It will be useful for young mothers to find out how good it is for the health of their children ...
Municipal State Educational Institution Lyceum No. 4
Rossosh Rossoshansky municipal district Voronezh region.
Methodical development in biology to help students taking the exam.
"Tasks in genetics" for grade 11
biology teacher of the highest qualification category
2016
I. Tasks for monohybrid crossing (complete and incomplete dominance)
1. In rabbits, gray coat color dominates over black. A homozygous gray rabbit was crossed with a black rabbit. Determine the phenotypes and genotypes of rabbits?
2. Do guinea pigs black coat color dominates over white. We crossed two heterozygous males and a female. What will be the first generation hybrids? What law is manifested in this inheritance? 
3. When crossing two white pumpkins in the first generation, ¾ of the plants were white and ¼ were yellow. What are the genotypes of the parents if white is dominant over yellow? 
4. Black cow Nochka brought a red calf. A black calf was born to the red cow Zorka. These cows are from the same herd with one bull. What are the genotypes of all animals? Consider different options. ( The black gene 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. Watermelon fruit may be green or striped. All watermelons obtained from crossing plants with green and striped fruits had only the green color of the fruit rind. What color of watermelon fruits can be in F 2?
7. In snapdragons, plants with wide leaves, when crossed with each other, always give offspring with the same leaves, and when an narrow-leaved plant is crossed with a broad-leaved plant, plants with leaves of intermediate width appear. 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 syndrome, accompanied by an anomaly in the development of connective tissue, is inherited as an autosomal dominant trait (B). The genes are in different couples autosomes. One of the spouses suffers from glaucoma and had no ancestors with Marfan's syndrome, and the second is diheterozygous according to these signs. Determine the genotypes of the parents, the possible genotypes and phenotypes of the children, the probability of birth healthy child. Make a scheme for solving the problem. What law of heredity is manifested in this case?
2. We crossed undersized (dwarf) tomato plants with ribbed fruits and plants of normal height with smooth fruits. 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 offspring had a normal stem height and ribbed fruits. Create crossover patterns. Determine the genotypes of parents and offspring of tomato plants in two crosses. What law of heredity is manifested in this case? 
3. When crossing horned red cows with polled black bulls, calves of two phenotypic groups were born: horned black and polled black. With further crossing of these same horned red cows with other polled black bulls, the offspring were individuals of polled red and polled black. Write schemes for solving the problem. Determine the genotypes of the parents and offspring in two crosses. What law of heredity is manifested in this case? 
4. Tomato fruits can be red and yellow, naked and pubescent. It is known that the genes for yellow color and pubescence are recessive. Of the tomatoes harvested on the collective farm, there were 36 tons of red hairless and 12 tons of red hairy ones. How many (approximately) yellow fluffy tomatoes can be in a collective farm crop if the source material was heterozygous?
5. When crossing a motley crested (B) chicken with the same rooster, eight chickens were obtained: four motley crested chickens, two white (a) crested and two black crested. Make a scheme for solving the problem. Determine the genotypes of parents, offspring, explain the nature of the inheritance of traits and the appearance of individuals with variegated color. What laws of heredity are manifested in this case? 
6. In mice, black coat color is dominant over brown (a). The long tail is determined by the dominant gene (B) and develops only in the homozygous state, in heterozygotes a short tail develops. Recessive genes that determine the length of the tail in the homozygous state cause the death of embryos. The genes of the two traits are not linked. When a female mouse with black hair and a short tail was crossed with a male with brown hair and a short tail, the following offspring were obtained: black mice with a long tail, black mice 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 offspring, the probability of death of the embryos. What law of heredity is manifested in this case? Justify the answer. 
7. A strawberry plant was crossed with mustachioed white-fruited plants with beardless red-fruited plants (B). All hybrids turned out to be mustachioed pink-fruited. When analyzing the crossing of F 1 hybrids, phenotypic splitting occurred in the offspring. Make a scheme for solving the problem. Determine the genotypes of the parental individuals, first-generation hybrids, as well as the genotypes and phenotypes of the offspring in analyzing crosses (F 2). Determine the nature of the inheritance of the color trait 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 descendants were obtained, among which 38 form white flowers with flat corollas, and 40 - cream flowers with flat corollas. When crossing phloxes with white flowers and funnel-shaped corollas 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. Draw two crosses. Determine the genotypes of the parents and offspring in two crosses. What law of heredity is manifested in this case?
9. There are two types of hereditary blindness, each of which is determined by recessive alleles of genes (a or b). Both alleles are on different pairs of homologous chromosomes. What is the probability of the birth of a blind grandson in a family in which maternal and paternal grandmothers are dihomozygous and suffer from various types blindness, and both grandfathers see well (do not have recessive genes). Make a scheme for solving the problem. Determine 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 IV blood type, the mother has I. Can a child inherit the blood type of his mother?
Two children were mixed up in the maternity hospital. The first pair of parents has I and II blood groups, the second pair - II and IV. One child has the II group, and the second - the I group. 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 her father has 00? Explain the answer.
