What are the early signs of pregnancy. When is the best time to get tested ...
Municipal state educational institution Lyceum No. 4
rossosh of the Rossoshansk municipal district of the Voronezh region.
Methodological development in biology to help students taking the exam.
"Problems in genetics" for grade 11
biology teacher of the highest qualification category
2016
I. Monohybrid crossing problems (complete and incomplete dominance)
1. In rabbits, the gray color of the coat dominates over black. A homozygous gray rabbit was crossed with a black rabbit. Identify the phenotypes and genotypes of the rabbits?
2. In guinea pigs, the black color of the coat dominates over the white. Two heterozygous male and female were crossed. What will the first generation hybrids be like? What law is manifested in this inheritance? 
3. When two white pumpkins were crossed in the first generation, ¾ of the plants were white and ¼ were yellow. What are the genotypes of the parents if white dominates over yellow? 
4. Black Cow Nochka brought a red calf. The red cow Zorka has a black calf. These cows are from the same herd, in which there is one bull. What are the genotypes of all animals? Consider different options. ( The gene for black coloration is dominant.)
5. How many dwarf pea plants can be expected when sowing 1200 seeds obtained by self-pollination of tall heterozygous pea plants? (Seed germination is 80%).
6. The fruit of a watermelon can be green or striped. All watermelons obtained by crossing plants with green and striped fruits had only a green fruit rind. What color of watermelon fruit can be in F 2?
7.In snapdragons, plants with wide leaves, when crossed, always produce offspring with the same leaves, and when a narrow-leaved plant is crossed with a broad-leaved plant, plants with leaves of intermediate width arise. What will be the offspring from crossing two individuals with leaves of intermediate width. 
1. In humans, glaucoma is inherited as an autosomal recessive trait (a), and Marfan's syndrome, accompanied by an abnormality in the development of connective tissue, is inherited as an autosomal dominant trait (B). Genes are found in different pairs of autosomes. One of the spouses suffers from glaucoma and did not have ancestors with Marfan's syndrome in the family, and the second is diheterozygous for these characteristics. Determine the genotypes of the parents, possible genotypes and phenotypes of children, the likelihood of having a healthy child. Make a scheme for solving the problem. What is the law of heredity manifested in this case?
2. Low-growing (dwarf) tomato plants with ribbed fruits and plants of normal height with smooth fruits were crossed. In the offspring, two phenotypic groups of plants were obtained: undersized with smooth fruits and normal height with smooth fruits. When crossing undersized tomato plants with ribbed fruits with plants having a normal stem height and ribbed fruits, all the offspring had a normal stem height and ribbed fruits. Make crossings schemes. Determine the genotypes of the parents and offspring of tomato plants in two crosses. What is the law of heredity manifested in this case? 
3. When crossing horned red cows with hornless black bulls, calves of two phenotypic groups were born: horned black and hornless black. With further crossing of the same horned red cows with other hornless black bulls, the offspring included hornless red and hornless black bulls. Write schemes for solving the problem. Determine the genotypes of the parents and offspring in two crosses. What is the law of heredity manifested in this case? 
4. Fruits of tomatoes can be red and yellow, bare and pubescent. Genes for yellow coloration and pubescence are known to be recessive. Of the tomatoes harvested on the collective farm, there were 36 tons of red pubescent and 12 tons of red pubescent. How many (approximately) in a collective farm harvest can there be yellow fluffy tomatoes if the source material was heterozygous?
5. When crossing a variegated crested (B) hen with the same rooster, eight chickens were obtained: four variegated crested chicks, two white (a) crested and two black crested chicks. Make a scheme for solving the problem. Determine the genotypes of parents, offspring, explain the character of inheritance of traits and the appearance of individuals with a variegated color. What laws of heredity are manifested in this case? 
6. In mice, black coat color dominates over brown (a). The long tail is determined by the dominant gene (B) and develops only in the homozygous state; the short tail develops in heterozygotes. Recessive genes that determine the length of the tail in a homozygous state cause the death of embryos. The genes of the two traits are not linked.When mating a female mouse with black hair and a short tail and a male with brown hair and a short tail, the following offspring were obtained: black mice with a long tail, black with a short tail, brown mice with a long tail and brown mice with a short tail. Make a scheme for solving the problem. Determine the genotypes of parents and offspring, the ratio of phenotypes and genotypes of the offspring, the probability of embryo death. What is the law of heredity manifested in this case? Justify the answer. 
7. We crossed a plant of whiskered white-fruited strawberries with beanless red-fruited plants (B). All hybrids turned out to be whiskered pink-fruited. When analyzing the crossing of F 1 hybrids, phenotypic cleavage occurred in the offspring. Make a scheme for solving the problem. Determine the genotypes of the parents, first generation hybrids, as well as the genotypes and phenotypes of the offspring in the analyzing crossing (F 2). Determine the nature of inheritance of the trait of color of the fetus. What laws of heredity are manifested in these cases? 
8. When crossing phlox plants with white flowers and a funnel-shaped corolla with a plant with cream flowers and flat corollas, 78 offspring were obtained, of which 38 form white flowers with flat corollas, and 40 - cream flowers with flat corollas. When phloxes with white flowers and funnel-shaped corollas were crossed with a plant with cream flowers and flat corollas, phloxes of two phenotypic groups were obtained: white with funnel-shaped corollas and white with flat corollas. Make a diagram of the two crosses. Determine the genotypes of the parents and offspring in two crosses. What is the law of heredity manifested in this case?
9. There are two types of hereditary blindness, each of which is determined by recessive gene alleys (a or b). Both alleles are on different pairs of homologous chromosomes. What is the probability of the birth of a blind grandchild in a family in which maternal and paternal grandmothers are dihomozygous and suffer from various types of blindness, and both grandfathers can see well (do not have recessive genes). Make a scheme for solving the problem. Identify the genotypes and phenotypes of grandparents, their children and possible grandchildren. 
III... Tasks for the inheritance of blood groups of the AB0 system
The boy has group I, his sister has group IV. What are the genotypes of the parents?
The father has blood type IV, the mother has I. Can a child inherit the blood type of his mother?
In the maternity hospital, two children were confused. The first pair of parents has I and II blood groups, the second pair - II and IV. One child has Group II, and the second has Group I. Identify the parents of both children.
Is it possible to transfuse blood to a child from a mother if her blood type is AB and the father is 00? Explain the answer.
5. Blood group and Rh factor - autosomal unlinked signs. The blood group is controlled by three alleles of one gene - i 0, I A, I B. Alleles IA and IB are dominant with respect to allele i 0. The first group (0) is determined by recessive genes i 0, the second group (A) is determined by the dominant allele IA, the third group (B) is determined by the dominant allele IB, and the fourth (AB) is determined by two dominant allele IAIB. A positive Rh factor dominates over a negative r. The father has a third blood group and is Rh positive (diheterozygote), the mother has a second group and is Rh positive (digomozygote). Determine the genotypes of the parents. What blood group and Rh factor can children in this family have, what are their possible genotypes and the ratio of phenotypes? Make a scheme for solving the problem. What is the law of heredity manifested in this case?
