Continuation. Beginning see No. 4/2004

Solving problems in genetics using the laws of G. Mendel

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 during the interaction of genes ... This allows students to choose the laws, patterns, rules, relationships necessary to solve the problem. To this end, you can give the text of any task and offer to determine which section it belongs to. Students should remember that genes are inherited, not traits.

Exercise 6 One of the breeds of chickens is distinguished by shortened legs (such chickens do not tear gardens). This feature is dominant. Managing it gene causes shortening of the beak at the same time. At the same time, in homozygous chickens, the beak is so small that they are not able to pierce eggshell 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?

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

Answer: both tasks for monohybrid crossing, since we are talking about one gene. ( Key words are highlighted in the text of tasks.)

Task types

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

In the condition calculation problem information must contain:
- about the nature of the trait inheritance (dominant or recessive, autosomal or sex-linked, etc.);
– directly or indirectly (through the phenotype) the genotypes of the parental 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-fingering) 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 the birth of a polydactyl child in this family.

The solution of this problem begins with the recording of its conditions and the designation of genes. Then the (presumably) genotypes of the parents are determined. The husband's genotype is known, the wife's genotype is easy to establish by phenotype - she is a carrier of a recessive trait, which means she is homozygous for the corresponding gene. The next step is writing gamete values. Attention should be paid to the fact that a homozygous organism forms one type of gametes, so the often occurring spelling in this case of two identical gametes does not make sense. A heterozygous organism forms two types of gametes. The connection of gametes is random, so the appearance of two types of zygotes is equally probable: 1:1.

Solution.

R: aa X Ah
gametes: ( a) (A) (a)
F1: Ah,ah,
where: A- polydactyly gene, a- normal gene

Answer: The probability of having a polydactyl child is approximately 50%.

Pay your attention to the inadmissibility of giving an answer in this form: "One child in the family will be born normal and one polydactyl" or even worse: "The first child will be polydactyl, and the second normal." It is impossible to say exactly how many and what kind of children the spouses will have, therefore it is necessary to operate with the concept of probability.
In the condition tasks for determining the genotype should contain information:
- about the nature of the 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, the brown color of the fur dominates over the blue. 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, introducing the notation of genes. We start the solution by drawing up a crossover scheme. The female has a dominant trait. She can be like a homo-( AA), and heterozygous ( Ah). The uncertainty of the genotype is denoted A_. A male with a recessive trait is homozygous for the corresponding gene - aa. The offspring with brown fur color inherited this gene from the mother, and from the father - the blue color gene, therefore, their genotypes are heterozygous. It is impossible to determine the genotype of the mother from the genotype of brown puppies. A blue puppy from each of the parents received a gene for blue color. Therefore, the mother is heterozygous (not purebred).

Solution.

R: aa X aa
gametes: ( A) (a) (a)
F1: 1 aa: 1 aa,
Where: A- gene for brown fur color, a- the gene for the blue color of the fur.

Answer: the female genotype Ah, that is, it is unclean.

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

Task 4. They crossed a motley rooster and a chicken. Received 26 motley, 12 black and 13 white chickens. How is feather color inherited in chickens?

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

Solution.

R: A + A X A + A
variegated variegated
gametes: ( A+) (A) (A+) (A)
F1: 1A + A +: 2 A+A : 1AA
black mottled white

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

Using Illustrated Problems in Lessons

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

Task 5.

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

Answers.

1. Dark coat color dominates.
2. R: AA X aa; F1: aa.
3. We observe manifestations of the rules of trait dominance and uniformity of the first generation.

The drawings may be schematic.

Task 6.

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

Answers.

1. The spherical shape of the fruit dominates.
2. R: aa X AA; F1: Ah; F2: 25% AA, 50% Ah, 25% aa.
3. The laws of uniformity of hybrids of the first generation (I Mendel's law) and the law of splitting (II Mendel's law).

Task 7.