5. Blood group and Rh factor - autosomal unlinked traits. The blood group is controlled by three alleles of one gene - i 0, I A, I B. Alleles I A and I B are dominant with respect to the allele i 0. The first group (0) is determined by recessive genes i 0, the second group (A) is determined by the dominant allele I A, the third group (B) is determined by the dominant allele I B, and the fourth (AB) is determined by two dominant allele I A I B . The positive Rh factor R dominates the negative r. The father has the third blood type and positive Rh (diheterozygote), the mother has the second group and positive Rh (digomozygote). Determine the genotypes of the parents. What blood type and Rh factor can children in this family have, what are their possible genotypes and phenotype ratio? Make a scheme for solving the problem. What law of heredity is 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. Birds are homogametic male and heterogametic female. A crested green female was crossed with a male without a crest and green plumage (heterozygote). The offspring included chicks crested green, without crest green, crested brown and without crest brown. Make a scheme for solving the problem. Determine the genotypes of parents and offspring, their gender. What laws of heredity are manifested in this case?
2. The body color of Drosophila is determined by an autosomal gene. The eye color gene is located on the X chromosome. Drosophila is heterogametic in males. 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 obtained 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 offspring in F 2 . What proportion of females from the total number of offspring in the second crossing is phenotypically similar to the parental female? Specify their genotypes. 
3. In humans, the gene for brown eyes dominates blue eyes (A), and the gene for color blindness is recessive (color blindness - d) and is linked to the X chromosome. A brown-eyed, normal-sighted woman whose father had Blue eyes and suffered from color blindness, marries a blue-eyed man with normal vision. Make a scheme for solving the problem. Determine the genotypes of parents and possible offspring, the probability 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- the presence of melanin in skin cells, a- albinism), and hemophilia is sex-linked (X n - normal blood clotting, X h - hemophilia). Determine the genotypes of the parents, as well as the possible genotypes, sex and phenotypes of children from the marriage of a diheterozygous woman for both alleles and an albino man with normal blood clotting. Make a scheme for solving the problem. Explain the answer.
1. When corn plants with smooth colored grains were crossed with a plant producing wrinkled uncolored seeds, in the first generation all plants produced smooth colored grains. When analyzing crosses of hybrids from F 1, there were four phenotypic groups in the offspring: 1200 smooth colored, 1215 wrinkled uncolored, 309 smooth uncolored, 315 wrinkled colored. Make a scheme for solving the problem. Determine the genotypes of the parents and offspring in 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 a starchy endosperm and a plant with an uncolored seed and a waxy endosperm, the offspring obtained splitting by phenotype: 9 plants with a colored seed and a 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 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. Based on the pedigree shown in the figure, determine and explain the nature of the inheritance of the trait highlighted in black (dominant or recessive, sex-linked or not. Determine the genotypes of the offspring 1,3,4,5,6,7. Determine the probability of birth from parents 1 ,2 next child with a trait highlighted in black on the pedigree. 
2. According to the pedigree shown in the figure, determine and justify the genotypes of parents, offspring, indicated in the diagram by the numbers 1,6,7. Set the probability of giving birth to a child with an inherited trait in woman number 6, if this trait has never manifested itself in her husband's family.
Based on the pedigree shown in the figure, determine and explain the nature of the inheritance of the trait highlighted in black. Determine the genotypes of parents, offspring1,6 and explain the formation of their genotypes.
According to the pedigree shown in the figure, determine the nature of the inheritance of the trait (dominant or recessive, linked or not linked to sex), highlighted in black, the genotypes of parents and children in the first generation. Indicate which of them is a carrier of the gene, the trait of which is highlighted in black.
Tasks for the interaction of genes
1. complementarity
In parrots, feather color is determined by two pairs of genes. The combination of two dominant genes determines the green color. Individuals recessive for both pairs of genes 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, they received parrots of all colors. Determine the genotypes of parents and offspring.
2. epistasis
The coat color of rabbits (as opposed to albinism) is determined by the dominant gene. The color of the color is controlled by another gene located on a different chromosome, and the gray color dominates over black (in albino rabbits, color color genes do not show themselves). 
What characteristics will the hybrid forms obtained from crossing a gray rabbit born from an albino rabbit with an albino carrying the black color gene have?
3. pleiotropy.
One of the breeds of chickens is distinguished by shortened legs. This feature is dominant. The gene controlling it simultaneously causes the shortening of the beak. At the same time, in homozygous chickens, the beak is so small that they are not able to break through the egg shell and die without hatching from the egg. 3,000 chicks have been produced in the incubator of a farm that breeds only short-legged hens. How many of them are short-legged?
4. polymer
The son of a white woman and a black man married a white woman. Can this couple have a child darker than their father?
ANSWERS:
I. Tasks for monohybrid crossing (complete and incomplete dominance)
1. A- gray color
a- black color R: AA and aa.
F 1: Ah, all grey.
2. A - black flower a - white color
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 splitting by genotype 1aa 2Aa 1 aa
4. Night-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 - splitting 3 green: 1 striped
7. A- wide leaves
a narrow leaves
Aa - intermediate leaf width
F 1: splitting by phenotype: 1 - wide leaves, 2 - intermediate leaf width, 1 - narrow leaves.