IV
1. In canaries, the presence of a crest is a dominant autosomal trait (A); the sex-linked gene X B determines the green color of the plumage, and X b - brown. In birds, the homogametic sex is male, and the heterogametic sex is female. A crested green female was crossed with a male without a tuft and green plumage (heterozygote). The offspring contained green crested chicks, green without a tufted, brown tufted and brown without a tufted. Make a scheme for solving the problem. Determine the genotypes of parents and offspring, their sex. What laws of heredity are manifested in this case?
2. The body coloration of Drosophila is determined by the autosomal gene. The eye color gene is located on the X chromosome. Male sex is heterogametic in Drosophila. A female with a gray body and red eyes was crossed with a male with a black body and white eyes. All offspring had a gray body and red eyes. The males that turned out in F1 were crossed with the parent female. Make a scheme for solving the problem. Determine the genotypes of the parents and females F 1, the genotypes and phenotypes of the potmy in F 2. What part of females from the total number of offspring in the second cross is phenotypically similar to the parental female? Indicate their genotypes. 
3. In humans, the brown eyes gene dominates over blue eyes (A), and the color blindness gene is recessive (color blindness - d) and linked to the X chromosome. A brown-eyed woman with normal vision, whose father had blue eyes and was color-blind, marries a blue-eyed man with normal vision. Make a scheme for solving the problem. Determine the genotypes of the parents and possible offspring, the likelihood of birth in this family of color-blind children with brown eyes and their gender. Explain the answer.
4. In humans, the inheritance of albinism is not sex-linked (A is the presence of melanin in skin cells, a is albinism), and hemophilia is sex-linked (X n is normal blood clotting, X h is hemophilia). Determine the genotypes of the parents, as well as the possible genotypes, sex and phenotypes of children from a marriage of a diheterozygous woman and an albino man with normal blood clotting for both alleles. Make a scheme for solving the problem. Explain the answer.
1. When crossing maize plants with smooth colored grains with a plant producing wrinkled, unpainted seeds, in the first generation, all plants produced smooth colored grains. When analyzing crosses of hybrids from F1 in the offspring there were four phenotypic groups: 1200 smooth colored, 1215 wrinkled unpainted, 309 smooth unpainted, 315 wrinkled colored. Make a scheme for solving the problem. Determine the genotypes of the parents and offspring in the two crosses. Explain the formation of four phenotypic groups in the second cross.
2. When crossing a diheterozygous corn plant with a colored seed and starchy endosperm and a plant with an uncolored seed and a waxy endosperm in the offspring, a phenotype splitting was obtained: 9 plants with colored seed and starchy endosperm; 42 - with colored seed and waxy endosperm; 44 - with uncolored seed and starchy endosperm; 10 - with uncolored seed and waxy endosperm. Make a scheme for solving the problem. Determine the genotypes of the original individuals, the genotypes of the offspring. Explain the formation of the four phenotypic groups. 
3. A diheterozygous pea plant with smooth seeds and tendrils was crossed with a plant with wrinkled seeds without tendrils. It is known that both dominant genes (smooth seeds and the presence of antennae) are localized on the same chromosome; crossing over does not occur. Make a scheme for solving the problem. Determine the genotypes of the parents, the phenotypes and genotypes of the offspring, the ratio of individuals with different genotypes and phenotypes. What law is manifested in this case? 
VI... Pedigree tasks
1. Using the pedigree shown in the figure, determine and explain the inheritance pattern of the trait highlighted in black (dominant or recessive, linked or not linked to the sex. Determine the genotypes of offspring 1,3,4,5,6,7. Determine the probability of birth to parents 1 , 2 next children with the trait highlighted in black on the pedigree. 
2. Using the pedigree shown in the figure, determine and justify the genotypes of the parents, descendants, indicated on the diagram by the numbers 1,6,7. Establish the probability of having a child with an inherited trait in a woman at number 6, if this trait has never manifested itself in the family of her spouse.
Using the pedigree shown in the figure, determine and explain the inheritance pattern of the trait highlighted in black. Determine the genotypes of parents, offspring1,6 and explain the formation of their genotypes.
Using the pedigree shown in the figure, determine the nature of inheritance of the trait (dominant or recessive, linked or not linked to the sex), highlighted in black, the genotypes of parents and children in the first generation. Indicate which of them is the carrier of the gene, the trait of which is highlighted in black.
Gene interaction problems
1.complementarity
In parrots, feather color is determined by two pairs of genes. The combination of two dominant genes determines the color green. Individuals recessive in both gene pairs are white. The combination of the dominant gene A and the recessive gene b determines the yellow color, and the combination of the recessive gene a with the dominant gene B determines the blue color. 
When two green individuals were crossed with each other, they received parrots of all colors. Determine the genotypes of the parents and offspring.
2.epistasis
Rabbit coat color (as opposed to albinism) is determined by the dominant gene. The color of the color is controlled by another gene located in another chromosome, and the gray color dominates over black (in albino rabbits, the color genes do not show themselves). 
What are the characteristics of hybrid forms obtained from crossing a gray rabbit born from an albino rabbit with an albino carrying the gene for black color?
3. pleiotropy.
One of the breeds of chickens is distinguished by short legs. This sign is dominant. The gene that controls it simultaneously causes the shortening of the beak. Moreover, in homozygous chickens, the beak is so small that they are unable to break through the eggshell and die without hatching from the egg. In the incubator of the farm, which breeds only short-legged chickens, 3000 chickens were received. How many are short-legged among them?
4.polymer
The son of a white woman and a black man married a white woman. Could this couple have a child darker than their father?
ANSWERS:
I... Monohybrid crossing problems (complete and incomplete dominance)
1. A - gray color
a - black color P: AA and aa.
F 1: Aa, all gray.
2. A - black floral - white
R: Aa and Aa.
F 1: AA 2Aa, aa. Splitting law.
3. A- white color
a - yellow
Both parents have the Aa genotype.
F 1 genotype cleavage 1aa 2Aa 1 aa
4.Nochka-Aa, her calf-aa. Dawn- aa, her calf- Aa, bull- Aa.
5. A - high
a-dwarf
F 1 1AA 2Aa 1aa
240 dwarf plants.
6.A - green color
a - striped color
F2 1AA 2Aa 1aa
F 2 - cleavage 3 green: 1 striped
7. A- wide leaves
a- narrow leaves
A- intermediate leaf width
F 1: splitting by phenotype: 1- wide leaves, 2- intermediate leaf width, 1- narrow leaves.
By genotype: AA 2Aa aa
II... Problems for dihybrid crossing1. A - norm, and - glaucoma.
B - Marfan syndrome, b - normal.
