When crossing with each other. Solving problems in molecular biology and genetics. Solving problems in general genetics
Tasks for incomplete dominance
Task number 5
\ When purebred white chickens are crossed with each other, the offspring turns out to be white, and when black chickens are crossed, they are black. The offspring from white and black individuals turns out to be motley. What kind of plumage will the descendants of a white rooster and a motley hen have?
Answer: Half of the chickens will be white and half will be variegated.
Task number 6
Plants of red-fruited strawberries, when crossed with each other, always give offspring with red berries, and plants of white-fruited strawberries - with white ones. As a result of crossing these varieties with each other, pink berries are obtained. What offspring will arise when hybrids with pink berries are crossed?
Answer: half of the offspring will be with pink berries and 25% with white and red.
Tasks for the inheritance of blood groups
Task number 7
What blood types can children have if both parents have an IV blood group?
Answer: the probability of having children with IV blood group is 50%, with II and III - 25% each.
Task number 8
Is it possible to transfuse blood to a child from a mother if her blood type is AB, and her father has 0?
Answer: you can't.
Task number 9
The boy has an IV blood type, and his sister has an I. What areblood types of their parents?
Answer: II and III.
Task number 10
Two boys (X and Y) were mixed up in the maternity hospital. X has I blood type, Y has II. Parents of one of them with I and IV blood groups, and the other with I and III blood groups. Who is whose son?
Answer: X has parents with I and III groups, Y has parents with I and IV.
Sex-linked inheritance problems
Task number 11
In parrots, the sex-linked dominant gene determines the green color of the plumage, and the recessive gene determines brown. A green heterozygous male is crossed with a brown female. What will the chicks be like?
Answer: half of the males and females will be green, half will be brown.
Task number 12
In Drosophila, the dominant gene for red eyes and the recessive gene for white eyes are located on the X chromosome. What eye color can be expected in first-generation hybrids if a heterozygous red-eyed female and a white-eyed male are crossed?
Answer: the probability of the birth of males and females with red and white eyes is 50% each.
Task number 13
Husbands and wives who are healthy for color blindness have:
– a color-blind son with a healthy daughter;
– a healthy daughter who has 2 sons: one is colorblind and the other is healthy;
– a healthy daughter who has five healthy sons.
What are the genotypes of this husband and wife?
Answer: genotypes of parents XD Xd, XD Y.
Task number 14
A tortoiseshell cat brought kittens in black, red and tortoiseshell. Is it possible to determine whether a black or red cat was the father of these kittens?
Answer: you can't.
Combined tasks
Task number 15
At a large cattle the polled gene dominates the horned gene, and the roan coat color is formed as an intermediate trait when white and red animals are crossed. Determine the probability of the birth of calves similar to the parents from crossing a heterozygous roan bull with a white horned cow.
Answer: the probability of having calves similar to their parents is 25% each.
Task number 16
From crossing two varieties of strawberries (one with a mustache and red berries, the other beardless with white berries), in the first generation all plants were with pink berries and a mustache. Is it possible to breed a beardless variety with pink berries by backcrossing?
Answer: you can, with a probability of 25% when crossing hybrid plants with a beardless parent plant that has white berries.
Task number 17
A man with Rh-negative blood of group IV married a woman with Rh-positive blood of group II (her father has Rh-negative blood of group I). There are 2 children in the family: with Rh-negative blood Group III and with Rh-positive blood of group I. Which child in this family is adopted if the presence of the Rh factor antigen in the erythrocytes is due to the dominant gene?
Answer: adopted child with I blood group.
Task number 18
In one family, four children were born to brown-eyed parents: two blue-eyed with I and IV blood groups, two - brown-eyed with II and IV blood groups. Determine the probability of the birth of the next child brown-eyed with I blood group.
Answer: the genotype of a brown-eyed child with blood type I is A-I0I0, the probability of the birth of such a child is 3/16, i.e. 18.75%.
Task number 19
A man with blue eyes and normal vision married a woman with brown eyes and normal vision (all of her relatives had Brown eyes and her brother was colorblind). What are the children of this marriage?
Answer: all children will be brown-eyed, all daughters will have normal vision, and the probability of having sons with color blindness is 50%.