Genotype: AA 2Aa aa
II. Tasks for dihybrid crossing1. A - normal, a - glaucoma.
B - Marfan's syndrome, b - normal.
One of the spouses suffers from glaucoma and had no ancestors with Marfan syndrome: aabb. The second spouse is diheterozygous: AaBb.
| norm. | norm | glaucoma | glaucoma |
Probability of having a healthy baby (normal/normal) = 1/4 (25%). In this case, the third law of Mendel (the law of independent inheritance) is manifested.
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 red heads. 3 red. Op., 3 females, 1 female op.4 tons op.
5. In this case, intermediate inheritance of color appears. AA - black, Aa - motley, aa - white. the parents of both the hen and the rooster have AaBB genotypes. And the gametes form the same: AB, aB. when they merge, the formation of genotypes occurs -AABB - black crested, AaBB - mottled crested, aaBB - white crested. ratio -1/2/1.
color gene:
A - black
a- brown
tail length gene:
B - long
c- short
vv - lethal
Вв - shortened
Solution:
1) AaBv x aaBv
black short x brown short
gametes - AB, AB, AB, AB AB AB
AaBB AABB AABB AABB AABB AABB AABB
h.d.h.u. k.d. to y. h. y letal k. y lethal
black with a long tail - 1/8
black with shortened - 2/8
brown with a long tail - 1/8
brown with shortened - 2/
lethal - 2/8. Law of Independent Succession
7. A- mustachioed
a- beardless
B - red
BB- pink
1) first crossing:
R AAvv * aaVV
mustache white used cr.
2) analyzing cross:
Aaaa * aaa
G AB AB AB AB AB
F 2 AaBb - mustachioed pink-fruited; Aabb - mustachioed white-fruited;
aaBb - beardless rose-fruited; aabb - beardless white-fruited;
3) the nature of the inheritance of the color trait 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 beaters,
a - funnel-shaped corollas.
B - white flowers,
b - cream flowers
first cross:
R aaVv x Aavv
G ab ab av av
F 1 AaBb aavb
second cross:
R aaBB x Aawb
G aB Av av
F 1 AaBb aaBv
In this case, Medel's third law is manifested - the law of independent inheritance.
9. A - normal, a - blindness No. 1.
B - normal, b - blindness No. 2.
Maternal grandmother AAbb, paternal grandmother aaBB. Grandpas - AABB.
| |
Probability of having a blind grandchild 0%
Tasks for the inheritance of blood groups of the AB0 system
1. Boy-j0j0. Sister - JAJB
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 2nd or 3rd blood groups.
3. first pair of parents:
P: j 0 j 0 x J A j 0 or j 0 j 0 x J A J A
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
P: 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.)
In the first pair of parents, the son has 1 gr. and he received gene 0 from both parents. The second couple are the parents of a boy with the 2nd 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 4 blood group
4. It is impossible, because in children, blood types are possible: A0 (II) or B0 (III), therefore, the blood of the fourth group, which the mother has, cannot be transfused.
5. Digeterozygous father I B i 0 Rr, dihomozygous mother I A I A RR.
| IV group | IV group | II group | II group |
Children in this family may have IV or II blood group, all are Rh-positive. The proportion of children with IV blood group is 2/4 (50%). The law of independent inheritance appears (Mendel's third law).
IV. Sex-linked and autosomal inheritance tasks
1. A - the presence of a crest, a - no crest.
X B - green plumage, X b - brown plumage.
A_X B Y-crested green female
aaX B X b - male without crest with green plumage (heterozygous)
Among the offspring were chicks without crest - aa. They got one gene a from their mother, one from their father. Therefore, the mother must have the gene a, hence the mother of Aa.
P AaX B Y x aaX B X b
| AaX V X V | aaX V X V | AaX B Y | aaX B Y |
|
| AaX B Xb | 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 B red eyes
X in - white eyes
R 1 AAX B X B * aaX in Y
gray body black body
red eyes white ch.
G AX B aX to 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 offspring have a gray body and red eyes.
Sex Ratio-50% : 50:%
3. A - brown eyes,
and blue eyes.
X D - normal vision,
X d - color blindness.
A_X D X _ brown-eyed woman with normal vision
aaX d Y is the woman's father, he could give his daughter only aX d, therefore, the brown-eyed woman is 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%), it is a boy.
4. A - normal, a - albinism.
X H - normal, 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. c.g. g.n. g.n. g.n. y..y.
Splitting by eye color - 1:1 by blood clotting - all daughters are healthy, boys - 1:1.
V. Problems on linked inheritance1 . A - smooth grains,
a - wrinkled grains.
B - colored grains,
b - uncolored grains.
R AABB x AABB
Since uniformity was obtained in the first generation (Mendel's first law), therefore, homozygotes were crossed, in F1 a diheterozygote was obtained, carrying dominant traits.
