home Auditorium 7 Child's eye color. Genetic possibility in %. Blue eyes are the result of a genetic mutation. Brown eyes are a dominant gene.

Child's eye color. Genetic possibility in %. Blue eyes are the result of a genetic mutation. Brown eyes are a dominant gene.

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 experiencing the taste of 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.

What happens to a child who is heterozygous for eye color? The answer is: he will have brown eyes.

The fact is that the child has one gene that can form a large amount of tyrosinase, and a gene that can form a small amount of tyrosinase. However, a single gene can produce a relatively large amount of tyrosinase, and this may be enough to turn the eyes brown.

As a result, two parents, one of whom is homozygous for brown eyes and the other homozygous for blue eyes, have children who are heterozygous and at the same time have brown eyes. The gene for blue eyes does not appear.

When a person has two different genes for some physical characteristic at identical locations on a pair of chromosomes and only one gene is expressed, that gene is called dominant. A gene that is not expressed is recessive. In the case of eye color, the gene for brown eyes is dominant to the gene for blue eyes. The gene for blue eyes is recessive to the gene for brown eyes.

It is impossible to tell just by looking at a person whether he is homozygous or heterozygous for brown eyes. Either way, his eyes are brown. One way to say something definitive is to find out something about his parents. If his mother or his father had blue eyes, he must be heterozygous. Another way to know something is to see the color of his children's eyes.

We already know that if a man who is homozygous but has brown eyes marries a woman who is homozygous for brown eyes, they will have children who are homozygous for brown eyes. But what will happen if he marries a heterozygous girl? A homozygous male would only form sperm cells with brown eye genes. His heterozygous wife would produce two types of eggs. During meiosis. since her cells have both a brown eye gene and a blue eye gene, the brown eye gene will travel to one end of the cell and the blue eye gene to the other. Half of the formed eggs will contain the gene for brown eyes, and the other half will contain the gene for blue eyes.

You need to understand that eye color depends not only on the pigment produced. The iris consists of anterior and posterior layers. The color of the eye depends on the distribution of pigments in different layers. In addition, the vessels and fibers of the iris play a role. For example, green eye color is determined by the blue or gray color of the back layer of the iris, and light brown pigment is distributed in the front layer. The total is green.

The definition of gray and blue eyes is similar, only the density of the fibers of the outer layer is even higher and their shade is closer to gray. If the density is not so high, then the color will be gray-blue. The presence of melanin or other substances produces a small yellow or brownish impurity.

The structure of the black iris is similar to the brown one, but the concentration of melanin in it is so high that the light incident on it is almost completely absorbed.

The chance of a sperm cell fertilizing an egg with the gene for brown eyes or an egg with the gene for blue eyes is therefore 50/50. Half of the fertilized eggs will be homozygous for brown eyes, and half will be heterozygous. But all children will have brown eyes.

Now suppose that both father and mother are heterozygous. Both would have brown eyes, but both would also have the gene for blue eyes. The father would form two kinds of sperm cells, one with the gene for blue eyes and one with the gene for brown eyes. In the same way, the mother would form two types of eggs.

Several combinations of sperm and egg cells are now possible. Suppose one of the sperm cells with the brown eye gene fertilizes one of the eggs with the brown eye gene. The child in this case will be homozygous for brown eyes and will naturally have brown eyes. Suppose that a sperm cell with a gene for brown eyes fertilizes an egg cell with a gene for blue eyes, or a sperm cell with a gene for blue eyes fertilizes an egg cell with a gene for brown eyes. In either case, the child will be heterozygous and will still have brown eyes.

But there is another option. What if a sperm cell with the blue eye gene fertilizes an egg with the blue eye gene?

In this case, the child will be homozygous for blue eyes and will have blue eyes.

Thus, two brown-eyed parents can have a blue-eyed child. A gene that had seemed to disappear appeared again.

Besides, you can tell something about the parents by looking at the child. Although their eyes are brown, just like the homozygous person, you know that they both must be heterozygous, otherwise the gene for blue eyes would not express itself.

Chapter from the book " Races and peoples»

William Boyd "Tsentrpoligraf" 2005

Approximate map of the distribution of blue and green eyes in Europe.

Blue And blue eyes are most common among the European population, especially in the Baltics and Northern Europe. Eyes of these shades are also found in the Middle East, for example, in Afghanistan, Lebanon, Iran.

Grey eye color is most common in Eastern and Northern Europe. It is also found in Iran, Afghanistan, Pakistan and parts of Northwest Africa.

Purely green eye color is extremely rare. Its speakers are found in Northern and Central Europe, less often in Southern Europe.

Brown- the most common eye color in the world. It is widespread in Asia, Oceania, Africa, South America and Southern Europe.

Black the type is widespread primarily among the Mongoloid race, in South, Southeast and East Asia.

PSYCHOLOGY

Psychology of sympathy

Appearance is not the only factor on which our attitude towards people depends. When we get to know a person, in addition to his appearance, we immediately notice his other properties that enhance or, conversely, reduce the impression that his appearance made on us.

There are certain prevailing ideas about what a positive person should be. So, many of us are convinced that a girl should be beautiful and a man should be smart. If you look at it, the requirement is quite cruel: clearly, not all girls are beautiful, just as not all men are very smart (after all, when we say “smart,” we mean that he is smarter than others, smarter than the majority, stands out from the majority). It turns out that we are ready to recognize only some privileged part of our fellow citizens as worthy of attention, defining everyone else an order of magnitude lower. In everyday life, of course, we don’t think about it, we don’t analyze this stereotype so deeply, but it lingers in our consciousness, takes root, and it turns out that it’s not always easy to get rid of it.