Task number 20
In canaries, the sex-linked dominant gene determines the green color of the plumage, and the recessive gene determines brown. The presence of a crest depends on an autosomal dominant gene, its absence depends on an autosomal recessive gene. Both parents are green with tufts. They had 2 chicks: a green male with a crest and a brown female without a crest. Determine the genotypes of the parents.
Answer: R: ♀ HZU Aa; ♂ HZHK Aa
Task number 21
A man suffering from color blindness and deafness married a woman with good hearing and normal vision. They had a son who was deaf and colorblind, and a daughter with good hearing and colorblindness. Is it possible for this family to have a daughter with both anomalies if deafness is an autosomal recessive trait?
Answer: the probability of having a daughter with both anomalies is 12.5%.
Tasks for the interaction of genes
Task number 22
The shape of the comb in chickens is determined by the interaction of two pairs of non-allelic genes: the walnut comb is determined by the interaction of the dominant alleles of these genes; the combination of one gene in the dominant state and the other in the recessive state determines the development of either a pink or pisiform crest; individuals with a simple crest are recessive for both alleles. What will be the offspring when two heterozygotes are crossed?
Given: A * B * - walnut
A * bb - pink
aaB * - pea-shaped
aabb - simple
9/16 - with walnut,
3/16 - with pink,
3/16 - with peas,
1/16 - with simple combs.
Task number 23
The brown color of mink fur is due to the interaction of dominant alleles. Homozygosity for recessive alleles of one or two of these genes gives platinum coloration. What will be the hybrids from crossing two heterozygotes?
Given: A * B * - brown
A * bb - platinum
ааВ* – platinum
aabb - platinum
9/16 - brown,
7/16 - platinum minks.
Task number 24
In alfalfa, the inheritance of flower color is the result of a complementary interaction of two pairs of non-allelic genes. When crossing plants of pure lines with purple and yellow flowers in the first generation, all plants were with green flowers, in the second generation splitting occurred: 890 plants with green flowers, 306 with yellow flowers, 311 with purple flowers and 105 with white flowers grew. Determine the genotypes of the parents.
Answer: AAbb and aaBB.
Task number 25
In rabbits, the recessive gene for the absence of pigment suppresses the action of the dominant gene for the presence of pigment. Another pair of allelic genes affects the distribution of the pigment, if any: the dominant allele determines the gray color (because it causes an uneven distribution of the pigment along the length of the hair: the pigment accumulates at its base, while the tip of the hair is devoid of pigment), the recessive allele determines black ( since it does not affect the distribution of the pigment). What will be the offspring from crossing two heterozygotes?
Given: A * B * - gray color
A * bb - black
aaB* - white
aabb - white
9/16 - gray,
3/16 - black,
4/16 - white rabbits.
Task number 26
In oats, grain color is determined by the interaction of two non-allelic genes. One dominant determines the black color of the grains, the other - gray. The black color gene suppresses the gene gray color. Both recessive alleles give a white color. When crossing black-grained oats, splitting was noted in the offspring: 12 black-grained: 3 gray-grained: 1 with white grains.
Determine the genotypes of the parent plants.
Given: A * B * - black color
A * bb - black
aaB* - gray
aabb - white
P: ♀ black
♂ black
in F1 - 12 black, 3 gray, 1 white.
Answer: AaBb and AaBb.
Task number 27
Human skin color is determined by the interaction of genes according to the type of polymer: the darker the skin color, the more dominant genes in the genotype: if 4 dominant genes - the skin is black, if 3 is dark, if 2 is swarthy, if 1 is light, if all genes are in recessive state - white. A black woman married a white-skinned man. What can their grandchildren be like if their daughter marries a mulatto (AaBb)?
Given: AABB - black leather
AaBB, AABb - dark skin
AaBb, AAbb, aaBB - dark skin
Aabb, aaBb - fair skin
aabb - white skin
P1: ♀ AABB × ♂ aabb
P2: ♀ AaBb × ♂ AaBb
Answer: the probability of having grandchildren with black skin is 6.25%, with dark skin - 25%, with swarthy skin - 37.5%, with light skin - 25%, with white skin - 6.25%.
Task No. 28 Inheritance of springiness in wheat is controlled by one or two dominant polymeric genes, while winterness is controlled by their recessive alleles. What will be the offspring when two heterozygotes are crossed?
