Cataract and polydactyly are caused. The main provisions of the chromosome theory of heredity. Linked inheritance of genes

Problem 20. In lovebirds, feather color is determined by two non-allelic genes. The presence in the genotype of two dominant genes (at least one of each allele) determines green color, the presence of a dominant allele in the genotype in combination with recessive alleles of another gene determines the yellow or blue color. Recessive dihomozygotes are white. When crossing green parrots, offspring were obtained from 54 green, 18 yellow, 18 blue and 6 white. Determine the form of gene interaction and genotypes of parents and offspring.

Problem 21. Drosophila Brown color the eye is due to the presence of the dominant allele A in the genotype, bright red - to the dominant allele B, red - to the combined presence of A and B. White is observed in recessive dihomozygotes. Determine the form of gene interaction. What is the ratio of offspring in F 2 obtained by crossing pure allelic flies AAvv, aaBB?

Task 22. When crossing black mice (AAbv) with albinos (aaBB), the entire first generation will have a gray color due to the uneven distribution of the pigment over the hair. Determine the form of gene interaction. What ratios of offspring can be expected in F 2 if mice with the aavb genotype are albino?

Problem 23. When crossing a red-bulb plant with a white-bulb plant in F 1, all individuals had a red color, and in the second generation, splitting of 9/16 red-bulb, 4/16 white-bulb, 3/16 yellow-bulb was observed. Determine the form of gene interaction. What are the genotypes of the parent plants, F1 hybrids and yellow bulbs?

Problem 24. The color of sweet pea flowers is determined by two non-allelic genes. The purple allele is dominant over the white allele for both genes. Determine the form of gene interaction. What will be the offspring from crossing a double dominant homozygote with a double recessive homozygote? What will be the ratio of purple and white flowers in second generation hybrids?

Task 25. In humans, differences in skin color are mainly due to two pairs of genes that interact with each other according to the type of polymer: BBCC - black skin, bbcc - white skin. Any three alleles for black skin result in dark skin, any two alleles for dark skin, and one for light skin. A pair of dark-skinned parents had a white-skinned and a dark-skinned child. Is it possible to determine the genotypes of the parents?

9. Linked inheritance of genes

Example: Cataract A and polydactyly B are caused by dominant autosomal closely linked genes (crossing over is not possible). The woman inherited cataracts from her mother and polydactyly from her father. Her husband is healthy for both signs. What is more likely to be expected in their children: the simultaneous manifestation of cataracts and polydactyly, the absence of these anomalies, or the presence of only one of them?

Decision: It is known that a woman is diheterozygous, and received signs from different parents. This suggests that her A genotype is the mother's chromosome.

and B is the paternal chromosome.

Her husband is healthy, therefore, his aavb genotype is a recessive homozygous.

It is also known that crossing over between these genes is impossible, the genes are closely linked and are transmitted together. Thus, a woman can have only two types of gametes - Av and aB. Let's make a crossover scheme.

	Crossing scheme:
Human cataract and polydactyly
A - cataract	cat., polydact.
a - normal vision
B - polydactyly
c - normal hand
genotypes F1 – ?
Probability?	50% of children only suffer from cataracts			50% of children only have polydactyly

Task 26. Genes A and B are linked and located at a distance of 3.6 morganids. What types of gametes, and in what ratio, are formed in individuals with genotypes:

b) Av

Problem 27. Cataract and polydactyly are caused by dominant autosomal closely linked genes. What offspring can be expected in a family where the husband is healthy and the wife is heterozygous for both traits, if it is known that the wife's mother also suffered from both anomalies, and the father is healthy?

Problem 28. Cataract and polydactyly are caused by dominant autosomal closely linked genes. What offspring can be expected from the marriage of spouses heterozygous for both characteristics, if it is known that their mothers suffered only from cataracts, and their father from polydactyly?

Problem 29. In humans, the Rh factor is linked to the locus that determines the shape of red blood cells, and is located at a distance of 3 morganids from it (crossing over is possible in 3% of cases). Rh-positiveness and elliptocytosis (elliptical erythrocytes) are determined by dominant autosomal genes. One of the spouses is heterozygous for both traits. The second spouse is Rh-negative and has normal red blood cells. Determine the percentages of probable genotypes and phenotypes of children in this family.