One of the spouses suffers from glaucoma and did not have ancestors with Marfan syndrome: aabb. The second spouse is diheterozygous: AaBb.
| norm. | norm | glaucoma | glaucoma |
The probability of having a healthy child (norm / norm) \u003d 1/4 (25%). In this case, Mendel's third law is manifested (the law of independent inheritance).
a - dwarfism
B - smooth
in-ribbed
first cross- P: aabb and AaBB, got F 1 - aaBb and AaBb
second - P: aabb and AAbb, got F 1 - Aabb.
4.R-AaBv and aavv.F1: 9 cr. Head .. 3 cr. Op., 3 w., 1 w. op. 4 tons railway op.
5. In this case, intermediate color inheritance appears. AA - black, A - motley, aa - white. parents and hen and rooster have genotypes AaBB. And the gametes form the same: AB, aB. when they merge, the genotypes are formed - AABB - black crested, AaBB - variegated crested, aaBB - white crested. ratio -1/2/1.
color gene:
A - black
a- brown
tail length gene:
B- long
c- short
cc - lethal
BB - shortened
Decision:
1) AaBv x aaBv
black, shortened х brown, shortened
gametes - AB, Av, AV, AV AV AV
AaBB AaBv aaBB aaBv AaBv Aavv aaBv aavv
h. d. h. c. d. c. h lethal c. lethal
black with jlin tail - 1/8
black with shortened - 2/8
brown with a long tail - 1/8
brown with a shortened - 2 /
lethal - 2/8. Independent inheritance law
7. A - mustache
a- beardless
B- red
Bb - pink
1) the first crossing:
R AAvv * aaBB
mustache white b / must. cr.
2) analyzing crossing:
AaBv * aavv
G AV Av av av av
F 2 AaBb - mustachioed pink-fruited; Aabb - whiskered white-fruited;
aaBb - whiskerless pink-fruited; aabb - whiskerless white-fruited;
3) the nature of inheritance of the trait of color of the fetus - incomplete dominance. In the first crossing - the law of uniformity of hybrids, in the second (analyzing) - independent inheritance of traits.
8.A - flat rims,
a - funnel-shaped corollas.
B - white flowers,
b - cream flowers
first cross:
P aaBv x Aavv
G av av av av
F 1 AaBb aavb
second cross:
P aaBB x Aavb
G av av av
F 1 AaBb aaBb
In this case, Medel's third law is manifested - the law of independent inheritance.
9. A - the norm, and - blindness No. 1.
B - norm, b - blindness # 2.
Maternal grandmother is AAbb, paternal grandmother is aaBB. Grandfathers - AABB.
| |
The probability of having a blind grandchild 0%
Tasks for the inheritance of blood groups of the AB0 system
1. Boy-j0j0. Sister- JАJВ
P J A j 0 and J A J B
2. Father - J A J B
Mother-j 0 j 0.
No, because children can have either blood groups 2 or 3.
3.first pair of parents:
P: j 0 j 0 x J А j 0 or j 0 j 0 x J A J А
G: j 0 J A, j 0 j0 J A
F: J A j0 (2 gr.), J 0 j 0 (1 gr.) Or J A j 0 (2 gr.)
second pair of parents
R: J A J A x J A J B or J A j 0 x J A J B
G: J A; J A, J B J A j 0 J A, J B
F: J A J A (2) J A J B (4) J A J A (2) J A J B (4) J A j 0 (2) J in j 0 (3 gr.)
The first couple of parents, the son has 1 gr. and he received gene 0 from both parents. The second pair is the parents of a boy with a second blood group.
This problem can be solved orally, because a child with 1 gr. blood cannot be born to a couple in which there is a person with a 4 blood group
4. It is impossible, because children may have blood groups: A0 (II) or B0 (III), therefore, blood of the fourth group, which the mother has, cannot be transfused.
5. Father-diheterozygote I B i 0 Rr, mother-dihomozygote I A I А RR.
| IV group | IV group | II group | II group |
Children in this family may have blood group IV or II, all Rh positive. The proportion of children with IV blood group is 2/4 (50%). The law of independent inheritance is manifested (Mendel's third law).
IV... Sex-linked and autosomal inheritance problems
1. A - the presence of a tuft, and - no tuft.
X B - green plumage, X b - brown plumage.
A_X B Y-crested green female
aaX B X b - male without a tuft with green plumage (heterozygote)
Among the offspring were chicks without a crest-aa. They got one gene a from their mother, one from their father. Therefore, the mother must have the gene and, therefore, the mother of Aa.
P AaX B Y x aaX B X b
| AaX B X B | aaX B X B | AaX B Y | aaX B Y |
|
| AaX B X b | aaX B X b | AaX b Y | aaX b Y |
In this case, the law of independent inheritance (Mendel's third law) and sex-linked inheritance appeared.
2. A - gray body
a - black body
X In red eyes
X in - white eyes
P 1 AAX B X B * aaX in Y
gray body black body
red eyes white ch.
G AX B aX in aY
F 1 AaX B X b AaX B Y
P 2 AAX B X B * AaX B Y
G AX B AX B aX B AY aY
F 2 AAX B X B AaX B X B AAX B Y AaX B Y
F 2 all descendants have a gray body and red eyes.
Sex ratio -50%: 50:%
3. A - brown eyes,
a - blue eyes.
X D - normal vision,
X d - color blindness.
A_X D X _ a brown-eyed woman with normal vision
aaX d Y- the woman's father, he could give his daughter only aX d, therefore, a brown-eyed woman- AaX D X d.
AaX D Y. - woman's husband
P AaX D X d x aaX D Y
G AX D AX d aX D ax d aX D aY
F 1 AAX D X D AaX D X d aaX D X D aaX D X d AAX D Y Aa X d Y aaX D XY aaX d Y
The probability of having a color blind child with brown eyes is 1/8, (12.5%), this is a boy.
4. A - the norm, a - albinism.
X N - norm, X h - hemophilia.
Woman AaX H X h, man aaX H Y
G AX H AX h aX H aX h aX H aY
F1 AaX H X H AaX H Y AaX H X h AaX h Y aaX H X H aaX H Y aaX H X h aaX h Y
ph.D. Ph.D. Ph.D. since. Mr. Mr. Mr. yy
Cleavage according to eye color - 1: 1 according to blood clotting - all daughters are healthy, boys - 1: 1.
V. Tasks for chained inheritance1 . A - smooth grains,
a - wrinkled grains.
B - colored grains,
b - uncolored grains.
RAABB x aavv
Since uniformity was obtained in the first generation (Mendel's first law), therefore, homozygotes were crossed, in F1 a diheterozygote was obtained, bearing dominant characters.
Analyzing cross:
| normal gametes | |||||||
| recombinant gametes | |||||||
| smooth | wrinkled. | smooth | wrinkled. |
Since in the second generation, an unequal number of phenotypic groups turned out, therefore, there was linked inheritance. Those phenotypic groups that are represented in large numbers are not crossovers, but groups represented in small numbers are crossovers formed from recombinant gametes, in which the linkage was broken due to crossing over in meiosis.