Given: A * B * - rage
A * bb - rage
aaB * - rage
aabb - winter
15/16 - spring,
1/16 - winter crops.
Hurdles game
The game takes place in 5 stages, because. the ability to solve 5 types of problems is tested (for monohybrid crossing; incomplete dominance; dihybrid crossing; sex-linked inheritance; gene interaction).
I stage. Students receive cards with task number 1 (5 options in total) and solve the problem by writing down the answer.
II stage. The student chooses a card front side which printed the answer he received in the previous problem, and solves problem No. 2.
III–V stages. The student continues to choose cards with the answers he gets and solves problems No. 3-5.
The student reports the last answer to the teacher, who checks the answers using the “key”.
If the answer is correct, then the student has overcome all the “barriers” - he solved all the problems correctly.
If the answer is incorrect, it means that the student solved some problem incorrectly and switched to " treadmill» Another option - the teacher, using the answer key, checks all his tasks.
The score is based on the number of tasks solved correctly.
Task number 1 (for monohybrid crossing).
Card 1. A pink comb is a dominant trait in chickens, a simple one is recessive. What will be the offspring if heterozygous hens with pink combs and homozygous roosters with simple ones are crossed?
Card 2. A heterozygous black rabbit was crossed with the same rabbit. Determine the offspring by genotype and phenotype if black fur is dominant over gray.
Card 3. A heterozygous red-fruited tomato was crossed with a homozygous red-fruited tomato. Determine the offspring by genotype and phenotype if the red color of the fruit dominates over the yellow.
Card 4. In oats, smut resistance dominates susceptibility. A plant of a variety susceptible to smut is crossed with a plant homozygous for resistance to this disease. What will be the offspring?
Card 5. In beans, the black color of the skin dominates over the white. Determine the color of seeds obtained by crossing homozygous plants with a black color of the seed coat.
Task number 2 (incomplete dominance).
Answers to the first question are given in parentheses.
Card 6 (1/2 Aa; 1/2 aa). When purebred white hens and the same roosters are crossed, the offspring turns out to be white, and when black hens and black roosters are crossed, they are black. The offspring from white and black individuals turns out to be motley. What plumage will the offspring of motley chickens have?
Card 7 (AA; Aa; Aa; aa). When purebred white hens and the same roosters are crossed, the offspring turns out to be white, and when black hens and black roosters are crossed, they are black. The offspring from white and black individuals turns out to be motley. What plumage will the offspring of a white rooster and a motley hen have?
Card 8 (1/2 AA; 1/2 Aa). Plants of red-fruited strawberries, when crossed with each other, always give offspring with red berries, and plants of white-fruited strawberries - with white ones. As a result of crossing these varieties with each other, pink berries are obtained. What will be the offspring if you cross hybrids with pink berries?
Card 9 (Aa). Plants of red-fruited strawberries, when crossed with each other, always give offspring with red berries, and plants of white-fruited strawberries - with white ones. As a result of crossing these varieties with each other, pink berries are obtained. What offspring will result if red-fruited strawberries are pollinated with pollen from hybrid strawberries with pink berries?
Card 10 (AA). At snapdragon plants with wide leaves, when crossed with each other, give offspring also with wide leaves, and plants with narrow leaves - only offspring with narrow leaves. As a result of crossing broad-leaved and narrow-leaved individuals, plants with leaves of intermediate width appear. What will be the offspring from crossing two individuals with leaves of intermediate width?
Task number 3 (for dihybrid crossing).
Answers to the second problem are given in parentheses.
Card 11 (1/4 white; 1/2 variegated; 1/4 black). A blue-eyed right-handed young man (his father was left-handed) married a brown-eyed left-hander (all her relatives are brown-eyed). What will be the children from this marriage, if brown eyes and right-handedness are dominant features?
Card 12 (1/2 variegated; 1/2 white). Rabbits were crossed: a homozygous female with normal hair and hanging ears and a homozygous male with elongated hair and erect ears. What will be the hybrids of the first generation, if the usual coat and erect ears are dominant characters?
Card 13 (1/4 red; 1/2 pink; 1/4 white). At sweet pea high growth dominates over dwarf, green beans - over yellow. What will be the hybrids when crossing a homozygous tall plant with yellow beans and a dwarf with yellow beans?