Problem 30. A homozygous plant with purple flowers and a short stem was crossed with a homozygous plant with red flowers and a long stem. F 1 had purple flowers and a short stem. When analyzing the crossing of F 1 plants with a double homozygous for recessive genes, the following descendants were obtained:

52 with purple flowers and short stem,

47 with purple flowers and long stem,

49 with red flowers and short stem,

45 with red flowers and long stem.

Determine the frequency of recombination of genes for color and stem length.

1. Genes are located on chromosomes. Each chromosome is a linkage group of genes.

2. Each gene in the chromosome occupies a specific place - a locus. Genes are arranged linearly on chromosomes.

3. Between homologous chromosomes, an exchange of allelic regions between non-sister chromatids (crossing over) can occur.

4. The distance between genes in a chromosome is proportional to the frequency of crossing over between them.

According to the provisions of the chromosome theory of heredity, genes located on one chromosome form clutch group and are inherited in a linked fashion. The number of linkage groups is equal to the haploid number of autosomes plus the number of sex chromosome linkage groups and the mitochondrial DNA linkage group. Thus, 25 linkage groups have been identified in humans: 22 are autosomal, 2 are sex chromosome linkage groups (X and Y) and a linkage group of mitochondrial DNA.

Thus, according to modern concepts, any pair of homologous chromosomes always contains several alleles. Traits whose genes are on the same chromosome are called linked. Naturally, the linked traits are mostly transmitted together. But together linked signs are not always transmitted. The fact is that in the prophase of the first division of meiosis, homologous chromosomes closely approach each other (conjugate). Sometimes at the time of conjugation, a crossover of homologous chromosomes may occur. In the future, when they diverge at the point of intersection, a break in the chromosomes is possible, and then the reunion of the lost sections due to the partner's chromosome. As a result, paired chromosomes exchange homologous regions. The genes of one chromosome, as it were, pass into another, homologous to it. This phenomenon is called crossing over. During the exchange of homologous regions, linked genes diverge into different gametes. The frequency of divergence of traits during crossing over is directly proportional to the distance between genes, i.e. the farther apart the genes are in the chromosome, the more often the exchange occurs, the closer they are located to each other, the less often the divergence of signs.

When linking genes, the number of certain gametes depends on the distance between the genes. This distance is usually calculated in morganides (M). One morganide corresponds to one percent of the formation of gametes in which homologous chromosomes have exchanged their sites. The patterns of inheritance of linked traits are used by geneticists to compile chromosome maps. Experimentally, the frequency of divergence of certain signs is established, i.e. distance between genes. Then the chromosome is drawn, and the relative locations of the loci are marked on it.

Sample problem solving

A task: In man cataract and polydactyly determined by autosomal dominant closely linked alleles of different genes. A woman, normal in both ways, married a man with two anomalies. It is known that he inherited cataracts from his mother, and polydactyly from his father. What is the prognosis for offspring in this family?

Decision:

Let's denote the alleles of genes:

AND– cataract

a- normal vision

AT– polydactyly

b- normal number of fingers

Let's establish the genotypes of the parents: since the woman is phenotypically normal in two ways, her genotype is: aabb. A man has both anomalies, which means that his genotype must have dominant alleles AND and AT. A man inherited one anomaly from his mother, which means that he inherited a normal allele from his father on this basis, i.e. the male is heterozygous for the first trait (genotype Ah). He received the second anomaly (polydactyly) from his father, and from his mother, on this basis, he received the allele of the norm, i.e. he is heterozygous for the second gene: Вb. So his genotype is: AaBb. Since the genes of two traits are linked, the arrangement of alleles in the homologous chromosome in a man should be given:

A b

on the chromosome that he received from his mother, alleles of cataracts are located - AND and the allele of the normal number of fingers - b. The normal vision allele is on the paternal chromosome. a and polydactyly allele AT. Since the genes of the two traits are absolutely linked, in a diheterozygous male with the genotype AaBb two types of gametes are formed: Ab and aB. Let's write down the crossing scheme, write out the gametes of the parents and all possible genotypes in F 1:

P 1: ♀ aabb - ♂ AaBb

It turned out that two genotypes are possible in F 1: Aabb– corresponds in phenotype to a cataract and a normal number of fingers, and aaBb- Corresponds to normal vision and polydactyly. The probability of each of these genotypes is ½ or 50%.

Answer: In this family, the birth of children with only one anomaly is possible: a 50% chance of having children with cataracts and a normal number of fingers, and 50% with normal vision and polydactyly.