Analyzing cross:
| normal gametes | |||||||
| recombinant gametes | |||||||
| smooth | wrinkled. | smooth | wrinkled. |
Since in the second generation an unequal number of phenotypic groups turned out, therefore, linked inheritance took place. Those phenotypic groups that are presented in large numbers are not crossovers, but groups that are presented in small numbers are crossovers formed from recombinant gametes in which linkage was broken due to crossing over in meiosis.
2. A- colored seed a- not colored seed B- starchy endosperm b- waxy endosperm seed, starch endosperm 42- Aavv- okr. seed, wax endosperm 44- aaBv- unstained seed, starchy endosperm 10- aavb- unstained seed waxy endosperm The presence in the offspring of two groups (42 - with colored waxy endosperm; 44 - with unstained waxy endosperm) in approximately equal proportions - the result of linked inheritance of alleles A and B , a and B among themselves. The other two 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 only two types of gametes are formed in the diheterozygous parent (full linkage).
- A gray body
F 1 AaBv all gray with normal wings. law of uniformity)
P AaVv x AaVv
T.K. there is no expected Mendelian splitting, which means that a crossing-over has occurred:
linked
F 2 AABB AABB AABB AABB
In the previous article, we talked about the tasks of the C6 line in general. Starting from this post, specific tasks in genetics that were included in the test tasks of past years will be dealt with.
Having a good understanding of such a biological discipline as genetics - the science of heredity and variability - is simply necessary for life. Moreover, genetics in our country has such a long-suffering history ...
Just think, Russia from the leading country in the study of genetics at the beginning of the 20th century is turning into a dense monster in eradicating even just genetic terminology from the minds of people from the late 30s to the mid-50s.
Is it possible to forgive the regime for killing by torture and starvation the greatest geneticist, the noblest servant of the people and science, the founder of the All-Union Institute of Plant Growing in Leningrad, Academician (1887 - 1943) .
Let's start the analysis of the real tasks of the C6 line with tasks on dihybrid cross that require knowledge by inheritance of the traits of two pairs of allelic genes (but which are non-allelic in relation to each other), located in different pairs of homologous chromosomes, therefore inherited .
The most surprising thing is that the level of difficulty of these tasks varies greatly. , what we are with you now and we will be convinced by examples of solving several tasks.
Further studying the material of this article, thanks to my detailed explanations, I hope that more complex tasks will be understandable for you. And for a more successful mastering of tasks for dihybrid crossing, I bring to your attention my book:
Task 1. About pigsfor a dihybrid cross(the simplest)
In pigs, black coat color (A) dominates over red (a), long bristles (B) dominate over short ones (c). The genes are not linked. What offspring can be obtained by crossing a black with a long stubble of a diheterozygous male with a homozygous black female with a short stubble. Make a scheme for solving the problem. Determine the genotypes of parents, offspring, offspring phenotypes and their ratio.
First, I would like to draw your attention to such moments. :
First, why is this dihybrid cross task? In the task, it is required to determine the distribution in inheritance of two traits: coat color (A or a) and length (B or c). Moreover, it is indicated that genes are linked, that is, the studied traits are in different pairs of homologous chromosomes and are inherited independently of each other according to Mendel's law. This means that offspring will be formed from all possible random combinations of gametes formed by a male and a female.
Secondly, it is this dihybrid crossing task that is the simplest of this type of task. It stipulates in advance that the studied signs are not linked. In addition, we can immediately (without analyzing all possible combinations of the birth of offspring) according to the given phenotype of the parents write down their genotype.
Solution:
1) genotypes of parents :
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 there are : A - dominant black coat color and a - recessive red coat color; B - dominant long bristle and b - recessive short;
female AAbb- as it is said about her that she homo zygote according to the color of the coat, which is also black in her, so we only write down AA, but about the length of the wool is not said homo is she zygote or Goethe rosygous, since this information would be redundant!(and so it is clear that if the female has short hair, then she can also be only homo zygote for this recessive sign bb ).
Of course, such long arguments about the record genotype parents according to this in the condition of their task phenotype you don't have to bring it. The main thing is that the first item you should indicate correctly by no means mistaken, the genotypes of both parents.
2) gametes :
dihetero a zygote male will produce with equal probability four sperm varieties AB, Ab, aB, ab(according to the law of purity of gametes, as a result, each gamete can have only one allele of any gene. And since the inheritance of two traits at once is studied, we enter one allelic gene of each studied trait into each gamete);
digomo zygote female (AAbb- how we found out exactly this genotype in her ) 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 spermatozoa AB, Ab, aB and ab with equal probability, then the birth of offspring with such
four genotypes : AABb, AAbb, AaBb and Aabb in relation to 1: 1: 1: 1 (25%, 25%, 25%, 25%)
and two phenotypes : A-B-– black longhair – 50% and A-bb– black shorthair – 50% ( gaps are those places where it makes no difference for manifestations phenotype what second dominant or recessive gene in these pairs of allelic genes may be present ).
So, we fully answered the questions of the task : the decision was made according to the standard scheme (parents, gametes, offspring), the genotypes of parents and offspring were determined, the phenotypes of the offspring were determined, and the possible ratio of genotypes and phenotypes of the offspring was determined.
