Genetics, its tasks. Heredity and variability are properties of organisms. Basic genetic concepts. Chromosomal theory of heredity. Genotype as an integral system. Development of knowledge about the genotype. Human genome.
Patterns of heredity, their cytological basis. Mono- and dihybrid crossing. Patterns of inheritance established by G. Mendel. Linked inheritance of traits, disruption of gene linkage. T. Morgan's laws. Genetics of sex. Inheritance of sex-linked traits. Gene interaction. Solving genetic problems. Drawing up crossing schemes.
Variability of characteristics in organisms: modification, mutation, combination. Types of mutations and their causes. The meaning of variability in the life of organisms and in evolution. Norm of reaction. The harmful effects of mutagens, alcohol, drugs, nicotine on the genetic apparatus of the cell. Protection of the environment from contamination by mutagens. Identification of sources of mutagens in the environment (indirectly) and assessment of the possible consequences of their influence on one’s own body. Hereditary human diseases, their causes, prevention.
Selection, its tasks and practical significance. Teachings of N.I. Vavilov about the centers of diversity and origin of cultivated plants. The law of homological series in hereditary variability. Methods for breeding new plant varieties, animal breeds, and strains of microorganisms. The importance of genetics for selection. Biological principles of growing cultivated plants and domestic animals.
Biotechnology, cellular and genetic engineering, cloning. The role of cell theory in the formation and development of biotechnology. The importance of biotechnology for the development of breeding, agriculture, microbiological industry, and preservation of the planet’s gene pool. Ethical aspects of the development of some research in biotechnology (human cloning, targeted changes in the genome).
Problem 1
When crossing two varieties of tomato with red spherical and yellow pear-shaped fruits in the first generation, all the fruits are spherical and red. Determine the genotypes of parents, first-generation hybrids, and the ratio of second-generation phenotypes.
Solution:
Since when crossing peas, all offspring individuals have the trait of one of the parents, which means that the genes for red color (A) and the genes for the spherical shape of fruits (B) are dominant in relation to the genes for yellow color (a) and pear-shaped fruits (b). genotypes of the parents: red spherical fruits - AABB, yellow pear-shaped fruits - aabb.
To determine the genotypes of the first generation, the ratio of phenotypes of the second generation, it is necessary to draw up crossing schemes:
First crossing scheme:
The uniformity of the first generation is observed, the genotypes of individuals are AaBb (Mendel's 1st law).
Second crossing scheme:
The ratio of phenotypes of the second generation: 9 – red spherical; 3 – red pear-shaped; 3 - yellow spherical; 1 – yellow pear-shaped.
Answer:
1) genotypes of the parents: red spherical fruits - AABB, yellow pear-shaped fruits - aabb.
2) genotypes F 1: red spherical AaBb.
3) ratio of phenotypes F 2:
9 – red spherical;
3 – red pear-shaped;
3 - yellow spherical;
1 – yellow pear-shaped.
Problem 2
The absence of small molars in humans is inherited as a dominant autosomal trait. Determine the possible genotypes and phenotypes of parents and offspring if one of the spouses has small molars, while the other does not have them and is heterozygous for this trait. What is the likelihood of having children with this anomaly?
Solution:
Analysis of the problem conditions shows that the crossed individuals are analyzed according to one characteristic - molars, which is represented by two alternative manifestations: the presence of molars and the absence of molars. Moreover, it is said that the absence of molars is a dominant trait, and the presence of molars is a recessive trait. This task is on, and to designate alleles it will be enough to take one letter of the alphabet. The dominant allele is denoted by the capital letter A, and the recessive allele by the lowercase letter a.
A - absence of molars;
a - the presence of molars.
Let's write down the genotypes of the parents. We remember that the genotype of an organism includes two alleles of the studied gene “A”. The absence of small molars is a dominant trait, therefore a parent who lacks small molars and is heterozygous means his genotype is Aa. The presence of small molars is a recessive trait, therefore a parent who lacks small molars is homozygous for the recessive gene, which means its genotype is aa.
When a heterozygous organism is crossed with a homozygous recessive organism, two types of offspring are formed, both genotype and phenotype. Crossbreeding analysis confirms this statement.
Crossing scheme
Answer:
1) genotypes and phenotypes P: aa – with small molars, Aa – without small molars;
2) genotypes and phenotypes of the offspring: Aa – without small molars, aa – with small molars; the probability of having children without small molars is 50%.
Problem 3
In humans, the gene for brown eyes (A) dominates over blue eyes, and the gene for color blindness is recessive (color blindness - d) and linked to the X chromosome. A brown-eyed woman with normal vision, whose father had blue eyes and suffered from color blindness, marries a blue-eyed man with normal vision. Make a diagram for solving the problem. Determine the genotypes of the parents and possible offspring, the likelihood of having color-blind children with brown eyes and their gender in this family.
Solution:
Since the woman is brown-eyed, and her father suffered from color blindness and was blue-eyed, she received the recessive blue-eyed gene and the color blindness gene from her father. Consequently, a woman is heterozygous for the eye color gene and is a carrier of the color blindness gene, since she received one X chromosome from a color blind father, her genotype is AaX D X d. Since the man is blue-eyed with normal vision, his genotype will be homozygous for the recessive gene a and the X chromosome will contain a dominant gene for normal vision, his genotype is aaX D Y.
Let's determine the genotypes of possible offspring, the probability of birth in this family of color-blind children with brown eyes and their gender, drawing up a crossing scheme:
Crossing scheme
Answer:
The scheme for solving the problem includes: 1) mother’s genotype – AaX D X d (gametes: AX D, aX D, AX d, aX D), father’s genotype – aaX D Y (gametes: aX D, aY);
2) genotypes of children: girls – AaX D X D, aaX D X D, AaX D X d, aaX D X d; boys – AaX D Y, aaXDY, AaX d Y, aaX D Y;
3) the probability of having color-blind children with brown eyes: 12.5% AaX d Y – boys.
Problem 4
When a pea plant with smooth seeds and tendrils was crossed with a plant with wrinkled seeds without tendrils, the entire generation was uniform and had smooth seeds and tendrils. When crossing another pair of plants with the same phenotypes (peas with smooth seeds and tendrils and peas with wrinkled seeds without tendrils), half of the plants with smooth seeds and tendrils and half of the plants with wrinkled seeds without tendrils were obtained. Make a diagram of each cross.
Determine the genotypes of parents and offspring. Explain your results. How are dominant traits determined in this case? What law of genetics is manifested in this case?