Task 2: AaBbDd? Distance between alleles A and b

Decision:

1. Let's write a schematic drawing of the genotype:

A b D

2. G: non-crossover AbD Abd, aBD, aBd,

Crossover abD, abd, ABD, ABd

3. To calculate the ratio of crossover and non-crossover gametes, we take into account that the number of crossover gametes is equal to the distance between alleles. Therefore: 20% - crossover, 80% non-crossover gametes.

Answer: 20% - crossover, 80% non-crossover gametes.

TASKS:

1. What types of gametes and in what ratio forms an individual with a genotype AaBbCs

a) with complete linkage of genes,

b) when genes are located in different pairs of homologous chromosomes? Explain the answer with a drawing.

2. What types of gametes and in what ratio forms an individual with a genotype AaBbCs

a) in the absence of gene linkage,

b) with complete linkage of genes A, b and c. Explain the answer with a drawing.

3. What types of gametes and in what ratio forms an individual with a genotype AaBbDd? Distance between alleles A and b equals 20 morganides, the D allele is on the other chromosome.

4. What types of gametes and in what ratio forms an individual with a genotype AaBBCc if alleles of genes are not linked?

5. What types of gametes and in what ratio forms an individual with a genotype NnCcPpDd? alleles P and D are absolutely linked, the distance between alleles N and With equals 10 morganides.

6. What types of gametes and in what ratio forms an individual with a genotype BbCcKkNn? alleles AT and With K and N equals 25 morganides.

7. What types of gametes and in what ratio are formed in an individual with a genotype MMNNOODdccPp provided that alleles are located in different pairs of homologous chromosomes.

8. What types of gametes and in what ratio forms an individual with a genotype MMNnCcDDEE with complete linkage of genes and their location in different pairs of chromosomes?

9. What types of gametes and in what ratio forms an individual with a genotype CcDd?Alleles C and D absolutely tied.

10. What types of gametes and in what ratio forms an individual with a genotype BbCcEe? alleles AT and FROM absolutely linked, allele E is on a different chromosome.

11. What types of gametes and in what ratio forms an individual with a genotype AaBbCsDd? alleles AND and AT absolutely linked, the distance between alleles With and D equals 40 morganides.

12. What types of gametes and in what ratio forms an individual with a genotype AaBbCsDd? alleles AND and AT absolutely linked, the distance between alleles With and D equals 8 morganides.

13. What types of gametes and in what ratio forms an individual with a genotype СсDdBb if the distance between alleles C and d is 18 morganides, allele B is located on another chromosome.

14. What types of gametes and in what ratio forms an individual with a genotype BbDd?Distance between alleles AT and d equal to 6 morganides?

15. What types of gametes and in what ratio forms an individual with a genotype DdCcВb? Distance between genes D and With equals 16 morganides. allele AT located on a different chromosome.

16. What types of gametes and in what ratio forms an individual with a genotype KkDd? Distance between genes To and d equals 8 morganides.

17. What types of gametes and in what ratio forms an individual with a genotype DdEeNnPp if the distance between alleles D and E is equal to 20 morganides, and alleles n and P absolutely tied.

18. In Drosophila flies, signs of body color and wing shape are linked. The dark coloration of the body is recessive in relation to the gray, short wings - to long ones. In the laboratory, gray long-winged females, heterozygous for both traits, were crossed with males with a black body and short wings. The offspring included 1394 gray long-winged individuals, 1418 black short-winged individuals, 287 black long-winged individuals, and 288 gray short-winged individuals. Determine the distance between the genes.

19. In rats, dark coat color dominates over light, pink color eye - over red. Both traits are linked. In the laboratory, offspring were obtained from crossing pink-eyed dark-haired rats with red-eyed light-haired rats: light red-eyed - 24, dark pink-eyed - 26, light pink-eyed - 24, dark red-eyed - 25. Determine the distance between the genes.

20. Alleles of genes for color blindness and night blindness are inherited through X-chromosome and are located at a distance of 50 morganids from each other (K. Stern, 1965). Both traits are recessive. Determine the probability of having children simultaneously with both anomalies in a family where the wife is heterozygous for both traits and inherited both anomalies from her father, and the husband has both forms of blindness.

21. Classic hemophilia and color blindness are inherited as X-linked recessive traits. The distance between genes is defined as 9.8 morganids. A girl whose father suffers from both hemophilia and color blindness, and whose mother is healthy and comes from a prosperous family for these diseases, marries a healthy man. Determine probable phenotypes children of this marriage.