The next circumstance on which the emergence of sympathy depends is the dissimilarity or similarity of the partners. They often say that these people got together because they are similar to each other. They say it no less often. that people got together precisely because they were very different. Depending on the situation, either one or the other is significant.

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 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 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.

To answer the question The gene for brown eyes is considered dominant. Does he always win if the other parent's eyes are not brown? given by the author old-timer the best answer is 3:1, unless I forgot the arithmetic. That is, such parents will have three children with brown eyes, and the fourth with blue eyes. But if the brown-eyed person himself has grandparents with different eye colors, the likelihood that the child will be blue-eyed increases MAYBE....

Answer from Alina[guru]
No not always. Mom's eyes are brown, father's are blue. I have brown ones, my sister has blue ones, and my second sister has green ones. My son's father has brown eyes, that is, both parents have brown eyes, but his son's eye color is green.

Answer from Neurosis[guru]
no, in my opinion this is nonsense... if he doesn’t always win, why is he dominant?

Answer from Accomplice[newbie]
No, no and NO! My mom and dad are brown-eyed, and I have blue eyes! My husband has brown ones, and our children have blue ones! Here!

Answer from Alexey 11[guru]
Unfortunately it is so

Answer from Larisa Pavlova[guru]
no, not always, my eyes are gray and my husband’s are brown, our daughter’s eyes are gray like mine

Answer from EYES GREEN KNEES BLUE[guru]
In our case, the brown one won))) as you know, I have green eyes))

Answer from Diver[active]
not always but often

Answer from N.[guru]
We went through this at school, there are calculations on the power of probability, open the textbook "General Biology".)

Answer from Yatyana[guru]
I have brown eyes, my husband has blue eyes, but my daughter still has blue eyes like her husband, which means that brown ones do not always dominate

Answer from Olka[newbie]
Not always!! ! A child does not necessarily inherit a dominant gene, but can also inherit a recessive one if it was present in previous generations). And we have brown-eyed parents and a blue-eyed daughter))

Answer from Anna Zilina[active]
won against us. My husband has brown eyes, my daughter also has brown eyes.

Answer from Kisa[expert]
Biology textbook for 9th grade. No not always.

Answer from ~Give me the Crown, Darling~[guru]
not always. my father is brown, and my mother is green

Answer from MarS[guru]
My wife and I have brown eyes, and my daughter has blue eyes, like her grandfather’s. AND.. . such a paradox. Until I was seven years old, my eyes were blue, and then they suddenly turned brown, that is, they became caried-dominated:.... (my father had blue eyes, my mother had brown eyes).

Answer from Irina I[guru]
Not always.

Answer from Hyperv strat[newbie]
We cannot isolate all alleles. Compared to all common alleles, it wins. But what if there is such a gray-brown-crimson color that its allele is more dominant? It's impossible to say for sure.

Answer from Vita[guru]
not always

Answer from Lyubov Semyonova[guru]
My eyes are green, my husband's are brown, and my daughter's are blue.

Answer from [email protected] [guru]
No. I have brown eyes, but our dad has blue eyes and our daughter has blue eyes.

Question: In humans, the gene for brown eyes dominates over blue eyes, and the ability to use predominantly the right hand dominates over left-handedness. The genes are not linked. A blue-eyed, right-handed man married a brown-eyed, right-handed woman. They had two children: a brown-eyed left-hander and a blue-eyed right-hander. Determine the probability of birth in this family of blue-eyed children who control predominantly the left hand.

In humans, the gene for brown eyes dominates over blue eyes, and the ability to use predominantly the right hand dominates over left-handedness. The genes are not linked. A blue-eyed, right-handed man married a brown-eyed, right-handed woman. They had two children: a brown-eyed left-hander and a blue-eyed right-hander. Determine the probability of birth in this family of blue-eyed children who control predominantly the left hand.

Answers:

Phenotype gene genotype brown A AA, Aa blue a aa right-handed B BB, BB left-handed in BB solution P mother AaBv*father aaBv G AB, Av, aB, av, aB, av F1 AaBB, AaBv, aaBB, aaBv, AaBv, Aavv ,ааВв,аавв kar kar gol kar kar gol gol pr pr pr pr pr right left right left Answer: the probability of giving birth to a blue-eyed left-hander is 1/8 or 12.5%

Given: A --- brown eyes AA, Aa a --- blue eyes aa B --- right-handedness BB, Bb b --- left-handedness bb Find: the probability of the birth in this family of blue-eyed children who control predominantly their left hand --- ? Solution: P: ♀AaBb x ♂aaBb G: AB aB Ab ab aB ab F₁: AaBB --- brown eyes; right-handedness; AaBb --- brown eyes; right-handedness; aaBB --- blue eyes; right-handedness; aaBb --- blue eyes; right-handedness; AaBb --- brown eyes; right-handedness; Aabb --- brown eyes; left-handedness; aaBb --- blue eyes; right-handedness; aabb --- blue eyes; left-handedness; ----1/8*100%=12.5% Answer: the probability of birth in this family of blue-eyed children who control predominantly the left hand is ---12.5%.