Solution:
This task is for dihybrid crossing, since the crossed organisms are analyzed according to two pairs of alternative characteristics. The first pair of alternative characters: seed shape - smooth seeds and wrinkled seeds; the second pair of alternative characters: the presence of antennae - the absence of antennae. Alleles of two different genes are responsible for these traits. Therefore, to designate alleles of different genes, we will use two letters of the alphabet: “A” and “B”. Genes are located on autosomes, so we will designate them only using these letters, without using the symbols of the X and Y chromosomes.
Since when crossing a pea plant with smooth seeds and tendrils with a plant with wrinkled seeds without tendrils, the entire generation was uniform and had smooth seeds and tendrils, we can conclude that the trait smooth pea seeds and the sign of the absence of tendrils are dominant traits.
And the gene that determines the smooth shape of peas; a - gene that determines the wrinkled shape of peas; B - gene that determines the presence of antennae in peas; b - gene that determines the absence of tendrils in peas. Parental genotypes: AABB, aabb.
First crossing scheme
Since during the 2nd crossing there was a split in two pairs of characters in a 1:1 ratio, we can assume that the genes determining smooth seeds and the presence of tendrils (A, B) are localized on one chromosome and are inherited linked, a plant with smooth seeds and tendrils are heterozygous, which means the genotypes of the parents of the second pair of plants are: AaBb; aabb.
Crossbreeding analysis confirms these arguments.
Second crossing scheme
Answer:
1. The genes that determine smooth seeds and the presence of tendrils are dominant, since during the 1st crossing the entire generation of plants was the same and had smooth seeds and tendrils. Genotypes of the parents: smooth seeds and tendrils - AABB (ametes AB), wrinkled seeds and without tendrils - aabb (ametes - ab). The genotype of the offspring is AaBb. The law of uniformity of the first generation appears when crossing this pair of plants
2. When crossing the second pair of plants, the genes that determine smooth seeds and the presence of tendrils (A, B) are localized on one chromosome and are inherited linked, since during the 2nd crossing, splitting occurred in two pairs of characters in a 1:1 ratio. The law of linked inheritance appears.
Problem 5
The genes for coat color in cats are located on the X chromosome. Black coloring is determined by the X B gene, red coloring is determined by the X b gene, heterozygotes X B X b have a tortoiseshell coloration. From a black cat and a red cat were born: one tortoiseshell and one black kitten. Make a diagram for solving the problem. Determine the genotypes of parents and offspring, the possible sex of kittens.
Solution:
An interesting combination: the genes for black and red colors do not dominate each other, but in combination give a tortoiseshell coloration. Codominance (gene interaction) is observed here. Let's take: X B – the gene responsible for the black color, X b – the gene responsible for the red color; X B and X b genes are equivalent and allelic (X B = X b).
Since a black cat and a red cat were crossed, their gentypes will look like: cat - X B X B (gametes X B), cat - X b Y (gametes X b, Y). With this type of crossing, the birth of black and tortoiseshell kittens is possible in a 1:1 ratio. Crossbreeding analysis confirms this judgment.
Crossing scheme
Answer:
1) genotypes of the parents: cat X B X B (gametes X B), cat - X b Y (gametes X b, Y);
2) genotypes of kittens: tortoiseshell - X B X b, X B X b Y;
3) gender of kittens: female - tortoiseshell, male - black.
When solving the problem, we used the law of gamete purity and sex-linked inheritance. Gene interaction - codominance. The type of crossing is monohybrid.
Problem 6
Diheterozygous male Drosophila flies with a gray body and normal wings (dominant traits) were crossed with females with a black body and shortened wings (recessive traits). Make a diagram for solving the problem. Determine the genotypes of the parents, as well as the possible genotypes and phenotypes of the offspring F 1, if the dominant and recessive genes of these traits are pairwise linked, and crossing over does not occur during the formation of germ cells. Explain your results.
Solution:
The genotype of a diheterozygous male is AaBb, the genotype of a female homozygous for recessive traits is: aabb. Since the genes are linked, the male gives two types of gametes: AB, ab, and the female gives one type of gametes: ab, so the offspring exhibit only two phenotypes in a 1:1 ratio.
Crossbreeding analysis confirms these arguments.
Crossing scheme
Answer:
1) genotypes of the parents: female aabb (gametes: ab), male AaBb (gametes: AB, ab);
2) genotypes of the offspring: 1AaBb gray body, normal wings; 1 aabb black body, shortened wings;
3) since the genes are linked, the male gives two types of gametes: AB, ab, and the female gives one type of gametes: ab, therefore the offspring exhibit only two phenotypes in a 1:1 ratio. The law of concatenated inheritance appears.
Problem 7
Parents with a loose earlobe and a triangular dimple on the chin gave birth to a child with a fused earlobe and a smooth chin. Determine the genotypes of the parents, the first child, and the genotypes and phenotypes of other possible offspring. draw up a diagram for solving the problem. Traits are inherited independently.
Solution:
Given:
Each of the parents has a free earlobe and a triangular fossa and they gave birth to a child with a fused earlobe and a smooth chin, which means that a free earlobe and a triangular chin are dominant traits, and a fused earlobe and a smooth chin are recessive traits. From these considerations we conclude: the parents are diheterozygous, and the child is dihomozygous for recessive traits. Let's create a table of features:
Therefore, the genotypes of the parents: mother AaBb (gametes AB, Ab, Ab, ab), father AaBb (gametes AB, Ab, Ab, ab), genotype of the first child: aabb - fused lobe, smooth chin.
Crossbreeding analysis confirms this judgment.
Phenotypes and genotypes of offspring:
free lobe, triangular fossa, A_B_
loose lobe, smooth chin, A_bb
fused lobe, triangular fossa, aaB_
Answer:
1) genotypes of the parents: mother AaBb (gametes AB, Ab, Ab, ab), father AaBb (gametes AB, Ab, Ab, ab);
2) genotype of the first child: aabb - fused lobe, smooth chin;
3) genotypes and phenotypes of possible descendants:
loose lobe, smooth chin, A_bb;
free lobe, triangular fossa, A_B_;
fused lobe, smooth chin, aabb.
Problem 8
In chickens, a sex-linked lethal gene (a) is found that causes the death of embryos; heterozygotes for this trait are viable. Draw up a scheme for solving the problem, determine the genotypes of the parents, sex, genotype of possible offspring and the probability of embryo death.
Solution:
According to the problem:
X A - development of a normal embryo;
X a - death of the embryo;
X A X a - viable individuals.