22. Cataract and polydactyly in humans, they are due to dominant autosomal closely linked (i.e., not showing crossing over) alleles. What offspring can be expected in a family with parents who are heterozygous for both traits, if it is known that the mothers of both spouses suffered only from cataracts, and the fathers only from polydactyly?

23. In humans, the Rh factor locus is linked to the locus that determines the shape of erythrocytes, and is located at a distance of 3 morganids from it (K. Stern, 1965). Rh-positive and elliptocytosis determined by dominant autosomal alleles. One of the spouses is heterozygous for both traits. At the same time, he inherited Rh-positiveness from his father, elliptocytosis from his mother. The second spouse is Rh-negative and has normal red blood cells. Determine the percentages of probable genotypes and phenotypes of children in this family.

24. Recessive alleles a and b in a person cause the presence of diabetes and a tendency to hypertension. Some pedigrees suggest that these genes are linked. What types of gametes are formed in a woman with a genotype AaBb(linked alleles A and B) and in a man with the genotype AaBb(linked A and b)? What is the probability for each of them to pass both diseases to the same child in the offspring at once?

25. Alleles AND and AT belong to the same linkage group, the distance between genes is 40 morganids. Estimate the probability of having children with both dominant alleles in a marriage where both parents are diheterozygous, while the woman received dominant alleles from the father, and the man - one from the father, the other from the mother.

26. Gene alleles color blindness and night blindness are inherited through X-chromosome and are located at a distance of 40 morganids. Determine the probability of having children with both anomalies in a family where the wife has normal vision, although her mother suffered from night blindness, and her father was color blind, and her husband is normal in relation to both signs.

27. A woman received from her mother an autosome with a dominant allele of the Pat gene, which causes a defect in the patella, and with an allele of the gene, which determines the II blood group. From her father, she received the pat allele, which determines the development of a normal patella, and the allele that determines blood type I. The distance between these genes is 10 morganids. Her husband has a normal patella and type I blood. Determine the probability of having a child with the characteristics of the father.

SECTION IV

VARIABILITY

Variation is common property living organisms acquire new traits. Change happens:

1.phenotypic (non-hereditary)

2. genotypic (hereditary)

a) combinative

b) mutational

Modification variability occurs under the influence of environmental factors without changes in the genotype. Due to modification variability, organisms adapt to changing environmental conditions.

Combination variability is the variability in which the combination of parental genes leads to the appearance of new traits in individuals. It is provided by crossing over in prophase I of meiosis, independent divergence of chromosomes in anaphase of meiosis I, and a random combination of gametes with a different set of chromosomes during fertilization.

Mutational variability is due to mutations - a change in the genetic material. Mutations occur under the influence of external or internal environmental factors. The process of formation of mutations is called mutagenesis, and the factors that cause mutations are called mutagenic. According to the nature of the change in the genetic material, mutations are divided into three groups:

1. Gene mutations are changes within a single gene

a) nucleotide insertions or deletions

b) substitution of one nucleotide for another

2. Chromosomal mutations (aberrations) are mutations caused by changes in the structure of chromosomes. They can be intrachromosomal (deletions, duplications, inversions), as well as interchromosomal (translocations)

3. Genomic mutations are mutations caused by a change in the number of chromosomes: polyploidy and heteroploidy (aneuploidy). Polyploidy is a multiple of the haploid increase in the number of chromosomes. Polyploidy in humans is a lethal mutation. Heteroploidy is an increase or decrease in the number of individual chromosomes. In humans, several diseases have been well studied, the cause of which is a change in the number of chromosomes (Down's syndrome, Klinefelter's syndrome, etc.)

Skin color in mulattoes is inherited as a cumulative polymer. At the same time, two autosomal unlinked genes respond to this trait. The son of a white woman and a black man married a white woman. Can a child from this marriage be darker than his father?

GIVEN:

A 1 - melanin A 2 - melanin

a 1 - absence a 2 - absence

DEFINE:

Can a child from this marriage be darker than his father?