Task 2. About the Datura plant for dihybrid crossing, but with a more complex condition .
A Datura plant with purple flowers (A) and smooth bolls (b) was crossed with a plant with purple flowers and spiny bolls. The following phenotypes were obtained in the offspring: with purple flowers and spiny boxes, with purple flowers and smooth boxes, with white flowers and spiny boxes, with purple flowers and smooth boxes. Make a scheme for solving the problem. Determine the genotypes of parents, offspring and the possible ratio of phenotypes. Establish the nature of inheritance of traits.
Note, that in this task we can no longer immediately definitely answer the question about the genotype of the parents, and therefore straightaway 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 indicate character inheritance of traits (independently inherited traits or linked). This was given in the previous assignment.
Solution:
1) we first define let and ambiguous possible genotypes of parents
R: A - bb (purple, smooth) and A - B - (purple, prickly)
2) we also ambiguously write out information about the gametes they produce
G: Ab, - b and AB, A -, - B, - -
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 extract from all of the above:
a) since among the offspring there are plants with smooth boxes (and this is a recessive trait), then the genotypes both parents necessarily must have a gene b. That is, we can already enter into 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), then the genotypes both parents must have the gene a(small);
4) only now we can already completely write out the genotypes of both parents : .. … ………………….. Aabb and AaBb and produced by them ………………………………………….
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 prickly boxes,
………….. with purple flowers and smooth boxes,
………….. with white flowers and prickly boxes,
………….. with white flowers and smooth boxes”,
then this is possible only when 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 make all possible combinations of crosses of the available gametes. It is most convenient to record using the Punnett lattice. In our problem, it will be, thank God, not classical (4 x 4 = 16), but only 2 x 4 = 8 :
G : AB Ab aB ab
Ab AABb AAbb AaBb Aabb
………….. purple prickly purple smooth purple prickly purple smooth
av AaBb Aabb aaBb aabb
…………. purple prickly purple smooth white prickly white smooth
7) the distribution in the 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. Very simple, if you understand the meaning of genetic terminology
From crossing 2 varieties of barley, one of which has a two-row dense ear, and the other has a multi-row loose ear, F 1 hybrids were obtained, with a two-row loose ear. What results in terms of phenotype and genotype will be obtained in backcrosses if the inheritance of traits is independent? Create crossover patterns.
Since it is said that they crossed varieties barley (yes, anything, the word variety “appears”), which means we are talking about homozygous organisms for both traits studied. 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 we can apply the calculations following from Mendel's 3rd law for dihybrid crossing.
It is also said what features the hybrids in F 1 had. They were with a two-row loose spike, which means that these features are dominant over the multi-row and density of the spike. Therefore, we can now introduce the designations of the alleles of the genes of these two studied traits and will not be mistaken in the correct use of capital and small letters of the alphabet.
Denote:
allelic gene for double-row spike BUT, and multi-row - a;
loose ear allelic gene AT, and dense - b,
then the genotypes of the original two varieties of barley will look like this: AAbb and aaBB. From their crossing in F 1, 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, right?
Task 4. “Not red, I’m not red at all, I’m not red, but golden”
A woman with brown eyes and red hair married a man with non-red hair and blue eyes. It is known that the woman's father had brown eyes, and her mother had blue eyes, and both had red hair. The father of the man had non-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 BUT(it is well known to everyone that brown eye color dominates blue color), and the allelic gene for blue eyes, respectively, will be a. Be sure to use the same letter of the alphabet, since this is one sign - eye color.
The allelic gene for non-red hair (hair color is the second trait studied) is denoted AT, 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 down incompletely at first, 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 BUT could only be a, that is, its genotype will be Aabb.
The genotype of a blue-eyed male with non-red hair can first be written as : aaB-. But since his mother had red hair, that is bb, then the second allelic gene at AT a man could only b. In this way, the genotype of a man will be written aaBb. The genotypes of his parents: father - aaB-; mothers - A-bb.
Children from the marriage of the analyzed spouses Ааbb x aaBb(and gametes, respectively : Ab, ab and aB, ab) will have equally likely 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 yourself see what tasks can be unequal in complexity. Unfair, yes unfair, I answer, as a tutor of the Unified State Examination in biology. You need luck, yes you need luck!
But you must admit that luck will be useful only for those who are really "in the know." Without knowledge of the laws of heredity of Gregor Mendel, it is impossible to solve the first task, so the conclusion can be one : .
In the next article by a biology tutor on Skype, we will analyze tasks on inheritance that are correctly solved by an even smaller number of students.
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 biology tutor via skype, contact in the comments. On my blog you can buy answers to all tests OBZ FIPI for all years of examinations and .
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 Punnett lattice, illustrate crossings for one or two traits and indicate what numerical ratios of genotypes and phenotypes should be expected in the offspring from these crossings.
- Outline the rules of inheritance, segregation, 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.
- Specify the possible genotypes of people with blood groups A; AT; AB; O.
- Give examples of polygenic traits.
- Indicate the chromosomal mechanism of sex determination and the types of inheritance of sex-linked genes in mammals, use this information in solving problems.
