase and lowercase letters in a gene, the organism is heterozygous for this trait and exhibits a dominant color, and a hidden recessive trait can be inherited by a child. A “suppressed” trait will appear in a baby when an absolute recessive allele is inherited from two parents. That is, parents may well have a child with blue eyes or with.
Using Latin letters, brown eye color, which is determined by the HERC2 gene, can be designated AA or Aa; blue eyes correspond to the set aa. When a trait is inherited, the child receives one letter from each parent. Thus, if dad has a homozygous trait of brown eyes, and mom has blue eyes, then the calculations look like this: AA+aa=Aa, Aa, Aa, Aa, i.e. a child can only achieve the Aa set, which manifests itself according to the dominant, i.e. the eyes will be brown. But if the father is heterozygous and has the Aa set, and the mother is blue-eyed, the formula looks like: Aa+aa=Aa, Aa, aa,aa, i.e. there is a 50% chance that a child with a blue-eyed mother will have the same eyes. For blue-eyed parents, the eye inheritance formula looks like: aa+aa=aa,aa, aa, aa, in this case the baby inherits only the recessive allele aa, i.e. his eye color will be blue.
In the EYCL 1 allele, eye color is inherited in the same way as in the HERC2 gene, but only the letter A indicates green. Nature arranges it in such a way that the existing dominant trait of brown eyes in the HERC2 gene “defeats” the existing green trait in the EYCL 1 gene.
Thus, a child will always inherit brown eye color if one of the parents has a homozygous dominant set of AA in the HERC2 gene. If a parent with brown eyes passes on the recessive gene a to the child, i.e. a sign of blue eyes, then the color of the eyes determines the presence of a green dominant trait in the EYCL 1 gene. In cases where a parent with green eyes does not transmit the dominant trait A, but “gives” the recessive allele a, the child is born with blue eyes.
Since eye color is determined by two genes, its shades are obtained from the presence of undetected characteristics. If a child has the AA genetic set in the HERC2 allele, then the eyes will be dark brown. The presence in the HERC2 gene of the trait of brown eyes of type Aa, and in the EYCL 1 gene of the recessive trait aa, causes light brown eyes. The homozygous trait of green eyes AA at the EYCL 1 locus determines a more saturated color than the heterozygous set Aa.