Determine the genotypes and phenotypes of the offspring
Crossing scheme
Answer:
1) genotypes of the parents: X A Y (gametes X A, Y), X A X A (gametes X A, X A);
2) genotypes of possible offspring: X A Y, X A X A, X A X a, X a Y;
3) 25% - X and Y are not viable.
luck 9
When a plant with long striped fruits was crossed with a plant with round green fruits, the offspring were plants with long green and round green fruits. When the same watermelon (with long striped fruits) was crossed with a plant that had round striped fruits, all the offspring had round striped fruits. Determine the dominant and recessive traits, genotypes of all parent watermelon plants.
Solution:
A - gene responsible for the formation of a round fruit
a - gene responsible for the formation of a long fruit
B - gene responsible for the formation of green color of the fruit
b - gene responsible for the formation of a striped fetus
Since when crossing a plant with long striped fruits with a plant with round green fruits, the F 1 offspring produced plants with long green and round green fruits, we can conclude that the dominant traits are round green fruits, and the recessive traits are long striped ones. The genotype of a plant with long striped fruits is aabb, and the genotype of a plant with round green fruits is AaBB, because in the offspring all individuals have green fruits, and 1/2 each have round and long fruits, which means that this plant is heterozygous for the dominant trait of shape fetus and homozygous for the dominant trait of fruit color. Genotype of offspring F 1: AaBb, aaBb. Considering that when crossing a parent watermelon with long striped fruits (digomozygous for recessive traits) with a plant having round striped fruits, all F 2 offspring had round striped fruits, the genotype of the parent plant with green striped fruits taken for the second crossing is: AAbb. The genotype of the offspring F 2 is Aabb.
Analyzes of the crosses carried out confirm our assumptions.
First crossing scheme
Second crossing scheme
Answer:
1) dominant characters - fruits are round, green, recessive characters - fruits are long, striped;
2) genotypes of parents F 1: aabb (long striped) and AaBB (round green);
3) genotypes of parents F 2: aabb (long striped) and AAbb (round striped).
Problem 10
A Datura plant with purple flowers (A) and smooth capsules (b) was crossed with a plant with purple flowers and spiny capsules. The following phenotypes were obtained in the offspring: with purple flowers and spiny capsules, with purple flowers and smooth capsules, with white flowers and smooth capsules, with white flowers and spiny capsules. Make a diagram for solving the problem. Determine the genotypes of parents, offspring and possible relationships between phenotypes. Establish the nature of inheritance of traits.
Solution:
And the gene for purple flower color;
a - gene for white flower color;
B - gene that forms the spiny capsule;
b - gene that forms a smooth capsule.
This task is for dihybrid crossing (independent inheritance of traits during dihybrid crossing), since plants are analyzed according to two characteristics: flower color (purple and white) and capsule shape (smooth and spiny). These traits are caused by two different genes. Therefore, to designate genes, we will take two letters of the alphabet: “A” and “B”. Genes are located on autosomes, so we will designate them only using these letters, without using the symbols of the X and Y chromosomes. The genes responsible for the analyzed traits are not linked to each other, so we will use the gene record of the cross.
Purple coloring is a dominant trait (A), and white coloring that appears in the offspring is a recessive trait (a). Each parent has a purple flower, which means they both carry the dominant gene A. Since they have offspring with the aa genotype, each of them must also carry the recessive gene a. Consequently, the genotype of both parental plants for the flower color gene is Aa. The trait spiny capsule is dominant in relation to the trait smooth capsule, and since when crossing a plant with a spiny capsule and a plant with a smooth capsule, offspring with both a spiny capsule and a smooth capsule appeared, the genotype of the parent with the dominant trait for the shape of the capsule will be heterozygous ( Bb), and in recessive - (bb). Then the genotypes of the parents are: Aabb, aaBb.
Now let’s determine the genotypes of the offspring by analyzing the crossing of parental plants:
Crossing scheme
Answer:
1) genotypes of the parents: Aabb (gametes Ab, ab) * AaBb (gametes AB, Ab, aB, ab);
2) genotypes and ratio of phenotypes:
3/8 purple spiny (AABb and AaBb);
3/8 purple smooth (AAbb and Aabb);
1/8 white spiny (aaBb);
1/8 white smooth (aabb);
Problem 11
It is known that Huntington's chorea (A) is a disease that manifests itself after 35-40 years and is accompanied by progressive impairment of brain function, and a positive Rh factor (B) is inherited as unlinked autosomal dominant traits. The father is diheterozygous for these genes, and the mother has a negative Rh factor and is healthy. Draw up a scheme for solving the problem and determine the genotypes of the parents, possible offspring and the probability of having healthy children with a positive Rh factor.
Solution:
And the gene for Huntington's disease;
a - gene for normal brain development;
B - positive Rh factor gene;
b - negative Rh factor gene
This task is for dihybrid crossing (unlinked autosomal dominant inheritance of traits in dihydride crossing). According to the conditions of the problem, the father is diheterozygous, which means his genotype is AaBb. The mother is phenotypically recessive for both traits, which means her genotype is aabb.
Now let’s determine the genotypes of the offspring by analyzing the crossing of parents:
Crossing scheme
Answer:
1) genotypes of the parents: father - AaBb (gametes AB Ab, aB, ab), mother aabb (gametes ab);
2) genotypes of the offspring: AaBb, Aabb, aaBb, aabb;
3) 25% of offspring with the aaBb genotype are Rh positive and healthy.
The sixth building of the Unified State Exam in biology is tasks. For people just starting out in biology, or test prep in particular, they are terrifying. Very in vain. Once you figure it out, everything will become simple and easy. 🙂
Refers to the basic level, with a correct answer you can get 1 primary point.
To successfully complete this task, you should know the following topics given in the codifier:
Topics in the codifier for task No. 6
Genetics, its tasks. Heredity and variability are properties of organisms. Genetics methods. Basic genetic concepts and symbolism. Chromosomal theory of heredity. Modern ideas about the gene and genome
Patterns of heredity, their cytological basis. Patterns of inheritance established by G. Mendel, their cytological basis (mono- and dihybrid crossing). T. Morgan's laws: linked inheritance of traits, disruption of gene linkage. Genetics of sex. Inheritance of sex-linked traits. Gene interaction. Genotype as an integral system. Human genetics. Methods for studying human genetics. Solving genetic problems. Drawing up crossing schemes.
“Solve the Unified State Exam” divides tasks into two large groups: monohybrid crossing and dihybrid crossing.
Before solving problems, we suggest compiling a small dictionary of terms and concepts in order to understand what is required of us.
Theory for crossing tasks
Traits are divided into two types: recessive and dominant.