- Explain the difference between sex-linked and sex-dependent traits; give examples.
- Explain how human genetic diseases such as hemophilia, color blindness, sickle cell anemia are inherited.
- Name the features of plant and animal breeding methods.
- Indicate the main directions of biotechnology.
- To be able to solve the simplest genetic problems using this algorithm:
Problem solving algorithm
- Determine the dominant and recessive trait based on the results of crossing the first generation (F1) and the second (F2) (according to the condition of the problem). Enter the letter designations: A - dominant and - recessive.
- Write down the genotype of an individual with a recessive trait or an individual with a genotype known by the condition of the problem and gametes.
- Write down the genotype of F1 hybrids.
- Make a diagram of the second crossing. Write the gametes of the F1 hybrids in the Punnett grid horizontally and vertically.
- Write down the genotypes of the offspring in the gamete crossing cells. Determine the ratio of phenotypes in F1.
Task design scheme.
Letter designations:
a) dominant trait _______________
b) recessive trait _______________
Gametes
F1(first generation genotype)
| gametes | ||
| ? | ? |
Punnett lattice
| F2 |
|
Phenotype ratio in F2: _____________________________
Answer:_________________________
Examples of solving problems for monohybrid crossing.
A task."There are two children in the Ivanov family: 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 eye color inherited in humans? What are the genotypes of all family members? Eye color is a monogenic autosomal trait."
The eye color trait is controlled by one gene (by condition). The mother of these children is blue-eyed, and her parents had brown eyes. This is possible only in the THAT case if both parents were heterozygous, therefore, brown eyes dominate over blue ones. Thus, grandmother, grandfather, father and daughter had the genotype (Aa), and mother and son - aa.
A task."A rooster with a pink comb is crossed with two hens that also have a pink comb. The first gave 14 chickens, all with a pink comb, and the second - 9 chickens, of which 7 with a pink comb and 2 with a leaf comb. The shape of the comb is a monogenic autosomal trait. What are genotypes of all three parents?
Before determining the genotypes of the parents, it is necessary to find out the nature of the inheritance of the comb shape in chickens. When a rooster was crossed with a second hen, 2 chickens with a leaf-shaped comb appeared. This is possible when the parents are heterozygous, therefore, it can be assumed that the pink-shaped comb in chickens dominates over the leaf-shaped one. Thus, the genotypes of the rooster and the second hen are Aa.
When the same rooster was crossed with the first hen, no splitting was observed, therefore, the first hen was homozygous - AA.
A task."In the family of brown-eyed right-handed parents, fraternal twins were born, one of which is brown-eyed left-handed, and the other blue-eyed right-handed. What is the probability of birth next child like their 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 the better possession of the left hand compared to the right. Let's introduce allele designations: A - brown eyes, a - blue eyes, B - right-handed, c - left-handed. Let's determine the genotypes of parents and children:
| R | AaVv x AaVv |
| F, | A_vv, aaB_ |
A_vv - phenotypic radical, which shows that this child is left-handed with brown eyes. The genotype of this child can be - Aavv, AAvv.
Further solution of this problem is carried out in the traditional way, by constructing the Punnett lattice.
| AB | Av | aB | Av | |
| AB | AABB | AAVv | AaBB | AaVv |
| Av | AAVv | AAvv | AaVv | aww |
| aB | AaBB | AaVv | aaBB | AaVv |
| av | AaVv | aww | aawww | aww |
Underlined are 9 variants of descendants that we are interested in. There are 16 possible options, so the probability of having a child similar to their parents is 9/16.
Ivanova T.V., Kalinova G.S., Myagkova A.N. "General Biology". Moscow, "Enlightenment", 2000
- Topic 10. "Monohybrid and dihybrid crossing." §23-24 pp. 63-67
- Topic 11. "Genetics of sex." §28-29 pp. 71-85
- Topic 12. "Mutational 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) dominates over blue (b). Write down the genotypes of the parents, possible phenotypes and genotypes of children born from the marriage of a fair-haired, blue-eyed man and a heterozygous brown-eyed, fair-haired woman.
Answer
Blond-haired blue-eyed man aabb.
Heterozygous brown-eyed blonde female aaBb.
Congenital myopia is inherited as an autosomal dominant trait, the absence of freckles as an autosomal recessive trait. The traits are on different pairs of chromosomes. The father has congenital myopia and lack of freckles, the mother has normal vision and freckles. There are three children in the family, two are nearsighted without freckles, one with normal vision and freckles. Make a scheme for solving the problem. Determine the genotypes of parents and born children. Calculate the probability of having children with nearsightedness and freckles. Explain what is the law 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, so the second child is aaBb.
If the mother is aa, then all her children have a, so the first child is Aabb.
If the first child has bb, then he took one b from the mother and one from the father, so 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 had a child with a fused earlobe and a smooth chin. Determine the genotypes of the parents, the first child, the phenotypes and genotypes of other possible offspring. Make a scheme for solving the problem. Traits are inherited independently.
Answer
The offspring showed recessive traits that were latent in the parents.
A - free earlobe, a - fused earlobe.
B - triangular fossa on the chin, b - smooth chin.
