Blue and green, their inheritance results in two pairs of genes. The shades of these colors are determined individual characteristics The body distributes 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 fair 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 color, the second is 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 when there is a recessive trait in two genes. In genetics, it is customary to designate dominant capital letter Latin alphabet, the recessive trait is a lowercase letter. If a gene contains capital letters and lowercase letters- the organism is heterozygous for this trait and has a dominant color, but 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 blue-eyed child 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 the father has a homozygous trait of brown eyes, and the mother 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 denotes 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. Presence of the trait in the HERC2 gene brown eyes according to type Aa, and in the EYCL 1 gene of the recessive trait aa, causes light brown eyes. Homozygous trait green eyes AA in the EYCL 1 locus determines a more saturated color than the heterozygous set Aa.

In humans, the gene for brown eyes is dominant over the gene causing blue eyes. A blue-eyed man, one of whose parents had brown eyes,

married a brown-eyed woman whose father had brown eyes and whose mother had blue eyes. What kind of offspring can be expected from this marriage?

In humans, the gene for brown eyes is dominant over the gene causing blue eyes. A blue-eyed man, one of the parents who had brown eyes, got married

on a brown-eyed woman whose father had brown eyes and his mother had blue eyes. What kind of offspring can be expected from this marriage?

No. 2. What kind of offspring can you expect?

from crossing a cow and a bull,
heterozygous for coat color?
Problem No. 3. Guinea pigs have frizzy hair
wool is determined by a dominant gene, and
smooth - recessive.
1. Crossing two curled pigs between
produced 39 individuals with shaggy fur and
11 smooth-haired animals. How many among
individuals having a dominant phenotype,
must be homozygous for this
sign?
2. Guinea pig with shaggy fur
crossed with an individual with a smooth
wool, gave birth to 28 frizzy and 26
smooth-haired offspring. Define
genotypes of parents and offspring.
Task No. 4. An offspring was obtained at the fur farm
in 225 minks. Of these, 167 animals have
brown fur and 58 bluish-gray minks
coloring. Determine the genotypes of the original
forms, if it is known that the gene is brown
coloring dominates the gene,
defining bluish-gray color
wool
Problem No. 5. A person has a gene for brown eyes
dominates the gene causing
Blue eyes. Blue-eyed man, one
of whose parents had brown eyes,
married a brown-eyed woman who
the father had brown eyes, and the mother had blue eyes.
What kind of offspring can be expected from this
marriage?
Problem No. 6. Albinism is inherited in
human as a recessive trait. In the family
where one of the spouses is an albino, and the other has
pigmented hair, have two children.
One child is albino, the other is
dyed hair. What is the probability
birth of the next albino child?
Problems on di- and polyhybrid crossing
Problem No. 8. At a large cattle gene
polledness dominates the horned gene, and
gene for black coat color - above the gene
red color. Both pairs of genes are located
in different pairs of chromosomes.
1. What will the calves turn out to be like if they are crossed?
heterozygous for both pairs
signs of a bull and a cow?
2. What kind of offspring should be expected from
crossing a black polled bull,
heterozygous for both pairs of traits,
with a red horned cow?
Problem No. 9. Dogs have black fur
dominates the coffee, and the short coat
- over the long one. Both pairs of genes are in
different chromosomes.
1. What percentage of black shorthairs
puppies can be expected from crossing two
individuals heterozygous for both traits?
2. The hunter bought a black dog with a short
wool and wants to be sure that it is not
carries genes long hair coffee color.
Which partner according to phenotype and genotype do you need?
select for crossing to check
genotype of the purchased dog?
Problem No. 10. A person has a gene for brown eyes
dominates the gene that determines
development of blue eye color, and the gene
conditioning the ability to better master
right hand, predominates over the gene,
determining the development of left-handedness. Both
pairs of genes are located in different
chromosomes. What kind of children can be if
Are their parents heterozygous?

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 this characteristic is he heterozygous?

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 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%).

The color of the iris is formed by coloring pigments (melanocytes) and depends on their concentration in the stroma. If a small amount of coloring pigment is produced, the eyes will most likely be light (blue or gray). In people with brown and black eyes, melanocytes in the stroma are located in large quantities. The number of cells producing coloring pigment is determined at the level of genotype formation and is a hereditary factor.

Many newborns are born with a light blue iris. This happens due to the fact that the mechanism of melanin production has not yet been fully established. Around six months, the number of melanocytes increases and the baby's eyes may change color to a darker one. If a baby is born with brown eyes, the possibility that they will eventually become blue is zero, since brown is the predominant color, and blue is weaker (recessive).

