Continuation. Beginning see No. 4/2004

Solving problems in genetics using G. Mendel's laws

IV. And one more skill, without which it is impossible to solve the problem - determine which section it belongs to : mono-, di- or polyhybrid crossing; sex-linked inheritance, or inheritance of traits through the interaction of genes... This allows students to select the laws, regularities, rules, and relationships necessary to solve the problem. For this purpose, you can give the text of a task and offer to determine which section it belongs to. Students must remember that genes are inherited, not traits.

Exercise 6. One of the chicken breeds is distinguished by shortened legs (such chickens do not tear up vegetable gardens). This sign is dominant. Its manager gene simultaneously causes shortening of the beak. Moreover, homozygous chickens have such a small beak that they are unable to break through eggshells and die without hatching from the egg. The incubator of the farm, which breeds only short-legged chickens, produced 3,000 chickens. How many of them are short-legged?

Exercise 7. In medicine, the distinction between the four human blood groups is of great importance. Blood type is a hereditary trait that depends on one gene. This gene has not two, but three alleles, designated by the symbols A, IN, 0 . Persons with genotype 00 have the first blood group, with genotypes AA or A0- second, BB or B0- third, AB– fourth (we can say that alleles A And IN dominate the allele 0 , whereas they do not suppress each other). What blood types are possible in children if their mother has the second blood group, and their father has the first?

Answer: both problems involve monohybrid crossing, since we're talking about about one gene. ( Key words are highlighted in the task text.)

Types of tasks

All genetic problems, no matter what topic they concern (mono- or polyhybrid crossing, autosomal or sex-linked inheritance, inheritance of mono- or polygenic traits), are reduced to three types: 1) computational; 2) to determine the genotype; 3) to determine the nature of inheritance of a trait.

In condition calculation problem must contain information:
– about the nature of inheritance of a trait (dominant or recessive, autosomal or sex-linked, etc.);
– directly or indirectly (through phenotype) the genotypes of the parent generation must be indicated.
The question of the calculation problem concerns the prediction of the genetic and phenotypic characteristics of the offspring. Let us give an example of a calculation-type problem.

Task 2. In humans, the gene for polydactyly (multi-fingered) dominates the normal structure of the hand. The wife has a normal hand, the husband is heterozygous for the polydactyly gene. Determine the probability of having a multi-fingered child in this family.

The solution to this problem begins with writing down its conditions and designating genes. Then the (presumably) genotypes of the parents are determined. The husband's genotype is known, the wife's genotype can be easily determined by phenotype - she is a carrier of a recessive trait, which means she is homozygous for the corresponding gene. The next stage is writing the meanings of gametes. It should be noted that a homozygous organism produces one type of gametes, therefore the often encountered writing in this case of two identical gametes does not make sense. A heterozygous organism produces two types of gametes. The combination of gametes is random, so the appearance of two types of zygotes is equally probable: 1:1.

Solution.

R: ahh X Ahh
gametes: ( A) (A) (A)
F 1: Ahh,ah,
Where: A– polydactyly gene, A- normal gene.

Answer: The probability of having a multi-fingered child is approximately 50%.

Please note that it is inadmissible to give an answer in the following form: “One child in the family will be born normal and one will be multi-fingered,” or even worse: “The first child will be multi-fingered, and the second will be normal.” It is impossible to say exactly how many and what kind of children the spouses will have, so it is necessary to operate with the concept of probability.
In condition genotype determination tasks must contain information:
– about the nature of inheritance of the trait;
– about the phenotypes of the parents;
– about the genotypes of the offspring (directly or indirectly).
The question of such a task requires characterization of the genotype of one or both parents.

Task 3. In minks, brown fur color dominates over blue fur. A brown female was crossed with a blue male. Among the offspring, two puppies are brown and one is blue. Is the female purebred?

We write down the condition of the problem by introducing gene designations. We begin the solution by drawing up a crossing scheme. The female has a dominant trait. She can be like homo ( AA), and heterozygous ( Ahh). We denote the uncertainty of the genotype A_. A male with a recessive trait is homozygous for the corresponding gene - ahh. Descendants with brown fur color inherited this gene from their mother, and the blue color gene from their father, therefore, their genotypes are heterozygous. It is impossible to determine the genotype of the mother based on the genotype of brown puppies. The blue puppy received the blue gene from each parent. Therefore, the mother is heterozygous (not purebred).

Solution.

R: Aa X aa
gametes: ( A) (A) (A)
F 1: 1 Aa: 1 aa,
Where: A– gene for brown fur color, A– gene for blue fur color.

Answer: female genotype – Ahh, that is, she is not purebred.

In conditions tasks to establish the nature of inheritance sign:
– only phenotypes of successive generations are offered (that is, phenotypes of parents and phenotypes of offspring);
– contains quantitative characteristics of the offspring.
In such a task, it is necessary to establish the nature of inheritance of a trait.