Child aabb, parents A_B_.
The child aa received one a from his father, another from his mother; one b from the father, the other from the mother, hence the parents of 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_ loose earlobe, triangular fossa on chin
3 A_bb loose earlobe, smooth chin
3 aaB_ fused earlobe, triangular fossa on chin
1 aabb fused earlobe, smooth chin
A black crested rooster is crossed with a similar hen. From them 20 chickens were obtained: 10 black crested, 5 brown crested, 3 black without crest and 2 brown without crest. Determine the genotypes of parents, offspring and the pattern of inheritance of traits. The genes of the two traits are not linked, the dominant traits are black plumage (A), crestedness (B).
Answer
A - black plumage, and - brown plumage.
B - crested, b - without crest.
Rooster A_B_, Hen 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 pattern of inheritance of traits is the law of independent inheritance.
Genetics, its tasks. Heredity and variability are properties of organisms. Methods of genetics. Basic genetic concepts and symbolism. Chromosomal theory of heredity. Modern ideas about 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 speculate 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 of certain “pangens” in the cell nucleus that determine the hereditary inclinations of the organism, and the release into the protoplasm of only those of them that determine the cell type, could not change the situation, as well as the theory of "germ plasm" by A. Weisman, according to which the traits acquired in the process of ontogenesis are not inherited.
Only the works of the Czech researcher G. Mendel (1822-1884) became the foundation stone of modern genetics. However, despite the fact that his works were cited in scientific publications, contemporaries did not pay attention to them. And only the rediscovery of the patterns of independent inheritance by three scientists at once - E. Chermak, 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 on the present stage are the study of the qualitative and quantitative characteristics of hereditary material, analysis of the structure and functioning of the genotype, deciphering the fine structure of the gene and methods for regulating gene activity, the search for genes that cause the development of human hereditary diseases and methods for their “correction”, the creation of a new generation of drugs similar to DNA vaccines , designing with the help of genetic and cell engineering organisms with new properties that could produce the necessary for a person medications and food, as well as a complete decoding of the human genome.
Heredity and variability - properties of organisms
Heredity- is the ability of organisms to transmit their characteristics and properties in a number of generations.
Variability- the property of organisms to acquire new characteristics during life.
signs- these are any morphological, physiological, biochemical and other features of organisms in which some of them differ from others, for example, eye color. properties They also call any functional features of organisms, which are based on a certain structural feature or a group of elementary features.
Organisms can be divided into quality and quantitative. Qualitative signs have two or three contrasting manifestations, which are called alternative features, for example, blue and brown eyes, while quantitative ones (milk yield of cows, wheat yield) do not have clearly defined differences.
The material carrier of heredity is DNA. There are two types of heredity in eukaryotes: genotypic and cytoplasmic. Carriers of genotypic inheritance are localized in the nucleus and further we will talk it is about her, 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, but their change can have a significant impact on the development of the organism, 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, on the one hand, leads to a decrease in plant productivity, and on the other hand, such variegated organisms are valued in decorative gardening. Such specimens are reproduced mainly asexually, since ordinary green plants are more often obtained during sexual reproduction.
Genetic methods
1. The hybridological method, or the method of crosses, consists in the selection of parent individuals and the analysis of offspring. At the same time, the genotype of an organism is judged by the phenotypic manifestations of genes in offspring obtained by a certain crossing scheme. This is the oldest informative method of genetics, which was most fully applied 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 makes it possible to determine the content of various substances in the body, in particular 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 sections. They allow you to identify genes for hereditary diseases even in embryos, establish paternity, etc.
5. The population-statistical method makes it possible to determine the genetic composition of a population, the frequency of certain genes and genotypes, the genetic load, and also to outline the prospects for the development of a population.
6. The method of hybridization of somatic cells in culture allows you to determine the localization of certain genes in chromosomes when cells of various organisms merge, for example, mice and hamsters, mice and humans, etc.
Basic genetic concepts and symbolism
Gene- This is a section of a DNA molecule, or chromosome, that carries information about a certain trait or property of an organism.
Some genes can influence the manifestation of several traits at once. Such a phenomenon is called pleiotropy. For example, the gene that determines the development of the hereditary disease arachnodactyly (spider fingers) also causes the curvature of the lens, the 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), each of the paired chromosomes contains one copy of the gene responsible for a particular trait. Such genes are called allelic.
Allelic genes most often exist in two variants - dominant and recessive. Dominant 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 labeled capital letters Latin alphabet (A, B, C, etc.), and recessive - lowercase (a, b, c, etc.). recessive alleles can only be expressed if they occupy loci on both paired chromosomes.
An organism that has the same allele on both homologous chromosomes is called homozygous for that gene, or homozygous(AA, aa, AABB, aabb, etc.), and an organism that has different gene variants on both homologous chromosomes - dominant and recessive - is called heterozygous for that 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. Such a phenomenon is called multiple allelism. However, even in this case, each chromosome from a pair carries only one allele, that is, all three gene variants in one organism cannot be represented.