2. A- colored seed a- not colored seed B- starchy endosperm b- waxy endosperm P AaBv x aavv G AB Av a B av F1 9- AaBv- env. seed, starch. endosperm 42- Aavv- env. seed, wax. endosperm 44- aaBv- uncolored seed, starchy endosperm 10- aavv- uncolored seed waxy endosperm The presence of two groups in the offspring (42 - with colored waxy endosperm; 44 - with uncolored waxy endosperm) in approximately equal proportions - the result of linked inheritance of alleles , a and B between themselves. Two other phenotypic groups are formed as a result of crossing over.
A - smooth seeds,
a - wrinkled seeds
B - the presence of antennae,
b - without antennae
| smooth | wrinkled. |
||
If crossing over does not occur, then the diheterozygous parent develops only two species of gametes (full linkage).
- A - gray body
F 1 AaBb are all gray with normal wings. Law of uniformity)
P AaBv x AaBv
T.K. there is no expected Mendelian splitting, which means that a crossing over has occurred:
linked
F 2 AABB AaBv AaBv aavv
In the previous article, we talked about the tasks of the C6 line in general. Starting from this post, specific problems in genetics that were included in the test tasks of past years will be analyzed.
Having a good understanding of such a biological discipline as genetics - the science of heredity and variability - is essential for life. Moreover, genetics in our country has such a long-suffering history ...
Just think, Russia from a leading country in the study of genetics at the beginning of the twentieth century is turning into a dense monster by etching out of people's minds even just genetic terminology from the late 30s to the mid 50s.
Is it possible to forgive the regime for the death by torture and hunger of the greatest geneticist, the noblest servant of the people and science, the founder of the All-Union Institute of Plant Industry in Leningrad, academician (1887 - 1943) .
Let's start the analysis of real tasks of the C6 line with tasks for dihybrid crossingthat require knowledge by the inheritance of the traits of two pairs of allelic genes (but which are non-allelic to each other) located in different pairs of homologous chromosomes, therefore inherited .
The most amazing thing is that the difficulty level of these tasks is very different. , what we are with you now and see examples of solving several tasks.
Studying further the material of this article, thanks to my detailed explanations, I hope that more complex tasks will be clear for you. And for a more successful development of the problems of dihybrid crossing, I bring to your attention my book:
Problem 1. About pigs for dihybrid crossing (the simplest)
In pigs, black hair color (A) dominates over red hair (a), long hair (B) - over short hair (c). Genes are not linked. What offspring can be obtained by crossing a black with long bristle, diheterozygous male with a homozygous black female with short bristles. Make a scheme for solving the problem. Determine the genotypes of the parents, offspring, phenotypes of the offspring and their ratio.
First, I want to draw your attention a little to such moments. :
First, why is this problem for a dihybrid crossing? The task requires to determine the distribution in inheritance of two traits: coat color (A or a) and length (B or b). Moreover, it is indicated that the genes are not linked, that is, the studied characters are in different pairs of homologous chromosomes and are inherited independently of each other according to Mendel's law. This means that the offspring will be formed from all possible random combinations of gametes formed by the male and the female.
Secondly, it is this dihybrid crossing problem that is the simplest of this type of task. It stipulates in advance that the studied characteristics are not linked. In addition, we can immediately (without analyzing all possible combinations of the birth of offspring) for a given phenotype of the parents to write down their genotype.
Decision:
1) parental genotypes :
male AaBb - since it is said about the male in the condition of the problem that he is diheterozygous, that is, heterozygous for both studied traits, then in the record of his genotype for each trait are present : A - dominant black coat color and a - recessive red coat color; B - dominant long bristles and b - recessive short;
female AAbb - since it is said about her that she homozygotic for the color of the coat, which is also black, so we write only AA, but about the length of the coat is not said homois she zygotic or goetherozygous because this information would be superfluous !!!(and so it is clear that if the female has short hair, then it can only be homozygotic for this recessive sign bb ).
Sure, such a long reasoning about writing genotype parents for the given in the condition of the problem their phenotypeyou do not need to provide. The main thing is that the first point you should indicate correctly, in no way mistaken, the genotypes of both parents.
2) gametes :
digheteroa zygotic male will produce with equal probability four sperm varieties AB, Ab, aB, ab (according to the law of gamete purity, as a consequence, each gamete can have only one allele of any gene. And since the inheritance of two traits is being studied at once, we inscribe one allelic gene of each trait under study in each gamete);
digomozygotic female (AAbb- as we found out she has exactly such a genotype ) will have everything the same eggs - Ab.
3) offspring :
since all the same type of female eggs Ab can be fertilized by any four types of sperm AB, Ab, aB and ab with equal probability, it is possible the birth of descendants with such
four genotypes : AABb, AAbb, AaBband Aabb in relation to 1: 1: 1: 1 (25%, 25%, 25%, 25%)
and two phenotypes : A-B- - black long-haired - 50% and A-bb - black shorthair - 50% (spaces are written where there is absolutely no difference for manifestationsphenotype which second dominant or recessive gene in these pairs of allelic genes may be present ).
So, we fully answered the questions of the assignment : the solution was drawn up according to the standard scheme (parents, gametes, offspring), the genotypes of the parents and offspring were determined, the phenotypes of the offspring were determined, and the possible ratio of the genotypes and phenotypes of the offspring was determined.
Problem 2. About the Datura plant for dihybrid crossing, but with a more complex condition.
The Datura plant with purple flowers (A) and smooth bolls (b) was crossed with a plant having purple flowers and spiny bolls. In the offspring, the following phenotypes were obtained: with purple flowers and prickly bolls, with purple flowers and smooth bolls, with white flowers and prickly bolls, with purple flowers and smooth bolls. Make a scheme for solving the problem. Determine the genotypes of the parents, offspring and the possible ratio of phenotypes. Establish the character of traits inheritance.
Note, that in this task we can no longer immediately unequivocally answer the question about the genotype of the parents, and therefore at once write full information about gametes, produced by them. This can only be done by carefully analyzing the information. about phenotypes offspring.
In the answer, you still have to remember to be sure to indicate character inheritance of traits (traits are independently inherited or linked). This was given in the previous assignment.