Task 4. Crossed a motley rooster and a hen. We received 26 motley, 12 black and 13 white chickens. How is plumage color inherited in chickens?

When solving this problem, the logic of reasoning may be as follows. Cleavage in the offspring indicates heterozygosity of the parents. A ratio close to 1: 2: 1 indicates heterozygosity for one pair of genes. According to the resulting proportions (1/4 white, 1/2 mottled, 1/4 black), black and white chickens are homozygous, and mottled chickens are heterozygous.
The designation of genes and genotypes followed by drawing up a crossing scheme shows that the conclusion drawn corresponds to the result of the crossing.

Solution.

R: A + A X A + A
variegated variegated
gametes: ( A+) (A) (A+) (A)
F 1: 1A+A+: 2 A+A : 1A.A.
black motley white

Answer: The color of the plumage in chickens is determined by a pair of semi-dominant genes, each of which determines the color white or black, and together they control the development of variegated plumage.

Using illustrated problems in lessons

Problems on genetics can be divided into textual and illustrated. The advantage of illustrated problems over text ones is obvious. It is based on the fact that visual perception of images activates the attention and interest of students, contributes to a better understanding of the conditions of the task and the patterns being studied.

Task 5.

1. What is the dominant coat color in rabbits?
2. What are the genotypes of parents and first-generation hybrids based on coat color?
3. What genetic patterns appear during such hybridization?

Answers.

1. Dark coat color dominates.
2. R: AA X ahh; F 1: Aa.
3. We observe manifestations of the rules of dominance of traits and uniformity of the first generation.

Drawings may be schematic.

Task 6.

1. Which tomato fruit shape (spherical or pear-shaped) is dominant?
2. What are the genotypes of the parents and 1st and 2nd generation hybrids?
3. What genetic patterns discovered by Mendel appear during such hybridization?

Answers.

1. The spherical shape of the fruit dominates.
2. R: ahh X AA; F 1: Ahh; F 2: 25% AA, 50% Ahh, 25% ahh.
3. Laws of uniformity of first generation hybrids (I Mendel’s law) and the law of segregation (II Mendel’s law).

Task 7.

1. What are the genotypes of parents and hybrids F 1, if the red color and round shape of the tomato fruit are dominant traits, and the yellow color and pear-shaped shape are recessive traits?
2. Prove that with such a crossing the law of independent distribution of genes is manifested.

Answers.

1. R: AaBb X aaBb; F 1: AaBB, 2AaBb, Aabb, aaBB, 2aaBb, aabb.
2. Inheritance of the color trait of tomato fruits occurs regardless of their shape, namely the ratio of the number of red to yellow fruits is equal to:
(37% + 14%) : (37% + 12%) = 1: 1,
and round to pear-shaped:
(37% + 37%) : (14% + 12%) = 3: 1.

Solving typical problems using G. Mendel's laws

Monohybrid cross

Solving problems of monohybrid crossing with complete dominance usually does not cause difficulties.
Therefore, we will dwell only on the example of solving problems of inheritance of a separate trait with incomplete dominance.

Task 8. Red-fruited strawberry plants, when crossed with each other, always produce offspring with red berries, and white-fruited strawberry plants - with white berries. As a result of crossing both varieties with each other, pink berries are obtained. What offspring arises when hybrid strawberry plants with pink berries are crossed with each other? What kind of offspring will you get if you pollinate red strawberries with pollen from hybrid strawberries with pink berries?

Solution.

When plants with pink berries are crossed with each other, the result is 25% red-fruited, 50% with pink berries and 25% white-fruited.
Plants with pink berries ( KB) – hybrids F 1. When crossing KB X KB Two types of gametes are formed: TO bear a sign of red fruiting and B a sign of whiteness. Using the Punnett lattice, entering the designations of gametes, we determine the genotype and phenotype of the resulting plants.

Crossbreeding QC X KB gives splitting: 50% QC(red fruits) and 50% KB(with pink berries).
The above solution (see tutorial“Biology for those entering university. Methods for solving problems in genetics." Compiled by Galushkova N.I. - Volgograd. : Ed. Grinin Brothers, 1999, p. 7) in my opinion, creates certain difficulties for students, because Two letter symbols are used to denote one characteristic, which is typical for solving dihybrid crossing problems.
When solving this problem, you can use other variants of letter symbols. For example:

A- gen...
A- gen...

But the task conditions do not say which feature is dominant, and students experience certain difficulties when writing down the task conditions.
In my practice, when solving problems of inheritance of a single trait with incomplete dominance, I use the following symbols:

A+- gen...
A- gen...
(See Exercise 2 in No. 4/2004 and Problem 4.)

To be continued

At incomplete dominance Heterozygotes do not exhibit any of the characteristics present in their parents. At intermediate inheritance hybrids carry average expression of traits.