Card 14 (1/2 red; 1/2 pink). In a figured pumpkin, the white color of the fruit dominates over the yellow, the disc-shaped shape dominates over the spherical one. What will hybrids look like from crossing a homozygous yellow spherical pumpkin and a yellow discoid (heterozygous for the second allele).
Card 15 (1/4 narrow; 1/2 intermediate; 1/4 wide). In tomatoes, the red color of the fruit dominates over the yellow, normal growth over dwarf. What will be the hybrids from crossing homozygous yellow tomatoes normal growth and yellow dwarfs?
Task No. 4 (for sex-linked inheritance).
Answers to the third problem are given in parentheses.
Card 16 (AaBa; Aabb). In Drosophila, the dominant gene for red eyes and the recessive gene for white eyes are located on the X chromosome. What eye color will the first generation hybrids have if a heterozygous red-eyed female and a male with white eyes are crossed?
Card 17 (AaBb). The absence of sweat glands in humans is inherited as an X-linked recessive trait. A young man not suffering from this disease married a girl without sweat glands. What is the prognosis for this couple's children?
Card 18 (Aabb). What kind of vision can children have from the marriage of a man and a woman who normally distinguish colors, if it is known that their fathers were color blind?
Card 19 (aaBb; aabb). What can be the vision of children from the marriage of a man and a woman who normally distinguish colors, if it is known that the father of the man was color blind?
Card 20 (aaBb). Can the children of a colorblind man and a colorblind woman (whose father was colorblind) be colorblind?
Task number 5 (on the interaction of genes).
Answers to the fourth problem are given in parentheses.
Card 21 (♀1/2 cr.; 1/2 white; ♂ 1/2 cr.; 1/2 white). The shape of the comb in chickens is determined by the interaction of two pairs of non-allelic genes: the walnut comb is determined by the interaction of the dominant alleles of these genes; the combination of one of the genes in a recessive state and the other in a dominant state determines the development of either a rose-shaped or pea-shaped crest. Individuals with a simple crest are recessive for both genes. What will be the offspring from crossing two heterozygotes?
Card 22 (♀ healthy; ♂ sick). The coloration of mice depends, in the simplest case, on the interaction of two genes. In the presence of gene A, mice are colored, they produce pigment. In the presence of gene a, there is no pigment, and the mouse has White color. The specific color of the mouse depends on the second gene. Its dominant allele B determines the gray color of the mouse, and the recessive b allele determines the black color. Aabb black mice were crossed with aaBB white mice. What will F2 be like?
Card 23 (♀ healthy; ♂ 1/2 sick). In a pumpkin, the disc-shaped shape of the fetus is determined by the interaction of two dominant genes A and B. In the absence of any of them in the genotype, spherical fruits are obtained. The combination of recessive alleles of both genes gives an elongated fruit shape. Determine the phenotypes of the offspring obtained from crossing two varieties of pumpkin with disc-shaped fruits having AaBb genotypes.
Card 24 (♀ healthy; ♂ healthy). The brown color of the fur in minks is due to the interaction of two dominant genes A and B. Homozygosity for recessive alleles of one or two of these genes gives a platinum color. When crossing two platinum minks aaBB and AAbb, all hybrids of the new generation were brown. What will be the offspring of these brown minks?
Card 25 (♀ 1/2 sick; ♂ 1/2 sick). Gene A in chickens suppresses the action of the black color gene B. Chickens with the A genotype have white color. In the absence of the B gene, chickens are also white (i.e., homozygous for the recessive gene are white). What will be the second generation from crossing white leghorns (AABB) and white Wyandotes (aabb)?
The key to the game "Running with barriers"
Option 1
Option 2
Option 3
Option 4
Option 5
Task #1
Card 1
Card 2
Card 3
Card 4
Card 5
Task #2
Card 6
1/2 variegated
1/4 black
Card 7
1/2 variegated
Card 8
1/4 red
1/2 pink
Card 9
1/2 red
1/2 pink
Card 10
1/2 int.,
1/4 wide
Task #3
Card 11
Card 12
Card 13
Card 14
Card 15
Task #4
Card 16
1/2 red
1/2 red
Card 17
Card 18
1/2 sick
Card 19
Card 20
1/2 sick
1/2 sick
Task number 5
Card 21
op. roses. mountains etc.