1. What are the genotypes of parents and hybrids F1 if the red color and round shape of tomato fruits are dominant traits, and the yellow color and pear shape are recessive traits?
2. Prove that with such crossing, the law of independent distribution of genes is manifested.

Answers.

1. R: AaBb X aaBb; F1: AaBB, 2AaBb, Aabb, aaBB, 2aaBb, aabb.
2. The inheritance of the color trait of tomato fruits goes regardless of their shape, namely, the ratio of the number of red fruits to yellow ones is equal to:
(37% + 14%) : (37% + 12%) = 1: 1,
and round shape to pear-shaped:
(37% + 37%) : (14% + 12%) = 3: 1.

Solving typical problems using the laws of G. Mendel

monohybrid cross

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

Task 8. Plants of red-fruited strawberries, when crossed with each other, always give offspring with red berries, and plants of white-fruited strawberries - 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? What offspring will you get if you pollinate a red-fruited strawberry with the pollen of a hybrid strawberry with pink berries?

Solution.

When plants with pink berries are crossed with each other, 25% red-fruited, 50% with pink berries and 25% white-fruited are obtained.
Plants with pink berries KB) – hybrids F1. When crossing KB X KB two types of gametes are formed: TO bear the sign of redness and B a sign of whiteness. Using the Punnett lattice, introducing gamete designations, we determine the genotype and phenotype of the resulting plants.

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

A- gene...
a- gene...

But the condition of the problem does not say which feature is dominant, and students experience certain difficulties in writing the condition of the problem.
In my practice, when solving problems on the inheritance of a single trait with incomplete dominance, I use the following symbols:

A+- gene...
A- gene...
(See exercise 2 in No. 4/2004 and problem 4.)

To be continued

At incomplete dominance heterozygotes do not show any of the traits of the parents. At intermediate inheritance hybrids carry an average expression of traits.

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

Task 3-1

When crossing plants of red-fruited strawberries with each other, plants with red berries are always obtained, and white-fruited ones with white ones. As a result of crossing both varieties, pink berries are obtained. What offspring will result from the pollination of red-fruited strawberries with the pollen of a plant with pink berries?

1. Plants with red and white fruits, when crossed with each other, did not give 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 color of fruits). This indicates that in this case the phenomenon of intermediate inheritance is observed.
3. Thus, plants with pink fruits are heterozygous, while those 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.

Task 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 heterozygous organisms have pink flowers. When two plants were crossed, half of the hybrids had pink and half had white flowers. Determine the genotypes and phenotypes of the parents.

Task 3-3

The calyx shape of strawberries can be normal and leaf-shaped. In heterozygotes, the cups are intermediate in shape between normal and leaf-shaped. Determine possible genotypes and phenotypes of offspring from crossing two plants with an intermediate calyx shape.

Task 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 gives white offspring, and crossing dark minks always gives dark ones. What offspring will be obtained from crossing cochinur minks with each other? What offspring will result from crossing kohinoor minks with whites?

Task 3-5

They crossed a motley rooster and a chicken. The result was 26 variegated, 12 black and 13 white chickens. What trait is dominant? How is the color of plumage inherited in this breed of chickens?

Task 3-6

In one Japanese variety of beans, when self-pollinated by a plant grown from a light spotted seed, it was obtained: 1/4 - dark spotted seeds, 1/2 - light spotted and 1/4 - seeds without spots. What offspring will result from crossing a plant with dark spotted seeds with a plant that has seeds without spots?

1. The presence of segregation in the offspring indicates that the original plant was heterozygous.
2. The presence of three classes of phenotypes in the offspring suggests that in this case codominance takes place. Segregation by phenotype in a ratio of 1:2:1 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.

Task 3-7

In cows, the genes for red (R) and white (r) color are codominant to each other. Heterozygous individuals (Rr) - 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?

Task 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 inherited in radishes? What offspring will be obtained from crossing plants with oval and round roots?