Decision:
1) we first define let and ambiguous possible genotypes of parents
R: A - bb (purple, smooth) and A - B - (purple, prickly)
2) we also write down information about the gametes they produce ambiguously
G: Аb, - b and AB, А -, - В, - -
3) we write down based on the known phenotype of the offspring, their possible genotypes
F1 A - B - (purple, prickly) A - bb (purple, smooth)
……. aaB - (white, prickly) aabb (white, smooth)
Now, the most important information that we can learn from all of the above:
a) since among the offspring there are plants with smooth bolls (and this is a recessive trait), the genotypes both parents necessarily must have a gene b. That is, we can already write in the genotype of the second parent b (small): A-B b;
b) since among the offspring there are plants with white flowers (and this is a recessive trait), the genotypes both parents must have a gene and (small);
4) only now we can completely write out the genotypes of both parents : .. … ………………….. Aabb and AaBb and the ………………………………………….
gametes : …. Ab, ab and AB, Ab, aB, ab
5) since, according to the condition of the problem, all possible combinations of plant traits were found in the offspring :
…………. “With purple flowers and spiny bolls,
………… .. with purple flowers and smooth bolls,
………… .. with white flowers and spiny bolls,
………… .. with white flowers and smooth bolls ”,
then this is possible only with independent inheritance signs;
6) since we have determined that the characters are not linked and are inherited independently of each other, it is necessary to produce all possible combinations of crosses of existing gametes. The most convenient way to record is using the Punnett grid. In our problem, it will be, thank God, not classical (4 x 4 \u003d 16), but only 2 x 4 \u003d 8 :
G : AB Ab aB ab
Ab AABb AAbb AaBb Aabb
………… .. purple prickly purple smooth purple prickly purple smooth
aw AaBb Aabb aaBb aabb
…………. purple prickly purple smooth white prickly white smooth
7) distribution in offspring will be
by genotype:1 AABb: 1 AAbb: 2 AaBb: 2 Aabb: 1 aaBb: 1 aabb
by phenotype: 3/8 - purple prickly (A-Bb);
………………… .. 3/8 - purple smooth (A-bb);
………………… .. 1/8 - white prickly (aaBb);
………………… .. 1/8 - white smooth (aabb).
Problem 3. Quite simple, if you understand the sense of genetic terminology
From crossing 2 varieties of barley, one of which has a dense two-row spike, and the other has a loose, multi-rowed ear, we got F 1 hybrids with a loose two-row spike. What results in phenotype and genotype will be obtained in backcrossing if the inheritance of traits is independent? Make crossings schemes.
Since it is said that they crossed varieties barley (yes, whatever, the word variety "appears"), so we are talking about homozygous organisms for both studied characteristics. And what signs are considered here:
a) the shape of the ear and b) its quality. Moreover, it is said that the inheritance of traits is independent, which means that we can apply the calculations following from the 3rd Mendel's law for dihybrid crossing.
It is also said what traits the hybrids in F1 possessed. They were with a two-row loose ear, which means that these signs are dominant over the diversity and density of the ear. Therefore, we can now introduce the designations of the alleles of the genes of these two studied characters and we will not be mistaken in the correct use of upper and lower case letters of the alphabet.
We denote:
allelic gene of two-row spike AND, and multi-row - and;
loose spike allelic gene IN, and dense - b,
then the genotypes of the original two varieties of barley will look like this: AAbb and aaBB... From crossing them in F1, hybrids will be obtained: AaBb.
Well, now to carry out backcrossing of hybrids AaBb with each of the original parent forms separately with AAbb, and then with aaBB, I'm sure it won't be difficult for anyone, isn't it?
Problem 4. "Not red, I'm not red at all, I'm not red, but gold"
A woman with brown eyes and red hair married a man with not red hair and blue eyes. It is known that the woman's father had brown eyes, and the mother's blue, both had red hair. The man's father did not have red hair and blue eyes, the mother had brown eyes and red hair. What are the genotypes of all these people? What could be the eyes and hair of the children of these spouses?
The allelic gene responsible for the manifestation of brown eye color is denoted AND (it is well known to everyone that brown eye color dominates over blue), and the allelic gene for blue eyes, respectively, will be and... Be sure to have the same letter of the alphabet, since this is one sign - eye color.
The allelic gene of non-red hair (hair color is the second trait under study) is denoted IN, since it dominates the allele responsible for the manifestation of red hair color - b.
The genotype of a woman with brown eyes and red hair, we can write at first incompletely, and so A-bb... But since it is said that her father was brown-eyed with red hair, that is, with the genotype A-bb, and her mother was blue-eyed and also with red hair ( aabb), then the second female allele at AND could only be and, that is, its genotype will be Aabb.
The genotype of a blue-eyed man with non-red hair can first be written as follows : aaB-... But since his mother had red hair, that is bb, then the second allelic gene at IN the man could only have b... In this way, the genotype of a man will be written aaBb... Genotypes of his parents: father - aaB-; mothers - A-bb.
Children from the marriage of the spouses being analyzed Aab x aaBb (and gametes, respectively : Ab, ab and aB, ab) will be with equally probable genotypes AaBb, Aabb, aaBb, aabb or by phenotype: brown-eyed not red, brown-eyed red, blue-eyed not red, blue-eyed red in the ratio 1:1:1:1 .
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Yes, now you can see for yourself what tasks of unequal complexity can be. It is unfair, but unfair, I answer, as a tutor for the exam in biology. You need luck, but you need luck!
But you must agree that luck will be useful only for those who are really "in the subject." Without knowledge of the laws of heredity Gregor Mendel, it is impossible to solve the first task, so there can be only one conclusion : .
In the next article by a biology tutor on Skype, we will analyze the tasks for inheritance that even fewer students correctly solve.
For those who want to understand well how to solve problems in genetics for dihybrid crossing, I can offer my book: ““
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Who will have questions for biology tutor via Skype, contact in the comments. On my blog you can buy answers to all tests OBZ FIPI for all the years of examinations, etc.
Having worked through these topics, you should be able to:
- Give definitions: gene, dominant trait; recessive trait; allele; homologous chromosomes; monohybrid crossing, crossing over, homozygous and heterozygous organism, independent distribution, complete and incomplete dominance, genotype, phenotype.
- Using the Pennett lattice, illustrate crossing by one or two traits and indicate what numerical ratios of genotypes and phenotypes should be expected in the offspring from these crosses.
- To outline the rules of inheritance, splitting and independent distribution of traits, the discovery of which was Mendel's contribution to genetics.
- Explain how mutations can affect the protein encoded by a particular gene.
- Indicate the possible genotypes of people with blood types A; IN; AB; ABOUT.
- Give examples of polygenic traits.
- Indicate the chromosomal mechanism of sex determination and the types of inheritance of sex-linked genes of mammals, use this information in solving problems.
- Explain what is the difference between sex-linked traits and gender-specific traits; give examples.
- Explain how such human genetic diseases as hemophilia, color blindness, sickle cell anemia are inherited.
- Name the features of plant and animal breeding methods.
- Indicate the main areas of biotechnology.
- To be able to solve the simplest genetic problems using this algorithm:
Algorithm for solving problems
- Determine the dominant and recessive trait according to the results of crossing the first generation (F1) and the second (F2) (according to the problem statement). Enter the letters: A - dominant and - recessive.
- Write down the genotype of an individual with a recessive trait or an individual with the genotype and gametes known from the problem statement.
- Record the genotype of the F1 hybrids.
- Make a diagram of the second cross. Write the gametes of F1 hybrids horizontally and vertically in the Pennett lattice.
- Record the genotypes of the offspring in the gamete crossing cells. Determine the ratio of phenotypes in F1.
Scheme for the design of tasks.
Letter designations:
a) dominant feature _______________
b) recessive sign _______________
Gametes
F1 (genotype of the first generation)
| gametes | ||
| ? | ? |
Pennett Lattice
| F2 |
|
The ratio of phenotypes in F2:_____________________________
Answer:_________________________
Examples of solving problems for monohybrid crossing.