At co-dominance Heterozygotes exhibit both parental characteristics. An example of intermediate inheritance is the inheritance of the color of strawberries or night beauty flowers, codominance is the inheritance of roan color in cattle.

Problem 3-1

When crossing red-fruited strawberry plants with each other, you always get plants with red berries, and white-fruited ones with white ones. As a result of crossing both varieties, pink berries are obtained. What offspring will be produced when red-fruited strawberries are pollinated with pollen from a plant with pink berries?

1. Plants with red and white fruits, when crossed with each other, did not produce splitting in the offspring. This indicates that they are homozygous.
2. Crossing homozygous individuals that differ in phenotype leads to the formation of a new phenotype in heterozygotes (pink coloring of fruits). This indicates that in this case the phenomenon of intermediate inheritance is observed.
3. Thus, plants with pink fruits are heterozygous, and plants with white and red fruits are homozygous.

Crossing scheme

AA - red fruits, aa - white fruits, Aa - pink fruits.

50% of the plants will have red and 50% pink fruits.

Problem 3-2

In the “night beauty” plant, the inheritance of flower color is carried out according to an intermediate type. Homozygous organisms have red or white flowers, while heterozygotes have pink flowers. When two plants were crossed, half of the hybrids had pink flowers and half had white flowers. Determine the genotypes and phenotypes of the parents.

Problem 3-3

The cup shape of strawberries can be normal or leaf-shaped. In heterozygotes, the calyxes have an intermediate shape between normal and leaf-shaped. Determine the possible genotypes and phenotypes of the offspring from crossing two plants with an intermediate cup shape.

Problem 3-4

Kohinoor minks (light, with a black cross on the back) are obtained by crossing white minks with dark ones. Crossing white minks with each other always produces white offspring, and crossing dark ones always produces dark offspring. What kind of offspring will be obtained from crossing Kohinoor minks with each other? What kind of offspring will come from crossing Kohinoor minks with white ones?

Problem 3-5

Crossed a motley rooster and a hen. As a result, we received 26 motley, 12 black and 13 white chickens. Which trait is dominant? How is the color of the plumage inherited in this breed of chicken?

Problem 3-6

In one Japanese bean variety, self-pollination of a plant grown from a light spotted seed produced: 1/4 dark spotted seeds, 1/2 light spotted seeds and 1/4 seeds without spots. What offspring will result from crossing a plant with dark spotted seeds with a plant with seeds without spots?

1. The presence of segregation in the progeny indicates that the original plant was heterozygous.
2. The presence of three classes of phenotypes in the offspring suggests that codominance occurs in this case. The 1:2:1 phenotypic segregation confirms this assumption.

When a plant with dark spotted seeds is crossed with a plant without spots (both forms are homozygous), all offspring will be uniform and will have light spotted seeds.

Problem 3-7

In cows, the red (R) and white (r) color genes are codominant with each other. Heterozygous individuals (Rr) are roans. The farmer bought a herd of roan cows and decided to keep only them and sell the red and white ones. What color bull should he buy in order to sell as many calves as possible?

Problem 3-8

By crossing radish plants with oval roots, 68 plants with round, 138 with oval and 71 with long roots were obtained. How is the root shape of radishes inherited? What offspring will be obtained from crossing plants with oval and round roots?

Problem 3-9

When strawberries with pink fruits were crossed with each other, the offspring produced 25% of individuals producing white fruits and 25% of plants producing red fruits. The remaining plants had pink fruits. Explain your results. What is the genotype of the individuals examined?

Problem 100.
Red-fruited strawberry plants, when crossed with each other, always produce offspring with red berries, and white-fruited strawberry plants - with white berries. As a result of crossing both varieties with each other, pink berries are obtained. What offspring arises when hybrid strawberry plants with pink berries are crossed with each other? What kind of offspring will you get if you pollinate white-fruited strawberries with pollen from hybrid strawberries with pink berries?
Solution:
A - gene for red color of berries;
a - gene for white berry color;
AA - homozygote with the phenotype - red coloring of berries;
aa - homozygote with the phenotype - white color of berries;
Aa - heterozygote with the phenotype - pink color of berries.

1. Let’s determine the offspring that arises when crossing hybrid strawberry plants with pink berries.

Crossing scheme

R: Aa x Aa
G: A, a, a, a


Phenotypes:
25%(AA) - red color of berries;

25%(aa) - white color of berries.

In case of incomplete dominance, when crossing dominant and pure lines of homozygotes with each other, the appearance of offspring with an intermediate form of the trait is observed; in this case, all the offspring had pink berries. At the same time, uniformity is observed in the F 1st generation.
Thus, with incomplete dominance of a trait, the splitting by phenotype and genotype coincides and is 1:2:1.