DECISION:

1) P: ♀ a 1 a 1 a 2 a 2 x ♂ A 1 A 1 A 2 A 2

White Negro

G: a 1 a 2 A 1 A 2

F 1: A 1 a 1 A 2 a 2

Genotype: all heterozygous

Phenotype: 100%

medium mulattos

2) P: ♀ a 1 a 1 a 2 a 2 x ♂ A 1 a 1 A 2 a 2

G: a 1 a 2 A 1 A 2 ; A 1 a 2 ; a 1 A 2 ; a 1 a 2

F 2: A 1 a 1 A 2 a 2; A 1 a 1 a 2 a 2 ; a 1 a 1 A 2 a 2 ; a 1 a 1 a 2 a 2

Genotype: 1:1:1:1

Phenotype: 1:2:1

medium light white

ANSWER:

If the father is an average mulatto and the mother is white, the son cannot be darker than his father.

THEORETICAL RATIONALE:

This task is on the polymeric interaction of genes. In polymeric inheritance, the development of one trait is controlled by several pairs of non-allelic genes located in different pairs of homologous chromosomes. In this case, there is cumulative (quantitative, accumulative polymerization), in which the quality of a trait depends on the number of dominant genes: the more there are in the genotype, the more pronounced the phenotypic manifestation (in fact, many more genes are responsible for skin color than indicated in the task therefore, the skin color of people is very diverse).

Linked inheritance

No. 1. In humans, the locus of the Rh factor is linked to the locus that determines the shape of erythrocytes, and is located at a distance of 3 morganids from it (K. Stern, 1965). Rh-positiveness and elliptocytosis are determined by dominant autosomal genes. One of the spouses is heterozygous for both traits. At the same time, he inherited Rh-positiveness from one parent, elliptocytosis from the other. The second spouse is Rh-negative and has normal red blood cells. Determine the percentages of probable genotypes and phenotypes of children in this family.

No. 2. Cataract and polydactyly in humans are caused by dominant autosomal closely linked (i.e., not showing crossing over) genes. However, not necessarily the genes of these anomalies, but also the gene for cataract with the gene for the normal structure of the hand, and vice versa, can be linked.

1. A woman inherited cataract from her mother and polydacty from her father. Her husband is normal for both signs. What is more likely to be expected in their children of the simultaneous appearance of cataracts and polydacty, the absence of both of these signs, or the presence of only one anomaly - cataracts or polydactyly.

2. What offspring can be expected in a family where the husband is normal and the wife is heterozygous for both traits, if it is known that the wife's mother also suffered from both anomalies, and her father was normal.

3. What offspring can be expected in a family with parents heterozygous for both traits, if it is known that the mothers of both spouses suffered only from cataracts, and the fathers only from polydactyly.

No. 3. Classical hemophilia and color blindness are inherited as X-linked recessive traits. The distance between genes is definitely 9.8 morganids.

1. A girl whose father suffers from both hemophilia and color blindness, and whose mother is healthy and comes from a family prosperous for these diseases, marries a healthy man. Determine the likely phenotypes of children from this marriage.

2. A woman whose mother was colorblind and her father was hemophilia marries a man with both conditions. Determine the probability of having children in this family at the same time with both anomalies.

No. 4. The gene for color blindness and the gene for night blindness, inherited through the X chromosome, are located at a distance of 50 morganids from each other (K. Stern, 1965). Both traits are recessive.

1. Determine the probability of having children at the same time with both anomalies in a family where the wife has normal vision, but her mother suffered from night blindness, and her father suffered from color blindness, while the husband is normal in relation to both signs.

2. Determine the probability of having children simultaneously with both anomalies in a family where the wife is heterozygous for both traits and inherited both anomalies from her father, and the husband has both forms of blindness.

No. 5. The dominant genes for cataract and elliptocytosis are located in the first autosome. Determine the probable phenotypes and genotypes of children from the marriage of a healthy woman and a diheterozygous man. There is no crossover.

No. 6. The distance between genes C and D is 4.6 morganides. Determine the % of gametes of each type: CD, cd, Cd and cD produced by a diheterozygous organism.

No. 7. The genes that control sickle cell anemia and b-thalassemia in humans are recessive, closely linked on chromosome C. Husband and wife are diheterozygous and inherit both mutant alleles from different parents. Determine the relative likelihood of developing these hereditary diseases for their future children.

No. 8. The genes that control leukodystrophy, methemoglobinemia and one form of ocular albinism in humans are located on chromosome 22. Leukodystrophy and ocular albinism are recessive traits, methemoglobinemia is dominant. Determine the relative probability of development for hereditary diseases in future children of the following couples: a) husband and wife are healthy; the husband inherited the genes of leukodystrophy and ocular albinism from his father, the wife received the same genes from two parents; b) the husband inherited the gene of methemoglobinemia from his father, and the gene of ocular albinism from his mother; the wife is healthy and heterozygous for the genes of albinism and leukodystrophy.