A task. "The Ivanov family has two children: a brown-eyed daughter and a blue-eyed son. The mother of these children is blue-eyed, but her parents had brown eyes. How is human eye color inherited? What are the genotypes of all family members? Eye color is a monogenic autosomal trait."
The eye color trait is controlled by one gene (conditionally). The mother of these children is blue-eyed, and her parents had brown eyes. This is possible only if both parents were heterozygous, therefore, brown eyes dominate over blue ones. Thus, grandmother, grandfather, dad and daughter had the genotype (Aa), and mom and son had aa.
A task. "A rooster with a pink comb is crossed with two hens, which also have a pink comb. The first gave 14 chickens, all with a pink comb, and the second - 9 chickens, 7 of them with a pink comb and 2 with a leaf comb. The shape of the comb is a monogenic autosomal trait. the genotypes of all three parents? "
Before determining the genotypes of the parents, it is necessary to find out the nature of inheritance of the ridge shape in chickens. When a rooster was crossed with a second hen, 2 chickens with a leaf-shaped comb appeared. This is possible with heterozygosity of the parents; therefore, it can be assumed that the rosy ridge in chickens dominates over the leaf ridge. Thus, the genotypes of the rooster and the second chicken are Aa.
When the same cock was crossed with the first chicken, no cleavage was observed, therefore, the first chicken was homozygous - AA.
A task. "Fraternal twins were born in a family of brown-eyed right-handed parents, one of whom is brown-eyed left-handed and the other blue-eyed right-handed. What is the likelihood of having the next child similar to your parents?"
The birth of a blue-eyed child in brown-eyed parents indicates the recessiveness of the blue color of the eyes, respectively, the birth of a left-handed child in right-handed parents indicates the recessiveness of better control of the left hand compared to the right. Let's introduce the designations of the alleles: A - brown eyes, a - blue eyes, B - right-handed, b - left-handed. Let's define the genotypes of parents and children:
| R | AaBv x AaBv |
| F, | A_bv, aaB_ |
А_вв - phenotypic radical, which shows that this child is left-handed with brown eyes. The genotype of this child may be - Aavv, Aavv.
The further solution of this problem is carried out in the traditional way, by constructing a Punnet lattice.
| AB | Av | aB | Av | |
| AB | AABB | AABv | AaBB | AaBv |
| Av | AABv | AAvv | AaBv | Aavb |
| aB | AaBB | AaBv | aaBB | AaBv |
| aw | AaBv | Aavb | aaBv | Aavb |
9 variants of descendants that interest us are underlined. There are 16 possible options in total, so the probability of having a child similar to your parents is 9/16.
Ivanova T.V., Kalinova G.S., Myagkova A.N. "General biology". Moscow, "Education", 2000
- Topic 10. "Monohybrid and dihybrid crossing." §23-24 pp. 63-67
- Topic 11. "Genetics of sex." §28-29 pp. 71-85
- Topic 12. "Mutation and modification variability." §30-31 pp. 85-90
- Topic 13. "Selection." §32-34 pp. 90-97
In humans, dark hair color (A) dominates over light color (a), brown eye color (B) - over blue (b). Write down the genotypes of the parents, the possible phenotypes, and the genotypes of the children born from the marriage of a fair-haired, blue-eyed man to a heterozygous brown-eyed blonde woman.
Answer
Blond blue eyed male aabb.
Heterozygous brown-eyed blonde woman aaBb.
Congenital myopia is inherited as an autosomal dominant trait, the absence of freckles is inherited as an autosomal recessive trait. The signs are on different pairs of chromosomes. The father has congenital myopia and lack of freckles, the mother has normal vision and freckles. The family has three children, two are nearsighted without freckles, one with normal vision and freckles. Make a scheme for solving the problem. Determine the genotypes of the parents and children born. Calculate the probability of having children with myopia and freckles. Explain what the law is in this case.
Answer
A - congenital myopia, and - normal vision.
B - freckles, b - no freckles.
Father A_bb, mother aaB_.
Children A_bb, aaB_.
If the father is bb, then all his children have b, then the second child is aaBb.
If the mother is aa, then all her children have a, then the first child is Aabb.
If the first child has bb, then he took one b from the mother and one from the father, then the mother is aaBb.
If the second child has aa, then he took one a from the mother and one from the father, then the father is Aabb.
The probability of having myopic children with freckles is 25%, the law of independent inheritance works.
Parents with a free earlobe and a triangular fossa on the chin gave birth to a child with a fused earlobe and a smooth chin. Identify the genotypes of the parents, first child, phenotypes and genotypes of other possible offspring. Make a scheme for solving the problem. Traits are inherited independently.
Answer
The offspring showed recessive signs that were latent in the parents.
A - free earlobe, and - fused earlobe.
B - triangular fossa on the chin, b - smooth chin.
Child aabb, parents A_B_.
Child aa received one a from his father, the other from his mother; one b from the father, the other from the mother, hence the parents are AaBb.
| AB | Ab | aB | ab | |
| AB | AABB | AABb | AaBB | AaBb |
| Ab | AABb | AAbb | AaBb | Aabb |
| aB | AaBB | AaBb | aaBB | aaBb |
| ab | AaBb | Aabb | aaBb | aabb |
9 A_B_ free earlobe, triangular fossa on the chin
3 A_bb free earlobe, smooth chin
3 aaB_ fused earlobe, triangular fossa on the chin
1 aabb fused earlobe, smooth cone
A black crested rooster is crossed with a similar hen. 20 chickens were obtained from them: 10 black crested, 5 brown crested, 3 black without a crest and 2 brown without a crest. Determine the genotypes of parents, descendants and the pattern of inheritance of traits. The genes of the two traits are not linked, the dominant traits are black plumage (A), crested (B).
Answer
A - black plumage, and - brown plumage.
B - crested, b - without crest.
Rooster A_B_, chicken A_B_.
Chickens A_B_ 10 pcs., AaB_ 5 pcs., A_bb 3 pcs., Aabb 2 pcs.
If the child has aa, then he took one a from the mother and one from the father, then the parents are AaB_.
If the child has bb, then he took one b from the mother and one from the father, then the parents are AaBb.
| AB | Ab | aB | ab | |
| AB | AABB | AABb | AaBB | AaBb |
| Ab | AABb | AAbb | AaBb | Aabb |
| aB | AaBB | AaBb | aaBB | aaBb |
| ab | AaBb | Aabb | aaBb | aabb |
9 A_B_ black crested
3 A_bb black without crest
3 aaB_ brown crested
1 aabb brown without crest
The regularity of the inheritance of traits is the law of independent inheritance.