Card 22
ser. black white
Card 23
Card 24
Card 25
white black
Problems for analyzing crosses
Task number 29 The red color of the fox is a dominant trait, the black-brown color is recessive. Analyzing crossing of two red foxes was carried out. The first one had 7 fox cubs - all of the red color, the second - 5 fox cubs: 2 red and 3 black-brown. What are the genotypes of all parents?
Answer: the male is black-brown in color, the females are homo- and heterozygous.
\Problem No. 30 In spaniels, the black color of the coat dominates over the coffee, and the short coat dominates over the long one. The hunter bought a short-haired black dog and, in order to be sure that it was a purebred, conducted an analysis cross. 4 puppies were born: 2 shorthair black, 2 shorthair coffee. What is the genotype of the dog bought by the hunter?
Answer: the dog bought by the hunter is heterozygous for the first allele.
Crossover tasks
Task number 31 Determine the frequency ( percentage) and types of gametes in a diheterozygous individual, if it is known that genes A and B are linked and the distance between them is 20 Morganids.
Answer: crossover gametes - Aa and ab - 10% each, non-crossover gametes - AB and ab - 40% each.
Task number 32 In tomatoes, high growth dominates over dwarf, spherical shape fruits - above the pear-shaped. The genes responsible for these traits are linked at a distance of 5.8 Morganids. They crossed a diheterozygous plant and a dwarf plant with pear-shaped fruits. What will be the offspring? The basis of innovative educational program specialized secondary full) general education MOU secondary school No. 5 for 2009 - 2014. (hereinafter referred to as the Program) is the development of the individual from the standpoint of humanistic, competence, activity,
Lethal Genes in the homozygous state can cause the death of offspring even before birth. At the same time, other genotypes survive. As with codominance, in this case three phenotypic classes are formed, but one of the phenotypes does not appear, since individuals carrying lethal genes die. Therefore, the splitting in the offspring differs from Mendelian.
Task 8-1
One of the breeds of chickens is distinguished by shortened legs - a dominant feature (such chickens do not break gardens). This gene also affects the length of the beak. At the same time, in homozygous dominant chickens, the beak is so small that they cannot hatch from the egg and die. In the incubator of a farm that breeds only short-legged hens (long-legged hens are not allowed to breed and are sent for sale), 3,000 chicks were obtained. How many of them were short-legged?
- All hens in the incubator were heterozygous (since homozygous short-legged hens die before birth).
- When heterozygous individuals are crossed with each other, the following offspring are formed:
25% of individuals with the AA genotype - die before birth,
50% of individuals with the Aa genotype are short-legged, 25% of individuals with the aa genotype are long-legged.
That is, short-legged individuals were 2/3 of all surviving offspring - about 2000 pieces.
Task 8-2
When crossing black mice with each other, black offspring are always obtained. When yellow mice are crossed, one third turns out to be black, and two thirds turn out to be yellow. How can these results be explained?
- Black mice are homozygous because all their offspring are uniform.
- Yellow mice are heterozygous, as segregation is observed in their offspring. Since heterozygous individuals carry a dominant trait, the yellow color is dominant.
- Yellow mice, when crossed with each other, never give only yellow offspring. In addition, the segregation in their offspring differs from Mendelian. This suggests that individuals homozygous for the dominant do not survive. Crossing analysis confirms this assumption.
Crossing scheme
Task 8-3
What happens if we assume that an organism develops a lethal mutation in which only heterozygous individuals die, while homozygous individuals remain viable?
Task 8-4
In mice, the short-tail gene in the dominant state is lethal, causing the death of the embryo in the early stages of development. Heterozygotes have shorter tails than normal individuals. Determine the phenotypes and genotypes of offspring arising from crossing long-tailed and short-tailed mice.
Task 8-5
When mirror carps were crossed, splitting was observed already in the first generation: 152 offspring were mirror and 78 with normal scales. How to explain these results? What offspring will result from crossing a mirror carp with an ordinary one?