No. 9. The genes that affect the presence of the Rh factor and the shape of red blood cells are located in the same autosome at a distance of 3 morganids. The woman received from her father the dominant gene Rh, which determines Rh-positiveness, and the dominant gene E, which determines the elliptical shape of erythrocytes, and from the mother, the recessive genes for Rh-negativity rh and the normal shape of erythrocytes (e). Her husband is Rh-negative and has a normal form of red blood cells. Determine the probability of the birth of a child phenotypically similar in these signs: a) with the mother; b) with his father.

No. 10. A man received from his father a dominant Rh-positive gene Rh and recessive gene, which determines the normal shape of erythrocytes (e), and from the mother - a recessive Rh-negativity gene rh and a dominant gene E, which causes the formation of elliptical erythrocytes. His wife is Rh-negative, with normal red blood cells. What is the probability (in percent) that the child will be similar to the father in these characteristics?

No. 11. A woman received from her mother an autosome with a dominant gene that causes a defect in the nails and patella, and a gene that determines blood group A. The homologous chromosome contains a recessive gene that does not affect the patella, and a gene for blood type I. The distance between genes is 10 morganids. The husband has a normal patella and no nail defect and homozygous blood group III. Determine the possible phenotypes in the offspring of this family.

Algorithm for solving problem No. 2.

Cataracts and polydactyly in humans are caused by dominant autonomic closely linked (i.e., not showing crossing over) genes. However, the genes of these anomalies may not necessarily be linked, but also the cataract gene with the gene for the normal structure of the hand and vice versa:

a) the woman inherited cataracts from her mother, and
polydactyly - from the father, her husband is normal in relation to both signs.
What is more likely to be expected in her children: the simultaneous manifestation
cataracts and polydactyly, the absence of both of these signs or
the presence of only one anomaly?

b) what offspring can be expected in a family where the husband is normal, and
the wife is heterozygous for both traits, if it is known that the wife's mother also suffered from both anomalies, and her father was normal;

c) what offspring can be expected in the family of parents
heterozygous for both traits, if it is known that the mothers of both
spouses suffered only cataracts, and fathers - polydactyly?

DEFINE:

F 1 - in each family

DECISION:

1. Let's write down the scheme of marriage of a woman who inherited a cataract from her mother, polydactyly from her father with a man who is normal in both respects.

ANSWER:

If a woman inherits cataracts from one of her parents (from her mother), and polydactyly from her father, and her husband is normal in both signs, then from this marriage 50% of children can be with polydactyly and 50% with cataracts.

2. Let's write down the scheme of marriage of a woman, diheterozygous (both anomalies from the mother) with a man, normal in relation to both traits.

ANSWER:

If a woman is heterozygous for both traits, and she inherited both abnormal genes from her mother, and a man is normal for both traits, then 50% of their children from their marriage will have both anomalies (both polydactyly and cataracts), and 50% will have healthy in relation to these pathologies.

3. Let's write down the scheme of marriage of parents who are heterozygous for both traits, given that the mothers of both spouses suffered only from cataracts, and the fathers from polydactyly.

ANSWER:

If the parents are heterozygous for both traits, given that the mothers of both spouses suffered only cataracts, the fathers - polydactyly, then 25% of the children from this marriage will suffer from cataracts, 25% - polydactyly, and 50% of the children will show both anomalies.

THEORETICAL RATIONALE:

The task of Morgan's law of linked inheritance is that genes located on the same chromosome are inherited linked.

Algorithm for solving problem No. 11.

The woman received from her mother an autosome with a dominant gene that causes a defect in the nails and patella, and a gene that determines blood type A. The homologous chromosome contains a recessive gene that does not affect the patella and the nature of the nails, and the I blood group gene. The distance between genes is 10 morganids. The husband has a normal patella and no nail defect and homozygous blood group III. Determine possible phenotypes in the offspring of this family

GIVEN:

A - syndrome of defect of nails and patella

a - norm

i - the first blood group

J A - second blood type

J B - third blood type

J A, J B - fourth blood type

DEFINE:

Phenotypes F 1

DECISION:

Possible clutch groups:

I , J A , J B i , J A , J B

P: ♀ A a x ♂ a a