Genetics, its tasks. Heredity and variability are the properties of organisms. Genetic methods. Basic genetic concepts and symbols. Chromosomal theory of heredity. Modern understanding of the gene and genome
Genetics, its tasks
The successes of natural science and cell biology in the 18th-19th centuries allowed a number of scientists to make assumptions about the existence of certain hereditary factors that determine, for example, the development of hereditary diseases, but these assumptions were not supported by appropriate evidence. Even the theory of intracellular pangenesis, formulated by H. de Vries in 1889, which assumed the existence in the cell nucleus of certain "pangens" that determine the hereditary inclinations of the organism, and the exit into the protoplasm of only those of them that determine the cell type, could not change the situation, as well as the theory of "germplasm" A. Weismann, according to which the characters acquired in the process of ontogenesis are not inherited.
Only the works of the Czech researcher G. Mendel (1822-1884) became the founding stone of modern genetics. However, despite the fact that his works were cited in scientific publications, his contemporaries did not pay attention to them. And only the rediscovery of the patterns of independent inheritance by three scientists at once - E. Cermak, K. Correns and H. de Vries - forced the scientific community to turn to the origins of genetics.
Genetics Is a science that studies the laws of heredity and variability and methods of managing them.
The tasks of genetics at the present stage are the study of the qualitative and quantitative characteristics of the hereditary material, the analysis of the structure and functioning of the genotype, the deciphering of the fine structure of the gene and methods of regulation of gene activity, the search for genes that cause the development of hereditary human diseases and methods of their "correction", the creation of a new generation of drugs by type DNA vaccines, the construction of organisms with new properties using the means of genetic and cellular engineering that could produce drugs and food products necessary for a person, as well as a complete decoding of the human genome.
Heredity and variability - properties of organisms
Heredity - This is the ability of organisms to transmit their traits and properties in a series of generations.
Variability - the property of organisms to acquire new characteristics during life.
Signs - any morphological, physiological, biochemical and other features of organisms, by which some of them differ from others, for example, eye color. Properties any functional features of organisms are called, which are based on a certain structural feature or a group of elementary features.
The traits of organisms can be divided into quality and quantitative... Qualitative signs have two or three contrasting manifestations, which are called alternative signs, for example, blue and brown eye color, while quantitative (milk yield of cows, yield of wheat) do not have clear differences.
The material carrier of heredity is DNA. In eukaryotes, two types of inheritance are distinguished: genotypic and cytoplasmic... Carriers of genotypic heredity are localized in the nucleus and further we will talk about it, and the carriers of cytoplasmic heredity are circular DNA molecules located in mitochondria and plastids. Cytoplasmic inheritance is transmitted mainly with the egg, therefore it is also called maternal.
A small number of genes are localized in the mitochondria of human cells, however, their change can have a significant impact on the development of the body, for example, lead to the development of blindness or a gradual decrease in mobility. Plastids play an equally important role in plant life. So, in some parts of the leaf, chlorophyll-free cells may be present, which leads, on the one hand, to a decrease in the productivity of the plant, and on the other, such variegated organisms are valued in decorative gardening. Such specimens are reproduced mainly asexually, since during sexual reproduction, ordinary green plants are more often obtained.
Genetic methods
1. The hybridological method, or the method of crossing, consists in the selection of parental individuals and analysis of the offspring. At the same time, the genotype of an organism is judged by the phenotypic manifestations of genes in the offspring obtained with a certain crossing pattern. This is the oldest informative method of genetics, which was most fully used for the first time by G. Mendel in combination with the statistical method. This method is not applicable in human genetics for ethical reasons.
2. The cytogenetic method is based on the study of the karyotype: the number, shape and size of the body's chromosomes. The study of these features makes it possible to identify various developmental pathologies.
3. The biochemical method allows you to determine the content of various substances in the body, especially their excess or deficiency, as well as the activity of a number of enzymes.
4. Molecular genetic methods are aimed at identifying variations in the structure and deciphering the primary nucleotide sequence of the studied DNA regions. They make it possible to identify genes of hereditary diseases even in embryos, establish paternity, etc.
5. Population-statistical method allows to determine the genetic composition of the population, the frequency of certain genes and genotypes, genetic load, and also to outline the prospects for the development of the population.
6. The method of hybridization of somatic cells in culture makes it possible to determine the localization of certain genes in chromosomes during the fusion of cells of various organisms, for example, a mouse and a hamster, a mouse and a human, etc.
Basic genetic concepts and symbols
Gene - This is a section of a DNA molecule, or chromosome, which carries information about a certain trait or property of an organism.
Some genes can influence the manifestation of several traits at once. This phenomenon is called pleiotropy... For example, a gene that causes the development of a hereditary disease of arachnodactyly (spider fingers) also causes a curvature of the lens, pathology of many internal organs.
Each gene occupies a strictly defined place in the chromosome - locus... Since in the somatic cells of most eukaryotic organisms, the chromosomes are paired (homologous), then in each of the paired chromosomes there is one copy of the gene responsible for a certain trait. Such genes are called allelic.
Allelic genes most often exist in two variants - dominant and recessive. Dominant is called an allele that manifests itself regardless of which gene is on the other chromosome, and suppresses the development of a trait encoded by a recessive gene. Dominant alleles are usually indicated by uppercase letters of the Latin alphabet (A, B, C, etc.), and recessive ones - by lowercase (a, b, c, etc.). Recessive alleles can only appear if they occupy loci on both paired chromosomes.
An organism that has the same alleles in both homologous chromosomes is called homozygous for a given gene, or homozygote (AA, aa, AABB, aabb, etc.), and an organism in which there are different variants of the gene in both homologous chromosomes - dominant and recessive - is called heterozygous for a given gene, or heterozygote (Aa, AaBb, etc.).
A number of genes can have three or more structural variants, for example, blood groups according to the AB0 system are encoded by three alleles - I A, I B, i. This phenomenon is called multiple allelism. However, even in this case, each chromosome from a pair carries only one allele, that is, all three variants of a gene in one organism cannot be represented.
Genome - a set of genes characteristic of a haploid set of chromosomes.
Genotype - a set of genes characteristic of a diploid set of chromosomes.
Phenotype - a set of signs and properties of an organism, which is the result of the interaction of the genotype and the environment.
Since organisms differ from each other in many traits, it is possible to establish the patterns of their inheritance only by analyzing two or more traits in the offspring. Crossbreeding, in which inheritance is considered and an accurate quantitative accounting of offspring is carried out for one pair of alternative characters, is called monohybridm, in two pairs - dihybrid, for most of the signs - polyhybrid.
By the phenotype of an individual, it is far from always possible to establish its genotype, since both an organism homozygous for the dominant gene (AA) and a heterozygous one (Aa) will have a dominant allele in the phenotype. Therefore, to check the genotype of an organism with cross fertilization, they use analyzing cross - crossing, in which an organism with a dominant trait is crossed with a homozygous for a recessive gene. In this case, an organism homozygous for the dominant gene will not split in the offspring, while in the offspring of heterozygous individuals there is an equal number of individuals with dominant and recessive traits.