Genome- a set of genes characteristic of the 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. Crossing, in which inheritance is considered and an accurate quantitative account of offspring is carried out for one pair of alternative traits, is called monohybrid m, in two pairs - dihybrid, according to more signs - polyhybrid.
According to 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 heterozygous (Aa) will have a manifestation of the dominant allele in the phenotype. Therefore, to check the genotype of an organism with cross-fertilization, analyzing cross A cross 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 produce splitting in the offspring, while in the offspring of heterozygous individuals an equal number of individuals with dominant and recessive traits is observed.
The following conventions are most often used to write crossover schemes:
R (from lat. parent- parents) - parent organisms;
$♀$ (alchemical sign of Venus - a mirror with a handle) - maternal individual;
$♂$ (alchemical sign of Mars - shield and spear) - paternal individual;
$×$ is the cross sign;
F 1, F 2, F 3, etc. - hybrids of the first, second, third and subsequent generations;
F a - offspring from analyzing crosses.
Chromosomal theory of heredity
The founder of genetics G. Mendel, as well as his closest followers, had no 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 the chromosomes. These assumptions have become the cornerstone of the chromosome theory of heredity.
In 1906, the English geneticists W. Batson and R. Pennet discovered a violation of Mendelian splitting when crossing sweet peas, and their compatriot L. Doncaster, in experiments with the gooseberry moth butterfly, discovered sex-linked inheritance. The results of these experiments clearly contradicted Mendelian ones, but given that by that time it was already known that the number of known features for experimental objects far exceeded 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, the experiments of the T. Morgan group began on a new experimental object - the Drosophila fruit fly. The results of these experiments made it possible by the mid-20s of the 20th century to formulate the main provisions of the chromosome theory of heredity, to determine the order of arrangement of genes in chromosomes and the distance between them, i.e., to compile the first maps of chromosomes.
The main provisions of the chromosome theory of heredity:
- Genes are located on chromosomes. Genes on the same chromosome are inherited together, 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 are arranged linearly on chromosomes.
- 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 ideas about the gene and genome
In the early 40s of the twentieth century, J. Beadle and E. Tatum, analyzing the results of genetic studies conducted 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 Escherichia coli gene and study the regulation of its activity. For this discovery, they were awarded the Nobel Prize in Physiology or Medicine in 1965.
In the course of the study, 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 regions include promoter, operator, and terminator. promoter called the region of the gene to which the RNA polymerase enzyme is attached, which ensures the synthesis of mRNA during transcription. FROM operator, located 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 begin. The structure of the repressor is usually encoded in a regulatory gene located in another part of the chromosome. The reading of information ends at a section of the gene called terminator.
coding sequence structural gene contains information about the sequence of amino acids in the corresponding protein. The coding sequence in prokaryotes is called cistronome, and the set of coding and regulatory regions of the prokaryotic gene - 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 may also contain additional small circular or open DNA molecules called plasmids. Plasmids are able to integrate into chromosomes and be transferred from one cell to another. They can carry information about sexual 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 a terminator. In addition, significant regions in eukaryotic genes ( exons) alternate with insignificant ( introns), which are completely transcribed into mRNAs and then excised during their maturation. Biological role introns is to reduce the likelihood of mutations in significant areas. Eukaryotic gene regulation is much more complex than that described for prokaryotes.
The human genome. In each human cell, there are about 2 m of DNA in 46 chromosomes, densely packed into a double helix, which consists of about 3.2 $ × $ 10 9 nucleotide pairs, which provides about 10 1900000000 possible unique combinations. By the end of the 1980s, the location of about 1,500 human genes was known, but their total number was estimated at about 100,000, since only hereditary diseases in humans have about 10,000, not to mention the number of various proteins contained in cells .
In 1988, the international project "Human Genome" was launched, which ended by the beginning of the 21st century. full transcript nucleotide sequences. He made it clear that two different person 99.9% have similar nucleotide sequences, and only the remaining 0.1% define our individuality. In total, approximately 30-40 thousand structural genes were discovered, 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. However, these genes encode a much larger number of proteins, such as tens of thousands of protective proteins - immunoglobulins.
97% of our genome is genetic "garbage" that exists only because it can reproduce well (the RNA that is transcribed in these regions never leaves the nucleus). For example, among our genes there are not only "human" genes, but also 60% of genes similar to those of the fruit fly, and up to 99% of our genes are related to chimpanzees.
In parallel with the decoding of the genome, chromosome mapping also took place, as a result of which it was possible not only to detect, but also to determine the location of some genes responsible for the development of hereditary diseases, as well as drug target genes.
The deciphering of the human genome does not yet have a direct effect, since we have received a kind of instruction for assembling such a complex organism as a person, but have not learned how to make it or at least correct errors in it. Nevertheless, the era of molecular medicine is already on the threshold, all over the world there is a development of so-called gene preparations that 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 cell's hereditary information and do not even encode the complete set of proteins necessary for their own functioning, they can significantly affect some features of the body. Thus, variegation in plants of chlorophytum, ivy and others is inherited by an insignificant 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, so this heredity is called maternal, or cytoplasmic, in contrast to the genotypic, which is localized in the nucleus.