Task 1
One of the breeds of chickens is different with shortened legs - a dominant feature (such chickens do not break gardens). This gene also affects the length of the beak. At the same time, in homozygous dominant chickens, the beak is so small that they cannot hatch from the egg and die. In the incubator of a farm that breeds only short-legged hens (long-legged hens are not allowed to breed and are sent for sale), 3,000 chicks were obtained. How many of them were short-legged?
Decision:
1. All hens in the incubator were heterozygous (since homozygous short-legged hens die before birth).
2. When heterozygous individuals are crossed with each other, the following offspring are formed:
25% of individuals with the AA genotype - die before birth,
50% of individuals with the Aa genotype are short-legged,
25% of individuals with the aa genotype are long-legged.
That is, short-legged individuals were 2/3 of all surviving offspring - about 2000 pieces.
Answer:
There were 2000 short-legged individuals.
Task 2
When crossing black mice with each other, black offspring are always obtained. When yellow mice are crossed, one third turns out to be black, and two thirds turn out to be yellow. How can these results be explained?
Decision:
1. Black mice are homozygous, since all their offspring are uniform.
2. Yellow mice are heterozygous, since splitting is observed in their offspring. Since heterozygous individuals carry a dominant trait, the yellow color is dominant.
3. Yellow mice, when crossed with each other, never give only yellow offspring. In addition, the segregation in their offspring differs from Mendelian. This suggests that individuals homozygous for the dominant do not survive.
Crossing scheme:
Task 3
When mirror carps were crossed, splitting was observed already in the first generation: 152 offspring were mirror and 78 with normal scales. How to explain these results? What offspring will result from crossing a mirror carp with an ordinary one?
Decision:
1. Mirror carps are heterozygous, since splitting is observed in their offspring. Since heterozygous individuals carry a dominant trait, the mirror scale dominates.
3. Mirror carps, when crossed with each other, never give only mirror descendants. In addition, the segregation in their offspring differs from Mendelian. This suggests that individuals homozygous for the dominant do not survive. Crossing analysis confirms this assumption.
First cross scheme:
Since carp with mirror scales are heterozygous, and those with normal scales are homozygous for recessive traits, when a mirror carp is crossed with a normal carp, half of the offspring will have mirror scales, and the other half will be normal. Crossing analysis confirms this assumption.
Second cross scheme:
Answer:
The proportion of heredity in the development of schizophrenia is 0.645; the share of the environment in the development of this pathological feature is 0.355.
Task 4
In mice, the short-tailed gene in the dominant homozygous state is lethal, causing the death of the embryo in the early stages of development. Heterozygotes have shorter tails than normal individuals. Determine the phenotypes and genotypes of offspring arising from crossing long-tailed and short-tailed mice.
Decision:
Since the short-tailed mice are heterozygous, the long-tailed mice are homozygous for a recessive trait. Therefore, when crossing short-tailed mice with long-tailed mice, half of the offspring will be short-tailed and half long-tailed, since the heterozygote gives two types of gametes, and the homozygote one. Crossing analysis confirms this assumption.
Crossing scheme:
Answer:
Short-tailed mice - Aa, long-tailed mice - aa.
Task 5
An analysis of the offspring from crossing two fruit flies with curled wings and short bristles showed the presence of different phenotypes in the following ratio:
4 - with twisted. wings shortened with bristles;
2 - with twisted. wings, normal setae;
2 - with norms. wings shortened with bristles;
1 - with norms. wings, normal setae.
How to explain the obtained results? What is the genotype of the original flies?
Decision:
1. Among the descendants, splitting is observed according to both characteristics. This indicates that the crossed individuals were diheterozygous.
2. Splitting for each individual pair of features is carried out in a ratio of 2:1. Deviation from splitting in a ratio of 3:1 indicates that in both cases, individuals homozygous for the dominant trait die.
Crossing analysis confirms this assumption.
Crossing scheme:
Answer:
The results obtained can be explained on the basis of the assumption that in this case there is an independent inheritance of two traits encoded by genes that, in the homozygous state of any dominant gene, cause the death of organisms.
Task 6
In mice, the gene for black body color (A) dominates the gene for brown color (a). These genes are located on the same pair of autosomes. The length of the tail is determined by the B and b genes located on a different pair of chromosomes. Individuals with a normal tail length have the BB genotype, those with a shortened tail have the Bb genotype. Mice with the bb genotype die in the embryonic state. What offspring should be expected from crossing two animals that are diheterozygous for these traits?