« A dominant trait suppresses a recessive one" is a stable phrase. What does suppress mean? This means that in the choice between a dominant and recessive trait, the dominant one will necessarily appear. Anyway. A dominant trait is indicated by a capital letter, and a recessive trait is indicated by a small letter. Everything is logical. In order for a recessive trait to appear in the offspring, it is necessary that the gene carries the recessive trait from both the female and the male.
For clarity: let’s imagine a sign, for example, the color of a kitten’s fur. Let us have two options for the development of events:
- Black wool
- White wool
Black wool is dominant over white. In general, tasks always indicate what dominates what; applicants are not required to know everything, especially about genetics.
Black wool will then be indicated by a capital letter. The most commonly used are A, B, C and further in alphabetical order. White wool, respectively, in small letters.
A - black wool.
a- white wool.
If the fusion of gametes results in the following combinations: AA, Aa, aA, then this means that the fur of the descendants of the first generation will be black.
If the fusion of gametes results in the combination aa, then the wool will be white.
What kind of gametes the parents have will be stated in the task conditions.
Gametes, or germ cells, are reproductive cells that have a haploid (single) set of chromosomes and participate, in particular, in sexual reproduction.
Zygote- a diploid cell formed as a result of fertilization.
Heterozygote - two genes that determine one trait - identical (AA or aa)
Homozygote - two genes that determine one trait are different (Aa)
Dihybrid cross- crossing of organisms that differ in two pairs of alternative characteristics.
Monohybrid cross- crossing, in which the crossed organisms differ in only one characteristic.
Analysis cross- crossing a hybrid individual with an individual homozygous for recessive alleles.
Gregor Mendel - the “father” of genetics
So, how to distinguish these types of crossing:
When crossing a monohybrid, we are talking about one trait: color, size, shape.
In a dihybrid cross we are talking about a pair of traits.
In an analytical cross, one individual can be absolutely anything, but the other’s gametes must carry exclusively recessive traits.
Alleles- different forms of the same gene, located in the same regions of homologous chromosomes.
It doesn't sound very clear. Let's figure it out:
1 gene carries 1 trait.
1 allele carries one trait value (it can be dominant or recessive).
Genotype- the totality of genes of a given organism.
Phenotype- a set of characteristics inherent in an individual at a certain stage of development.
Problems often ask you to indicate the percentage of individuals with a certain genotype or phenotype or to indicate the breakdown by genotype or phenotype. If we simplify the definition of phenotype, then phenotype is the external manifestation of characteristics from the genotype.
In addition to all sorts of concepts, you need to know the laws of Gregor Mendel, the father of genetics.
Gregor Mendel crossed peas with fruits that differed in color and smoothness of the skin. Thanks to his observations, the three laws of genetics emerged:
I. Law of uniformity of first generation hybrids:
In a monohybrid crossing of different homozygotes, all descendants of the first generation will be identical in phenotype.
II. Law of splitting
When crossing descendants of the first generation, a splitting of 3:1 in phenotype and 1:2:1 in genotype is observed.
III. Law of independent cleavage
When a dihybrid crossing of two different homozygotes occurs in the second generation, phenotypic cleavage is observed in a ratio of 9:3:3:1.
When the skill of solving genetic problems is acquired, the question may arise: why do I need to know Mendel’s laws if I can already solve the problem perfectly well and find splitting in particular cases? Attention answer: in some tasks you may need to indicate by what law the splitting occurred, but this applies more to tasks with a detailed answer.
Having gained some grounding in theory, you can finally move on to the tasks. 😉
Analysis of typical tasks No. 6 of the Unified State Exam in biology
Types of gametes in an individual
How many types of gametes are produced in an individual with the aabb genotype?
We have two pairs of allelic chromosomes:
First pair: aa
Second pair: bb
These are all homozygotes. You can make only one combination: ab.
Types of gametes in crossing
How many types of gametes are formed in diheterozygous pea plants during dihybrid crossing (the genes do not form a linkage group)? Write down the number in response.
Since plants are diheterozygous, this means that for both traits, one allele is dominant and the other is recessive.
We obtain genotypes AaBb and AaBb.
Gametes in problems are designated by the letter G, without commas, in circles; the gametes of one individual are indicated first, then a semicolon (;) is placed, and the gametes of another individual are written, also in circles.
Crossing is indicated by an "x".
Let's write out the gametes; to do this, we'll go through all the combinations:
The gametes of the first and second individuals turned out to be the same, so their genotype was also the same. This means we have 4 different types of gametes:
Calculation of the proportion of diheterozygotes
When crossing individuals with genotypes AaBb with AaBb (genes are not linked), the proportion (%) of heterozygotes for both alleles (diheterozygotes) in the offspring will be….
Let's create a Punnett lattice. To do this, we write out the gametes of one individual in a column, the gametes of another in a row, and we get a table:
Let's find diheterozygotes in the table:
Total zygotes: 16
Diheterozygotes:4
Let's calculate the percentage: =
Application of Mendel's laws
The rule of uniformity of the first generation will appear if the genotype of one of the parents is aabb, and the other is
According to the rule of uniformity, monohybrid homozygotes must be crossed, one with a dominant trait, and the other with a recessive trait. This means that the genotype of the other individual must be AABB.
Answer: AABB.
Phenotype ratio
The genotype of one of the parents will be AaBb if, during an analyzing dihybrid crossing and independent inheritance of traits, a split in phenotype is observed in the offspring in the ratio. Write down the answer as a sequence of numbers showing the ratio of the resulting phenotypes, in descending order.
Analyzing dihybrid cross, which means that the second individual has a recessive dihomozygote: aabb.
Here you can do without a Punnett grid.
Generations are designated by the letter F.
F1: AaBb; Aabb; aaBb; aabb
All four variants of phenotypes are different, so they relate to each other as 1:1:1:1.
Answer: 1111
Genotype ratio
What is the ratio of genotypes in the offspring obtained from crossing individuals with genotypes AaBb x AABB?
AaBb x AABB
F1: AaBb; Aabb; aaBb; aabb
All 4 genotypes are different.
Inheritance of certain traits or diseases
What is the probability of having healthy boys in a family where the mother is healthy and the father is sick with hypertrichosis, a disease caused by the presence of a gene linked to the Y chromosome?
If a trait is linked to the Y chromosome, it means that it is not reflected in any way on the X chromosome.
The female sex is homozygous: XX, and the male is heterozygous: XY.
Solving problems with sex chromosomes is practically no different from solving problems with autosomes.
Let's make a table of genes and traits, which should also be compiled for problems about autosomal chromosomes, if the traits are indicated and this is important.