2. We will determine the offspring if we pollinate white-fruited strawberries with pollen from hybrid strawberries with pink berries

Crossing scheme

R: Aa x aa
G: Ah, ah, ah


Phenotypes:
50%(Aa) - pink color of berries;
50%(aa) - white color of berries.

Since when crossing the known recessive homozygous white-fruited strawberry with strawberry with pink berries, the F 1 offspring split in a 1:1 ratio (i.e. 50% of individuals will have a dominant and 50% recessive phenotype). Thus, in the analysis crossbreeding1 carried out, strawberries with pink berries are heterozygous for dominant and recessive traits.

Incomplete dominance based on feather color in chickens

Problem 101.
Crossed a motley rooster and a hen. As a result, we received 26 blue, 52 motley and 25 white chickens. Which trait is dominant? How is the color of the plumage inherited in this breed of chicken?
Solution:
The ratio of phenotypes based on plumage is characterized by the following ratio: blue: motley: white = 1:2:1 [(26/25):(54/25):(25/25) = 1.04:2.16:1], which corresponds to the classical cleavage according to phenotype when crossing two heterozygotes (according to G. Mendel’s II law under conditions of incomplete dominance).
A - black color gene;
a - white color gene;
AA - homozygote - black plumage;
aa - homozygote - white plumage;
Aa - heterozygote - variegated plumage color.

Crossing scheme

R: Aa x Aa
G: A, a, a, a
F 1: 1AA:2Aa:1aa; 25%(AA):50%(Aa):25%(aa)
Three types of genotype are observed. Genotype splitting is 1:2:1.
Phenotypes:
25%(AA) - black plumage;
50%(Aa) - variegated plumage color;
25%(aa) - white plumage color.
Three types of phenotype are observed. Phenotype splitting is 1:2:1.

Since the cleavage by genotype and phenotype is the same and is characterized by the ratio: 1:2:1, the color of the plumage in this breed of chickens is of the type of incomplete dominance.

Incomplete dominance based on color in horses

Problem 102
In one small farm, two bay mares produced 8 foals over the course of ten years. All foals were golden yellow in color. Both mares were artificially inseminated with the sperm of a white thoroughbred stallion. When crossing golden-yellow young mares with a bay stallion, white foals never appeared in their offspring. Which trait is dominant? How is color inherited in horses? What is the probability of producing white foals from a golden yellow mare and a bay stallion? What are the possible genotypes when crossing golden yellow horses with each other?
Solution:

1. Determining the type of inheritance of a trait, for this we consider the scheme of crossing a gray mare with a white stallion

R: AA x aa
G: Aa
F 1: 1Aa:1aa;
There is 1 type of genotype.
Phenotypes:
100%(Aa) - golden yellow color;
One type of phenotype is observed. All offspring are golden yellow

Since all foals have a different phenotype from their parents - a golden-yellow color with intermediate value coloration, then we can assume that the trait manifests itself under conditions of incomplete dominance.

2. Determine the probability of the appearance of bay foals from a golden yellow mare and a bay stallion

Crossing scheme

R: Aa x aa
G: Ah, ah, ah
F 1: 1Aa:1aa; 50%(Aa):50%(aa)
There are 2 types of genotype. Genotype splitting is 1:1.
Phenotypes:

0% - white suit;
50%(aa) - bay color.
Two types of phenotype are observed. Phenotype splitting is 1:1.

Crossbreeding shows that the probability of producing white foals from a golden yellow mare and a bay stallion is zero (0%). Since only bay and golden-yellow foals are born in the offspring of a golden-yellow mare and a bay stallion, the white color gene in horses is incompletely dominant over the bay color gene.

3. Determination of possible genotypes of offspring when crossing golden yellow horses with each other

Crossing scheme

R: Aa x Aa
G: A, a, a, a
F 1: 1AA:2Aa:1aa; 25%(AA):50%(Aa):25%(aa)
Three types of genotype are observed. Genotype splitting is 1:2:1.
Phenotypes:
25%(AA) - white suit;
50% (Aa) - golden yellow color;
25%(aa) - bay color.
Three types of phenotype are observed. Phenotype splitting is 1:2:1.

Since the splitting by genotype and phenotype is the same and is characterized by the ratio: 1:2:1, the color of this breed of horses is inherited according to the type of incomplete dominance.

Conclusions:
1) When horses with bay and white colors are crossed with each other, all offspring have a golden-yellow color, which indicates the inheritance of a trait with incomplete dominance.
2) The probability of white foals from a bay mare and a golden yellow stallion is 0%, which indicates the dominance of the white color trait over the bay.
3) When crossing golden-yellow horses with each other, three types of phenotype are observed in the offspring in the ratio:
white color: golden yellow color: bay color = 1:2:1.
4) The trait of the white color of horses is incompletely dominant in relation to the trait of the bay color.