Decision:
when crossing diheterozygous individuals with each other, 16 genotypes will be observed in the offspring:
Crossing scheme:
Answer:
The offspring will be black with short tails (AABb and AaBb), black with normal tails (AABB and AaBB), brown with short tails (aaBb) and brown with normal tails (aaBB) in a ratio of 6:3:2:1.
Task 7
In astrakhan sheep, the gene developmental Shirazi trait - gray skin, very expensive fur, dominant in relation to the black color gene, lethal in the homozygous dominant state. In the homozygous state, such genes caused the death of lambs due to their underdevelopment. gastrointestinal tract. Decorating a newborn lamb with thick, beautiful wool, the Shirazi gene killed him.
Specify the genotypes and phenotypes of offspring when gray sheep are crossed with black ones.
Decision:
Given:
A - the gene for the gray color of the fur of karakul-shirazi:
a - gene for black color of karakul fur.
Gray astrakhan sheep are heterozygous (Aa), while black sheep are homozygous (aa). When crossed, they will give viable offspring of gray and black in a ratio of 1: 1.
Crossing analysis confirms this assumption.
Crossing scheme:
Answer:
The gene that causes the development of the Shirazi trait is lethal in the homozygous state, therefore, in the litter of each gray sheep and black ram, 50% of Shirazi and 50% of black lambs are born. All of them are viable.
Task 8
Sickle cell anemia and thalassemia are inherited as two traits with incomplete dominance; The genes are not linked and are located on autosomes. In heterozygotes for sickle cell anemia, as well as in heterozygotes for thalassemia, the disease does not have a pronounced clinical picture. But in all cases, carriers of the gene for thalassemia or sickle cell anemia are resistant to malaria. Double heterozygotes (dihybrids for both pairs of analyzed traits) develop microdrepanocytic anemia.
Homozygotes for sickle cell anemia and thalassemia in the vast majority of cases die in childhood. Determine the probability of having perfectly healthy children in a family where one of the parents is heterozygous for sickle cell anemia, but normal for thalassemia, and the second is heterozygous for thalassemia, but normal for sickle cell anemia.
Decision:
We arrange the condition of the problem in the form of a table:
We determine the genotypes of the parents entering into marriage: Sstt and ssTt.
Crossing scheme:
Answer:
The probability of having perfectly healthy children (sstt) in this family is 25%.
Task 9
A heterozygous bull of the Holstein breed carries a recessive gene that causes hairlessness. In the homozygous state, this gene leads to the death of the calf.
Determine the probability of having a non-viable calf from crossing this bull with one of his daughters from a normal cow.
Decision:
A - a gene that causes the presence of a coat; a - a gene that causes the absence of a coat.
When a heterozygous bull (Aa) is crossed with a cow (AA) homozygous for the dominant gene, all offspring will be normal in phenotype, and 1/2 of the calves will be Aa and 1/2 AA in genotype.
When a heterozygous bull is crossed with a homozygous daughter, all offspring will be normal (1/2 Aa and 1/2 AA. When a heterozygous bull is crossed with a heterozygous daughter, the probability of having a non-viable calf is 1/4 (25%).
Crossing analysis confirms this assumption.
First cross scheme:
Second cross scheme:
Answer:
When crossing a heterozygous bull with a heterozygous heifer, the probability of the birth of a non-viable calf is 25%.
Task 10
In plants homozygous for the chlorophyll mutation, the synthesis of the chlorophyll molecule is impaired. Such plants develop as long as the reserves nutrients in the seed do not dry out, because they are not capable of photosynthesis. Such plants die at an early stage of development.
Determine the probability of the appearance of non-viable offspring when crossing heterozygous plants for this trait.
Decision:
H - gene, normal synthesis of chlorophyll; h - a gene that causes a violation of the synthesis of chlorophyll.
When heterozygous plants (Hh) are crossed with each other, the probability of non-viable offspring is 1/4.
Crossing analysis confirms this assumption.
Crossing scheme:
Answer:
When heterozygous plants (Hh) are crossed with each other, the probability of non-viable offspring is 1/4 (25%).