The letter above the Y indicates that the gene is linked to that chromosome. Traits can be dominant or recessive, they are designated by capital and small letters, they can relate to both the H chromosome and the Y chromosome, depending on the task.
♀ХХ x ХY a
F1: XX-girl, healthy
XY a - boy, sick
The boys born to this couple will be 100% sick, which means 0% healthy.
Blood groups
What ABO blood group does a person with genotype I B I 0 have? Write down the number in response.
Let's use the table: 
Our genotype contains agglutinogens B and 0. This pair gives the third blood group.
Working with the circuit
Using the pedigree shown in the figure, determine the probability (in percentage) of the birth of parents 1 and 2 of a child with the trait indicated in black, with complete dominance of this trait. Write your answer as a number.
So, let's learn to analyze such schemes.
We see that the trait appears in both men and women, which means it is not gender-linked.
It appears in every generation, which means it is dominant.
If one of the couple’s children does not display the trait, it means the parents are heterozygotes.
F1: AA- appears
Aa- manifests itself
Aa- manifests itself
aa - does not appear
3 - appears from 4
Municipal state educational institution Lyceum No. 4
Rossoshi city, Rossoshansky municipal district, Voronezh region.
Methodological development in biology to help students taking the Unified State Exam.
“Genetics problems” for 11th grade
biology teacher of the highest qualification category
2016
I. Problems for monohybrid crossing (complete and incomplete dominance)
1. In rabbits, gray coat color dominates over black coat color. A homozygous gray rabbit was crossed with a black rabbit. Determine the phenotypes and genotypes of rabbits?
2. In guinea pigs, black coat color dominates over white coat color. Two heterozygous males and a female were crossed. What will the first generation hybrids be like? What law is manifested in this inheritance? 
3. When crossing two white pumpkins in the first generation, ¾ of the plants were white and ¼ were yellow. What are the genotypes of the parents if white color is dominant over yellow? 
4. The black cow Nochka brought a red calf. The red cow Zorka gave birth to a black calf. These cows are from the same herd, which contains one bull. What are the genotypes of all animals? Consider different options. ( The gene for black coloring is dominant.)
5. How many dwarf pea plants can be expected when sowing 1200 seeds obtained by self-pollinating tall heterozygous pea plants? (Seed germination is 80%).
6. Watermelon fruits may be green or striped in color. All watermelons obtained from crossing plants with green and striped fruits had only a green fruit rind. What color can watermelon fruits be in F2?
7. In snapdragon, plants with wide leaves, when crossed with each other, always produce offspring with the same leaves, and when a narrow-leaved plant is crossed with a broad-leaved plant, plants with leaves of intermediate width appear. What will be the offspring from crossing two individuals with leaves of intermediate width. 
1. In humans, glaucoma is inherited as an autosomal recessive trait (a), and Marfan syndrome, accompanied by an anomaly in the development of connective tissue, is inherited as an autosomal dominant trait (B). The genes are located in different pairs of autosomes. One of the spouses suffers from glaucoma and had no ancestors with Marfan syndrome, and the second is diheterozygous for these characteristics. Determine the genotypes of the parents, possible genotypes and phenotypes of the children, and the probability of having a healthy child. Make a diagram for solving the problem. What law of heredity is manifested in this case?
2. Crossed short (dwarf) tomato plants with ribbed fruits and plants of normal height with smooth fruits. The offspring produced two phenotypic groups of plants: low-growing with smooth fruits and normal height with smooth fruits. When low-growing tomato plants with ribbed fruits were crossed with plants that had normal stem height and ribbed fruits, all offspring had normal stem height and ribbed fruits. Make crossbreeding schemes. Determine the genotypes of the parents and offspring of tomato plants in two crosses. What law of heredity is manifested in this case? 
3. When crossing horned red cows with polled black bulls, calves of two phenotypic groups were born: horned black and polled black. When these same horned red cows were further crossed with other polled black bulls, the offspring were polled red and polled black. Write schemes for solving the problem. Determine the genotypes of parents and offspring in two crosses. What law of heredity is manifested in this case? 
4. Tomato fruits can be red and yellow, bare and pubescent. It is known that the genes for yellow color and hairiness are recessive. Of the tomatoes harvested on the collective farm, there were 36 tons of red hairless tomatoes and 12 tons of red hairy ones. How many (approximately) yellow fluffy tomatoes can there be in a collective farm harvest if the starting material was heterozygous?
5. When crossing a pied crested (B) hen with the same rooster, eight chickens were obtained: four pied crested chickens, two white (a) crested chickens and two black crested chickens. Make a diagram for solving the problem. Determine the genotypes of parents and offspring, explain the nature of inheritance of traits and the appearance of individuals with variegated colors. What laws of heredity are manifested in this case? 
6. In mice, black coat color dominates over brown coat color (a). The long tail is determined by the dominant gene (B) and develops only in the homozygous state; heterozygotes develop a short tail. Recessive genes that determine tail length in a homozygous state cause the death of embryos. The genes for the two traits are not linked. When a female mouse with black fur and a short tail is crossed with a male mouse with brown fur and a short tail, the following offspring are obtained: black mice with a long tail, black mice with a short tail, brown mice with a long tail and brown mice with a short tail. Make a diagram for solving the problem. Determine the genotypes of parents and offspring, the ratio of phenotypes and genotypes of offspring, the probability of death of embryos. What law of heredity is manifested in this case? Justify your answer. 
7. The baleen-bearded, white-fruited strawberry plant was crossed with the moustached, red-fruited plant (B). All hybrids turned out to be pink-fruited with whiskers. During the analytical crossing of F 1 hybrids, phenotypic cleavage occurred in the offspring. Make a diagram for solving the problem. Determine the genotypes of parental individuals, first-generation hybrids, as well as the genotypes and phenotypes of offspring during analytical crossing (F 2). Determine the nature of inheritance of the fruit color trait. What laws of heredity are manifested in these cases? 
8. When crossing phlox plants with white flowers and a funnel-shaped corolla with a plant with cream flowers and flat corollas, 78 descendants were obtained, among which 38 produced white flowers with flat corollas, and 40 produced cream flowers with flat corollas. By crossing phloxes with white flowers and funnel-shaped corollas with a plant with cream flowers and flat corollas, phloxes of two phenotypic groups were obtained: white with funnel-shaped corollas and white with flat corollas. Draw up diagrams of two crosses. Determine the genotypes of parents and offspring in two crosses. What law of heredity is manifested in this case?