Mechanisms of inheritance of eye color in humans

It is impossible to predict with absolute accuracy what color the baby’s iris will be. Mendel's laws say that iris color is inherited in the same way as hair color. Programming genes dark colors, are considered stronger (dominant), and the genes that create a light color are considered weaker. When forming a phenotype dominant gene takes over the recessive giving the eyes a stronger and more saturated color

Basic situations of gene interaction

Brown-eyed parent and blue eyed parent(AA and aa)
The child of such a couple will have the genotype Aa, this means that the genotype of his brown-eyed father is AA, and his mother is aa. During the merger, the dominant genes interacted with the recessive ones and created the Aa pair, where the father's genes predominate. The probability that a child's eyes will be brown is 90%. There are exceptions, mostly, for some reason only girls fall under them, so the possibility of having a blue-eyed baby is also quite real, although only in 10% of general cases.

Brown-eyed parent (genotype Aa) and blue-eyed parent (aa)
In this situation, after the merger, the following four genotypes will appear: Aa, Aa, aa, aa. The chances of both genotypes are equal, since the probability of having a brown-eyed or blue-eyed child is the same (50 to 50).

Brown-eyed parents (genotype Aa)
In this case we see the formation of three pairs dominant genotype Ah, so in 75% of cases this couple will have a child with brown eyes.

Both parents with blue eyes (genotype aa)
In such a tandem, the probability of having a child with eyes like their parents is almost 100%, since the genotype of this couple absolutely does not contain the dominant gene “A”, which is responsible for dark color iris.

In relation to himself rare color eyes - green, blue will dominate, forming, so the probability of having a baby with green eyes will be 40%. If one parent has brown eyes and the other has green eyes, their child:

• in 50% of cases, be born with brown eyes;
• in 37% of cases with greens;
• and in 13% of cases will have blue eyes.

Recently, genetic scientists have identified 4 additional genes that affect the mechanism of inheritance of iris color. It was found that the genes of the ancestors, as well as their ancestors up to the 16th generation, are also responsible for the mechanism of formation of the color of the iris. Therefore, if parents with brown eyes give birth to a child with sky-colored eyes, this may be the result of a predominance of the grandparents' genes.

The mechanism of color inheritance is a rather complex genetic process, which is determined by the interaction of dominant and recessive genes. On the formation of the number of melanocytes and their location as well as others hereditary factors and geographic location of a person.

Eye color: how it is passed on from parents to child. Calculate the child's eye color.

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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:

(*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 ones come from? HERC2 green eye - explained below. In the meantime - some data from general theory genetics: brown HERC2 - dominant, and blue is recessive, so the carrier has one brown and one blue HERC2 eye color will be brown. However, to his children the bearer of one brown and one blue HERC2 with a 50x50 probability it can transmit both brown and blue HERC2 , that is, the dominance of brown has no effect on the transfer of the copy HERC2 children.

For example, a wife has eyes brown color, 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, since she will pass on one of her two to the children brown genes. But grandchildren can have eyes of any color:

So, for example:

HERC2 about the mother's t is brown (the mother, for example, has both HERC2 brown)

HERC2 from the father - blue (father, for example, has both HERC2 blue)

HERC2 The child has one brown and one blue. The eye color of such a child is always brown; at the same time your HERC2 he can pass on the blue color to his children (who can also receive it from the second parent HERC2 blue and then have eyes either blue or green).

Now let's move on to the gene gay: it comes in green and blue (blue, grey); each person also has two copies: a person receives one copy from his mother, the second from his father. Green gay is dominant gene, blue gay - recessive. A person thus has either 2 blue gene gay or 2 green genes gay or one blue and one green gene gay . At the same time, this affects the color of his eyes only if he has HERC2 from both parents - blue (if he received brown from at least one of the parents HERC2 , his eyes will always be brown).

So, if a person received blue from both parents HERC2 , depending on the gene gay his eyes might be following colors:

gay gene: 2 copies	Human eye color
Green and Green	green
Green and blue	green
blue and blue	blue

General table for calculating the eye color of a child, brown eye color is indicated by “K”, green eye color is indicated by “Z” and blue eye color is indicated by “G”:

HERC2	Gey	eye color
QC	ZZ	brown
QC	Zg	brown
QC	GG	brown
Kg	ZZ	brown
Kg	Zg	brown
Kg	GG	brown
yy	ZZ	green
yy	Zg	green
yy	GG