Task 3-9

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

Task 100.
Plants of red-fruited strawberries, when crossed with each other, always give offspring with red berries, and plants of white-fruited strawberries - 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? What offspring will be obtained if white-fruited strawberries are pollinated with the pollen of hybrid strawberries with pink berries?
Solution:
A - gene for red color of berries;
a - gene for white color of berries;
AA - homozygous with phenotype - red color of berries;
aa - homozygous with a phenotype - white color of berries;
Aa - heterozygote with a phenotype - pink color of berries.

1. Let's define the offspring that occurs when hybrid strawberry plants with pink berries are crossed with each other

Crossing scheme

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


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

25% (aa) - white color of berries.

With incomplete dominance, when the dominant and pure lines of homozygotes are crossed 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 1 generation.
Thus, with incomplete dominance of the trait, the splitting by phenotype and genotype coincide and is 1:2:1.

2. Let's determine the offspring if we pollinate white-fruited strawberries with pollen from hybrid strawberries with pink berries

Crossing scheme

R: Aa x aa
G: A, a a


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

Since when crossing a known recessive homozygous white-fruited strawberry with a strawberry with pink berries, a 1:1 split occurred in the F 1 offspring (i.e. 50% of individuals will have a dominant and 50% recessive phenotype). Thus, when analyzing crosses1, strawberries with pink berries are heterozygous for dominant and recessive traits.

Incomplete dominance on the basis of feather color in chickens

Task 101.
They crossed a motley rooster and a chicken. The result was 26 blue, 52 motley and 25 white chickens. What trait is dominant? How is the color of plumage inherited in this breed of chickens?
Solution:
The ratio of phenotypes on the basis of plumage is characterized by the following ratio: blue: variegated: white = 1:2:1 [(26/25):(54/25):(25/25) = 1.04:2.16:1], which corresponds to the classical splitting by phenotype when two heterozygotes are crossed (according to the II law of G. Mendel under conditions of incomplete dominance).
A - black color gene;
a - white color gene;
AA - homozygous - black plumage;
aa - homozygous - white plumage;
Aa - heterozygous - 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)
There are three types of genotype. Splitting by genotype - 1:2:1.
Phenotypes:
25% (AA) - black plumage;
50% (Aa) - variegated plumage;
25% (aa) - white plumage.
Three types of phenotype are observed. Splitting by phenotype - 1:2:1.

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

Incomplete color dominance in horses

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

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

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

Since all foals have a phenotype different from their parents - a golden yellow color that has intermediate value color, it can be assumed that the trait manifests itself under conditions of incomplete dominance.

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

Crossing scheme

R: Aa x aa
G: A, a a
F 1: 1Aa:1aa; 50%(Aa):50%(Aa)
There are 2 types of genotype. Splitting by genotype - 1:1.
Phenotypes:

0% - white suit;
50% (aa) - bay suit.
Two types of phenotype are observed. Splitting by phenotype - 1:1.

Crossing shows that the probability of having 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 incompletely dominates the bay color gene.

3. Determination of possible offspring genotypes when golden-yellow horses are crossed

Crossing scheme

R: Aa x Aa
G: A, a A, a
F 1: 1AA:2Aa:1aa; 25%(AA):50%(Aa):25%(Aa)
There are three types of genotype. Splitting by genotype - 1:2:1.
Phenotypes:
25% (AA) - white suit;
50% (Aa) - golden yellow suit;
25% (aa) - bay suit.
Three types of phenotype are observed. Splitting by phenotype - 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 a bay and white color are crossed, all offspring have a golden yellow color, which indicates the inheritance of a trait with incomplete dominance.
2) The probability of the appearance of foals with a white color from a bay mare and a golden yellow stallion is 0%, which indicates the dominance of the trait of a white color over a bay.
3) When golden-yellow horses are crossed among themselves, three types of phenotype are observed in the offspring in the ratio:
white suit: golden yellow suit: bay suit = 1:2:1.
4) The sign of the white color of the horses is incompletely dominant in relation to the sign of the bay color.