9. There are two types of hereditary blindness, each of which is determined by recessive gene alleys (a or b). Both alleles are found on different pairs of homologous chromosomes. What is the probability of having a blind grandson in a family in which the maternal and paternal grandmothers are dihomozygous and suffer from various types of blindness, and both grandfathers see well (do not have recessive genes). Make a diagram for solving the problem. Determine the genotypes and phenotypes of grandparents, their children and possible grandchildren. 
III. Problems related to the inheritance of blood groups of the AB0 system
The boy has group I, his sister has group IV. What are the genotypes of the parents?
The father has blood type IV, the mother has blood type I. Can a child inherit his mother’s blood type?
Two children were mixed up in the maternity hospital. The first pair of parents have blood groups I and II, the second pair – II and IV. One child has group II, and the second has group I. Identify the parents of both children.
Is it possible to transfuse blood to a child from the mother if she has blood type AB and the father has 00? Explain your answer.
5. Blood type and Rh factor are autosomal unlinked traits. Blood group is controlled by three alleles of one gene - i 0, I A, I B. Alleles I A and I B are dominant in relation to allele i 0. The first group (0) is determined by recessive genes i 0, the second group (A) is determined by the dominant allele I A, the third group (B) is determined by the dominant allele I B, and the fourth (AB) is determined by two dominant allele I A I B. Positive Rh factor R dominates negative r. The father has the third blood group and positive Rh (diheterozygote), the mother has the second group and positive Rh (dihomozygote). Determine the genotypes of the parents. What blood type and Rh factor can the children in this family have, what are their possible genotypes and the ratio of phenotypes? Make a diagram for solving the problem. What law of heredity is manifested in this case?
IV
1. In canaries, the presence of a crest is a dominant autosomal trait (A); The sex-linked gene X B determines the green color of the plumage, and X b - the brown color. In birds, the homogametic sex is male and the heterogametic sex is female. A tufted green female was crossed with a male without a tuft and green plumage (heterozygote). The offspring included green crested chicks, green crestless chicks, brown crested chicks and brown chicks without crested chicks. Make a diagram for solving the problem. Determine the genotypes of parents and offspring, their gender. What laws of heredity are manifested in this case?
2. Drosophila body color is determined by an autosomal gene. The gene for eye color is located on the X chromosome. The male sex is heterogametic in Drosophila. A female with a gray body and red eyes was crossed with a male with a black body and white eyes. All offspring had a gray body and red eyes. The resulting F1 males were crossed with the parent female. Make a diagram for solving the problem. Determine the genotypes of parents and females of F 1, genotypes and phenotypes of offspring in F 2. What proportion of females from the total number of offspring in the second cross is phenotypically similar to the parent female? Indicate their genotypes. 
3. In humans, the gene for brown eyes dominates over blue eyes (A), and the gene for color blindness is recessive (color blindness - d) and linked to the X chromosome. A brown-eyed woman with normal vision, whose father had blue eyes and suffered from color blindness, marries a blue-eyed man with normal vision. Make a diagram for solving the problem. Determine the genotypes of the parents and possible offspring, the likelihood of having color-blind children with brown eyes and their gender in this family. Explain your answer.
4. In humans, the inheritance of albinism is not sex-linked (A - the presence of melanin in skin cells, a - albinism), and hemophilia is sex-linked (X n - normal blood clotting, X h - hemophilia). Determine the genotypes of the parents, as well as the possible genotypes, sex and phenotypes of children from the marriage of a woman diheterozygous for both alleles and an albino man with normal blood clotting. Make a diagram for solving the problem. Explain your answer.
1. When corn plants with smooth, colored kernels were crossed with a plant that produced wrinkled, uncolored seeds, in the first generation all plants produced smooth, colored kernels. When analyzing the crossing of hybrids from F 1, there were four phenotypic groups in the offspring: 1200 smooth colored, 1215 wrinkled uncolored, 309 smooth uncolored, 315 wrinkled colored. Make a diagram for solving the problem. Determine the genotypes of parents and offspring in two crosses. Explain the formation of four phenotypic groups in the second cross.
2. When crossing a diheterozygous corn plant with a colored seed and starchy endosperm and a plant with an uncolored seed and waxy endosperm, the offspring resulted in splitting according to phenotype: 9 plants with colored seeds and starchy endosperm; 42 – with colored seed and waxy endosperm; 44 – with uncolored seed and starchy endosperm; 10 – with uncolored seed and waxy endosperm. Make a diagram for solving the problem. Determine the genotypes of the original individuals and the genotypes of the offspring. Explain the formation of four phenotypic groups. 
3. A diheterozygous pea plant with smooth seeds and tendrils was crossed with a plant with wrinkled seeds without tendrils. It is known that both dominant genes (smooth seeds and the presence of tendrils) are localized on the same chromosome; crossing over does not occur. Make a diagram for solving the problem. Determine the genotypes of the parents, phenotypes and genotypes of the offspring, the ratio of individuals with different genotypes and phenotypes. What law is manifested in this case? 
VI. Problems on pedigrees
1. Based on the pedigree shown in the figure, determine and explain the nature of inheritance of the trait highlighted in black (dominant or recessive, sex-linked or not. Determine the genotypes of offspring 1,3,4,5,6,7. Determine the probability of birth from parents 1 ,2 next children with the trait highlighted in black on the pedigree. 
2. Based on the pedigree shown in the figure, determine and justify the genotypes of the parents and descendants, indicated in the diagram by numbers 1,6,7. Establish the probability of having a child with hereditary trait in a woman number 6, if this trait has never appeared in her husband’s family.
Based on the pedigree shown in the figure, determine and explain the nature of inheritance of the trait highlighted in black. Determine the genotypes of parents and offspring1,6 and explain the formation of their genotypes.
Using the pedigree presented in the figure, determine the nature of inheritance of the trait (dominant or recessive, sex-linked or not), highlighted in black, and the genotypes of parents and children in the first generation. Indicate which of them is the carrier of the gene whose characteristic is highlighted in black.
Gene interaction problems
1. complementarity
In parrots, feather color is determined by two pairs of genes. The combination of two dominant genes determines the color green. Individuals that are recessive for both pairs of genes are white. The combination of the dominant gene A and the recessive gene b determines the yellow color, and the combination of the recessive gene a with the dominant gene B determines the blue color. 
When two green individuals were crossed with each other, parrots of all colors were obtained. Determine the genotypes of parents and offspring.
2. epistasis
The coloration of a rabbit's fur (as opposed to albinism) is determined by a dominant gene. Color color is controlled by another gene located on another chromosome, with gray color dominating over black (in albino rabbits, color color genes do not manifest themselves). 
What characteristics will the hybrid forms obtained from crossing a gray rabbit born from an albino rabbit with an albino carrying the black color gene have?