To record crossing schemes, the following conventions are most often used:
P (from lat. parent - parents) - parental organisms;
$ ♀ $ (alchemical sign of Venus - a mirror with a handle) - a mother;
$ ♂ $ (alchemical sign of Mars - shield and spear) - father;
$ × $ - crossing sign;
F 1, F 2, F 3, etc. - hybrids of the first, second, third and subsequent generations;
F a - offspring from the analyzing cross.
Chromosomal theory of heredity
The founder of genetics G. Mendel, as well as his closest followers, did not have the slightest idea about the material basis of hereditary inclinations, or genes. However, already in 1902-1903, the German biologist T. Boveri and the American student W. Setton independently suggested that the behavior of chromosomes during cell maturation and fertilization makes it possible to explain the splitting of hereditary factors according to Mendel, i.e., in their opinion, genes must be located on chromosomes. These assumptions became the cornerstone of the chromosomal theory of heredity.
In 1906, English geneticists W. Batson and R. Pennett discovered a violation of Mendelian cleavage when crossing sweet peas, and their compatriot L. Doncaster discovered sex-linked inheritance in experiments with the goose moth butterfly. The results of these experiments clearly contradicted Mendelian, but if we consider that by that time it was already known that the number of known characters for experimental objects was much higher than the number of chromosomes, and this suggested that each chromosome carries more than one gene, and the genes of one chromosomes are inherited together.
In 1910, T. Morgan's group began experiments on a new experimental facility - the fruit fly Drosophila. The results of these experiments made it possible by the middle of the 20s of the XX century to formulate the main provisions of the chromosomal theory of heredity, to determine the order of arrangement of genes in chromosomes and the distance between them, that is, to draw up the first maps of chromosomes.
The main provisions of the chromosomal theory of heredity:
- Genes are located on chromosomes. Genes of one chromosome are inherited jointly, or linked, and are called clutch group... The number of linkage groups is numerically equal to the haploid set of chromosomes.
- Each gene occupies a strictly defined place in the chromosome - a locus.
- Genes in chromosomes are arranged linearly.
- Disruption of gene linkage occurs only as a result of crossing over.
- The distance between genes on a chromosome is proportional to the percentage of crossing over between them.
- Independent inheritance is characteristic only for genes of non-homologous chromosomes.
Modern understanding of the gene and genome
In the early 40s of the twentieth century, J. Beadle and E. Tatum, analyzing the results of genetic studies carried out on the neurospore fungus, came to the conclusion that each gene controls the synthesis of an enzyme, and formulated the principle "one gene - one enzyme" ...
However, already in 1961 F. Jacob, J. L. Monod and A. Lvov managed to decipher the structure of the E. coli gene and study the regulation of its activity. For this discovery, he was awarded the Nobel Prize in Physiology or Medicine in 1965.
In the process of research, in addition to structural genes that control the development of certain traits, they were able to identify regulatory ones, the main function of which is the manifestation of traits encoded by other genes.
The structure of the prokaryotic gene. The structural gene of prokaryotes has a complex structure, since it includes regulatory regions and coding sequences. Regulatory sites include a promoter, an operator, and a terminator. Promoter The name of the region of the gene to which the enzyme RNA polymerase is attached, which ensures the synthesis of mRNA during transcription. FROM operatorlocated between the promoter and the structural sequence can bind repressor protein, which does not allow RNA polymerase to start reading hereditary information from the coding sequence, and only its removal allows transcription to start. The structure of the repressor is usually encoded in a regulatory gene located in another region of the chromosome. Reading information ends at a section of the gene called terminator.
Coding sequence a structural gene contains information about the amino acid sequence in the corresponding protein. The coding sequence in prokaryotes is called cistron, and the set of coding and regulatory regions of the prokaryotic gene is operon... In general, prokaryotes, which include E. coli, have a relatively small number of genes located on a single ring chromosome.
The cytoplasm of prokaryotes can also contain additional small circular or unclosed DNA molecules called plasmids. Plasmids are able to integrate into chromosomes and be transmitted from one cell to another. They can carry information about sex characteristics, pathogenicity, and antibiotic resistance.
The structure of the eukaryotic gene. Unlike prokaryotes, eukaryotic genes do not have an operon structure, since they do not contain an operator, and each structural gene is accompanied only by a promoter and terminator. In addition, in the genes of eukaryotes, significant regions ( exons) alternate with insignificant ( introns), which are completely rewritten into mRNA and then excised during their maturation. The biological role of introns is to reduce the likelihood of mutations at significant sites. The regulation of genes in eukaryotes is much more complex than that described for prokaryotes.
The human genome. Each human cell contains about 2 m of DNA in 46 chromosomes, tightly packed in a double helix, which consists of approximately $ 3.2 × $ 10 9 nucleotide pairs, which provides about 10.19 billion possible unique combinations. By the end of the 80s of the twentieth century, the location of about 1500 human genes was known, but their total number was estimated at about 100 thousand, since only hereditary diseases in humans are about 10 thousand, not to mention the number of various proteins contained in cells ...
In 1988, the international project "Human Genome" was launched, which by the beginning of the XXI century ended with a complete decoding of the nucleotide sequence. He made it possible to understand that two different people have 99.9% similar nucleotide sequences, and only the remaining 0.1% determine our individuality. In total, about 30-40 thousand structural genes were found, but then their number was reduced to 25-30 thousand. Among these genes there are not only unique, but also repeated hundreds and thousands of times. Nevertheless, these genes encode a much larger number of proteins, for example, tens of thousands of protective proteins - immunoglobulins.
97% of our genome is genetic "garbage" that exists only because it is able to reproduce well (RNAs that are transcribed in these regions never leave the nucleus). For example, among our genes there are not only "human" genes, but also 60% of genes similar to those of the Drosophila fly, and up to 99% of genes have in common with chimpanzees.
In parallel with the decoding of the genome, the mapping of chromosomes also took place, as a result of which it was possible not only to discover, but also to determine the location of some genes responsible for the development of hereditary diseases, as well as target genes of drugs.
Deciphering the human genome has not yet given a direct effect, since we received a kind of instruction for assembling such a complex organism as a human, but we did not learn how to make it or even correct errors in it. Nevertheless, the era of molecular medicine is already on the verge, the development of so-called gene drugs is underway all over the world, which can block, remove or even replace pathological genes in living people, and not just in a fertilized egg.
We should not forget that in eukaryotic cells, DNA is contained not only in the nucleus, but also in mitochondria and plastids. Unlike the nuclear genome, the organization of mitochondrial and plastid genes has much in common with the organization of the prokaryotic genome. Despite the fact that these organelles carry less than 1% of the hereditary information of the cell and do not even encode a complete set of proteins necessary for their own functioning, they can significantly affect some of the characteristics of the organism. Thus, variegation in plants of chlorophytum, ivy and others is inherited by a small number of descendants, even when two variegated plants are crossed. This is due to the fact that plastids and mitochondria are transmitted mostly with the cytoplasm of the egg, therefore this inheritance is called maternal, or cytoplasmic, as opposed to genotypic, which is localized in the nucleus.