3. pleiotropy.
One of the chicken breeds is distinguished by shortened legs. This sign is dominant. The gene that controls it simultaneously causes the beak to shorten. At the same time, homozygous chickens have such a small beak that they are unable to break through the eggshell 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?
4. polymer
The son of a white woman and a black man married a white woman. Could this couple have a child darker than the father?
ANSWERS:
I. Problems for monohybrid crossing (complete and incomplete dominance)
1. A - gray color
a- black color R: AA and aa.
F 1: Ah, all gray.
2. A - black flower color and - white color
R: Aa and Aa.
F 1: AA 2Aa, aa. The law of splitting.
3. A - white color
a-yellow color
Both parents have the Aa genotype.
F 1 segregation according to genotype 1aa 2Aa 1 aa
4.Nochka-Aa, her calf-aa. Dawn - ah, her calf - ah, bull - ah.
5. A- high
a-dwarf
F 1 1АА 2Аа 1аа
240 dwarf plants.
6.A - green color
a- striped color
F2 1АА 2Аа 1аа
F 2 - split 3 green: 1 striped
7. A- wide leaves
a- narrow leaves
Aa - intermediate leaf width
F 1: splitting according to phenotype: 1 - wide leaves, 2 - intermediate leaf width, 1 - narrow leaves.
By genotype: AA 2Aa aa
II. Dihybrid crossing problems1. A – normal, a – glaucoma.
B – Marfan syndrome, b – normal.
One of the spouses suffers from glaucoma and had no ancestors with Marfan syndrome: aabb. The second spouse is diheterozygous: AaBb.
| norm. | norm | glaucoma | glaucoma |
Probability of having a healthy baby (normal/normal) = 1/4 (25%). In this case, Mendel's third law (the law of independent inheritance) appears.
a- dwarfism
B- smooth
v-ribbed
first crossing - P: aabb and AaBB, received F 1 - aaBb and AaBb
second - P: aabb and AAbb, received F 1 - Aabb.
4.P-AaBv and aavv.F1: 9 cr. goals.. 3 cr. Op., 3 female heads, 1 female op.4 tons w. op.
5. In this case, intermediate inheritance of color appears. AA - black, Aa - motley, aa - white. the parents, both the hen and the rooster, have AaBB genotypes. And the gametes form the same: AB, aB. when they merge, the formation of genotypes occurs: AABB - black crested, AaBB - motley crested, aaBB - white crested. ratio -1/2/1.
color gene:
A - black
a- brown
tail length gene:
B- long
c- short
bb - letal
BB - shortened
Solution:
1) AaBv x aaBv
black, short x brown, short
gametes - AB, Av, aV, av aV av
AaBB AaBv aaBB aaBv AaBv Aavv aaBv aavv
h.d.h.u. k.d. to u. h. y letal k. y letal
black with long tail - 1/8
black with shortened - 2/8
brown with long tail - 1/8
brown with shortened - 2/
lethal - 2/8. Law of independent inheritance
7. A- mustachioed
a- mustacheless
B- red
BB- pink
1) first crossing:
R AAbb * aaBB
mustache white used cr.
2) analyzing crossing:
AaVv * aavv
G AB Av aV av av
F 2 AaBb - moustached pink-fruited; Aabb - whiskered white-fruited;
aaBb - beardless pink-fruited; aabb - beardless white-fruited;
3) the nature of inheritance of the fruit color trait is incomplete dominance. In the first crossing - the law of uniformity of hybrids, in the second (analyzing) - the independent inheritance of characteristics.
8. A - flat rims,
a - funnel-shaped corollas.
B - white flowers,
b - cream flowers
first crossing:
R aaVv x Aavv
G aw aw aw aw
F 1 AaBb aavv
second crossing:
Raat
G aB Av aw
F 1 AaBb aaBb
In this case, Medel’s third law is manifested - the law of independent inheritance.
9. A is the norm, a is blindness No. 1.
B - normal, b - blindness No. 2.
Maternal grandmother is AAbb, paternal grandmother is aaBB. Grandfathers - AABB.
| |
The probability of having a blind grandson is 0%
Problems related to the inheritance of blood groups of the AB0 system
1. Boy-j0j0. Sister-JАJВ
P J A j 0 and J A J B
2. Father - J A J B
Mother-j 0 j 0.
No, because Children can have either 2nd or 3rd blood groups.
3. first pair of parents:
P: j 0 j 0 x J A j 0 or j 0 j 0 x J A J A
G: j 0 J A , j 0 j0 J A
F: J A j0 (2 g), j 0 j 0 (1 g) or J A j 0 (2 g)
second pair of parents
R: J A J A x J A J B or J A j 0 x J A J B
G: J A ; J A , J B J A j 0 J A , J B
F: J A J A (2) J A J B (4) J A J A (2) J A J B (4) J A j 0(2) J in j 0(3 gr.)
The first pair of parents has a son with 1 gram. and he received gene 0 from both parents. The second couple are the parents of a boy with blood type 2.
This problem can be solved orally, because a child with 1 gr. blood cannot be born to a couple that has a person with blood type 4
4. It is impossible, because Children may have blood groups: A0 (II) or B0 (III), therefore blood of the fourth group that the mother has cannot be transfused.
5. Diheterozygous father I B i 0 Rr, digomozygous mother I A I A RR.
| IV group | IV group | Group II | Group II |
Children in this family may have blood group IV or II, all are Rh positive. The proportion of children with blood group IV is 2/4 (50%). The law of independent inheritance appears (Mendel's third law).
IV. Problems with sex-linked and autosomal inheritance
1. A – the presence of a crest, and – no crest.
X B – green plumage, X b – brown plumage.
A_X B Y-crested green female
ааX B X b- male without crest with green plumage (heterozygote)
Among the offspring there were chicks without a crest - aa. They received one gene a from their mother, one from their father. Therefore, the mother must have the gene a, hence the mother is Aa.
P АаX В Y x ааX B X b
| AaX B X B | aaX B X B | AaX B Y | aaX B Y |
|
| AaX B X b | aaX B X b | AaX b Y | aaX b Y |
In this case, the law of independent inheritance (Mendel's third law) and sex-linked inheritance appeared.
2. A - gray body
a- black body
X B red eyes
X in - white eyes
P 1 AAX B X B * aaX in Y
gray body black body
red eyes white eyes
G AX B aX to aY
F 1 AaХ B X b AaХ B Y
P 2 AAX B X B * AaX B Y
G AX B AX B AX B AY aY
F 2 AAX B X B Aax B X B AAX B Y Aax B Y
F 2 all offspring have a gray body and red eyes.
Gender ratio: 50%: 50:%
3. A – brown eyes,
a – blue eyes.
X D – normal vision,
X d – color blindness.
A_X D X _ brown-eyed woman with normal vision
aaX d Y is the woman’s father, he could only give aX d to his daughter, therefore, the brown-eyed woman is AaX D X d.
AaX D Y. - woman’s husband
P AaX D X d x aaX D Y
G AX D AX d AX D AX d AX D aY
F 1 ААХ D Х D АаХ D Х d ааХ D Х D ааХ D Х d ААХ D Y Аа Х d Y ааХ D ХY ааХ d Y
The probability of having a colorblind child with brown eyes is 1/8 (12.5%) and it is a boy.
4. A - normal, a - albinism.
X H - normal, X h - hemophilia.
Woman АаХ Н Х h, man ааХ H Y
G AX H AX h aX H aX h aX H aY
F1 AaX H X H AaX H Y AaX H X h AaX h Y aaX H X H aaX H Y aaX H X h aaX h Y
Ph.D. Ph.D. Ph.D. k.g. g.n. g.n. g.n. g..g.
Splitting according to eye color is 1:1 according to blood clotting - all daughters are healthy, boys - 1:1.
V. Problems involving chained inheritance1 . A - smooth grains,
a - wrinkled grains.
B - colored grains,
b - uncolored grains.
R AABB x AABB
Since uniformity was obtained in the first generation (Mendel’s first law), therefore, homozygotes were crossed, and in F1 a diheterozygote was obtained, carrying dominant traits.
Analysis cross:
| normal gametes | |||||||
| recombinant gametes | |||||||
| smooth | wrinkled | smooth | wrinkled |
Since in the second generation there was an unequal number of phenotypic groups, therefore, linked inheritance took place. Those phenotypic groups that are represented in large numbers are not crossovers, but groups that are represented in small numbers are crossovers formed from recombinant gametes in which linkage was disrupted due to crossing over in meiosis.
2. A- colored seed a- non-colored seed B- starchy endosperm b- waxy endosperm P AaBv x aavv G AB Av a B av F1 9- AaBv- env. seed, starch endosperm 42- Aavv- env. seed, wax endosperm 44- aaBv- uncolored seed, starchy endosperm 10- aabv- uncolored seed waxy endosperm The presence in the offspring of two groups (42 - with colored waxy endosperm; 44 - with uncolored waxy endosperm) in approximately equal proportions is the result of linked inheritance of alleles A and b , a and B among themselves. The other two phenotypic groups are formed as a result of crossing over.
A - smooth seeds,
a - wrinkled seeds
B - presence of antennae,
b - without antennae
| smooth | wrinkled |
||
If crossing over does not occur, then the diheterozygous parent produces only two types of gametes (full linkage).
- A - gray body
F 1 AaBv are all gray with normal wings. Law of Uniformity)
R AaBv x AaBv
T.K. there is no expected Mendelian splitting, which means crossing over has occurred:
linked
F 2 AABB AaBv AaBv aavv
Dihybrid crossing. Examples of solving typical problems
Task 1. In humans, complex forms of myopia dominate over normal vision, and brown eye color dominates over blue. A brown-eyed, myopic man, whose mother had blue eyes and normal vision, married a blue-eyed woman with normal vision. What is the percentage chance of having a child with the mother's characteristics?
Solution
Gene Trait
A development of myopia
a normal vision
B Brown eyes
b Blue eyes
P ♀ aabb x ♂ AaBb
G ab, AB, Ab aB, ab
F 1 AaBb; Aabb; aaBb; aabb
Answer: a child with the aabb genotype has blue eyes and normal vision. The probability of having a child with these signs is 25%.
Problem 2. In humans, red hair color dominates over light brown hair, and freckles dominate over their absence. A heterozygous red-haired man without freckles married a fair-haired woman with freckles. Determine the % probability of having a red-haired child with freckles.
Solution
Gene Trait
A red hair
a brown hair
B presence of freckles
b absence of freckles
P ♀ Aabb x ♂ aaBB
F 1 AaBb; aaBb
A red-haired child with freckles has the genotype AaBb. The probability of having such a child is 50%.
Answer: There is a 50% chance of having a red-haired baby with freckles.
Problem 3. A heterozygous woman with a normal hand and freckles marries a six-fingered heterozygous man who does not have freckles. What is the probability of having a child with a normal hand and no freckles?
Solution
Gene Trait
A six-fingered (polydactyly),
a normal hand
B presence of freckles
b no freckles
P ♀ aaBb x ♂ Aabb
G aB, ab, Ab, ab
F 1 AaBb; Aabb; aaBb; aabb
Answer: the probability of having a child with the aabb genotype (with a normal hand, without freckles) is 25%.
Problem 4. The genes that determine a predisposition to cataracts and red hair are located on different pairs of chromosomes. A red-haired woman with normal vision married a fair-haired man with cataracts. What phenotypes can they have children with if the man’s mother has the same phenotype as his wife?
Solution
Gene Trait
A blonde hair,
a Red hair
B development of cataracts
b normal vision
P ♀ aabb x ♂ AaBb
G ab, AB, Ab, aB, ab
F 1 AaBb; Aabb; aaBb; aabb
Answer: phenotypes of children – fair-haired with cataracts (AaBb); fair-haired without cataracts (Aabb); red-haired with cataracts (aaBb); red-haired without cataracts (aabb).
Task 5. What is the percentage probability of having a child with diabetes mellitus if both parents are carriers of the recessive gene for diabetes mellitus. In this case, the mother’s blood Rh factor is positive, and the father’s is negative. Both parents are homozygous for the gene that determines the development of the Rh factor. What Rh factor will the children of this couple have?
Solution
Gene Trait
A normal carbohydrate metabolism
a development of diabetes mellitus
Rh+ Rh positive blood
rh- Rh negative blood.
P♀ AaRh + Rh + x ♂ Aarh - rh -
G ARh + , aRh + , Arh - ,arh -
F 1 AARh + rh - ; AaRh + rh - ; AaRh + rh - ; aaRh + rh-
Answer: the probability of having a child with diabetes is 25%; all children in this family will have a positive Rh factor.
Problem 6. Normal growth in oats dominates over gigantism, early ripening over late ripening. The genes for both traits are located on different pairs of chromosomes. What percentage of late-ripening plants of normal growth can be expected from crossing plants heterozygous for both traits?
Solution
P♀AaBb x ♂AaBb
G AB, Ab, AB, Ab,