The proposed problem book includes problems on the following topics: molecular genetics (the role of nucleic acids in plastic metabolism), inheritance of traits in monohybrid crossing (I and II Mendel's laws), inheritance of traits in dihybrid crossing (III Mendel's law), inheritance of sex-linked traits. Tasks are arranged by difficulty, with asterisks (*) marking tasks of increased complexity.
The problem book contains methodological recommendations, the purpose of which is to help in independent development of methods for solving problems. The manual provides examples of typical problems with a detailed explanation of the design, symbols and solutions. Each type of task is preceded by a brief theoretical material. To consolidate the acquired knowledge, test tasks are offered (Appendix 5), which can be solved both in class and at home (with subsequent credit for this test).
The manual can be used both for in-depth study of biology at school and in preparation for entering universities.
Solving problems in molecular genetics
A gene is a section of DNA that codes for a specific protein. The slightest change in the structure of DNA leads to changes in the protein, which in turn changes the chain of biochemical reactions with its participation, which determine a particular trait or series of traits.
The primary structure of a protein, i.e. sequence of amino acid residues encoded in DNA as a sequence of nitrogenous bases adenine (A), thymine (T), guanine (G) and cytosine (C). Each amino acid is encoded by one or more sequences of three nucleotides - triplets. Protein synthesis is preceded by the transfer of its code from DNA to messenger RNA (mRNA) - transcription. During transcription, the principle of addition, or complementarity, is fulfilled: A, T, G and C in DNA correspond to U (uracil), A, C and G in mRNA. Direct protein synthesis, or broadcast, occurs on the ribosome: amino acids brought to the ribosome by their transfer RNAs (tRNAs) are combined into a polypeptide chain of a protein corresponding to triplets of mRNA bases.
An unambiguous relationship between the sequences of nucleotides in DNA and amino acids in the polypeptide chain of a protein allows one to determine the other. Knowing the changes in DNA, we can tell how the primary structure of the protein will change.
Problem 1. A fragment of a DNA molecule consists of nucleotides arranged in the following sequence: TAAAATGGCAACC. Determine the composition and sequence of amino acids in the polypeptide chain encoded in this part of the gene.
Solution
We write out the DNA nucleotides and, breaking them into triplets, we get the codons of the DNA molecule chain:
TAA–ATG–GCA–ACC.
We make triplets of mRNA that are complementary to the DNA codons and write them in the line below:
DNA: TAA–ATG–GCA–ACC
mRNA: AUU–UAC–CGU–UTT.
Using the codon table (Appendix 6), we determine which amino acid is encoded by each mRNA triplet:
Ile–Tir–Arg–Trp.
Problem 2. A fragment of the molecule contains the amino acids: aspartic acid–alanine–methionine–valine. Define:
a) what is the structure of the section of the DNA molecule encoding this sequence of amino acids;
b) the number (in%) of different types of nucleotides in this part of the gene (in two chains);
c) the length of this gene section.
Solution
a) Using the codon table (Appendix 6), we find the mRNA triplets encoding each of the indicated amino acids.
Protein: Asp-Ala-Met-Val
mRNA: GAC–GCA–AUG–GUU
If an amino acid corresponds to several codons, then you can choose any of them.
We determine the structure of the DNA chain that encoded the structure of mRNA. To do this, under each codon of the mRNA molecule we write the complementary codon of the DNA molecule.
1st DNA strand: CTG–CGT–TAC–CAA.
b) To determine the number (%) of nucleotides in this gene, it is necessary, using the principle of complementarity (A–T, G–C), to complete the second strand of DNA:
2nd DNA strand: GAC–GCA–ATG–GTT
We find the number of nucleotides (ntd): in two chains - 24 ntd, of which A = 6. We make up the proportion:
24 ntd – 100%
6 ntd – x%
x = (6x100) : 24 = 25%
According to Chargaff's rule, the amount of adenine in a DNA molecule is equal to the amount of thymine, and the amount of guanine is equal to the amount of cytosine. That's why:
T = A = 25%
T + A = 50%, therefore
C + G = 100% – 50% = 50%.
C = G = 25%.
c) The DNA molecule is always double-stranded, its length is equal to the length of one chain. The length of each nucleotide is 0.34 nm, therefore:
12 ntd x 0.34 = 4.08 nm.
Problem 3. The molecular weight of protein X is 50 thousand daltons (50 kDa). Determine the length of the corresponding gene.
Note. The average molecular weight of one amino acid can be taken to be 100 Da, and that of one nucleotide – 345 Da.
Solution
Protein X consists of 50,000: 100 = 500 amino acids.
One of the strands of the gene encoding protein X must consist of 500 triplets, or 500 x 3 = 1500 ntd.
The length of such a DNA chain is 1500 x 0.34 nm = 510 nm. This is the length of the gene (double-stranded DNA section).
Solving problems in general genetics
Basic concepts and symbols
Gene– a section of a DNA molecule in a chromosome containing information about the primary structure of one protein; genes are always paired.
Genotype- the totality of all the genes of an organism.
Phenotype- the totality of all external signs of an organism.
Hybrid- an organism formed as a result of the crossing of individuals that differ in a number of characteristics.
Alternative signs– contrasting features (white – black, yellow – green).
Locus– location of the gene on the chromosome.
Allelic genes– two genes that occupy identical loci on homologous chromosomes and determine alternative traits.
Non-allelic genes– genes occupying different loci on chromosomes.
Traits inherited according to Mendel– the most common ones in solving problems (Appendix 7).
Allelic genes can be in two states: dominant, denoted by a capital letter of the Latin alphabet ( A, IN, WITH etc.), or recessive, denoted by a lowercase letter ( A, b, With etc.).
Organisms that have identical alleles of one gene, for example dominant ( AA) or recessive ( ahh), are called homozygous. They produce one type of gametes ( A) or ( A).
Organisms that have different alleles of the same gene ( Ahh), are called heterozygous. They produce two types of gametes ( A And A).
Symbols:
x – crossing of organisms;
P – parents;
F – children; the index means the generation number: F 1, F 2, F n, etc.;
G – parental gametes or sex cells;
– “shield and spear of Mars”, male gender;
– “mirror of Venus”, female.
Stages of problem solving
1.
Recording the genotypes and phenotypes of the parents.
2.
Recording possible gamete types for each parent.
3.
Record possible types of zygotes.
4.
Calculation of the ratio of genotypes and phenotypes of offspring.
1. Graphic method:
2. Algebraic method:
F 1 ( IN + b) (IN + b)
F 2 = BB + 2Bb + bb
3. Punnett grid:
Monohybrid cross
Solving monohybrid crossing problems involves analyzing the inheritance of traits determined by only one pair of allelic genes. Mendel determined that when homozygous individuals are crossed, differing in one pair of characters, all offspring are phenotypically uniformly (I Mendel's law).
With complete dominance, first generation hybrids have the characteristics of only one of the parents, since in this case the expression of the gene A in an allelic pair does not depend on the presence of another gene A(allele A does not manifest itself, so it is called recessive), and heterozygotes ( Ahh) are not phenotypically different from homozygotes ( AA).
When crossing monohybrids in the second generation, the characteristics are split into the original parental ones in a ratio of 3: 1 by phenotype and 1: 2: 1 by genotype (Mendel’s II law): 3/4 of the descendants have characteristics determined by the dominant gene, 1/4 - a trait of the recessive gene.
Task 1. Determine the genotypes and phenotypes of the offspring of brown-eyed heterozygous parents.
Given:
A– brown eyes
A- Blue eyes
Define: F 1
Solution
Heterozygous brown-eyed parents Aa
Character splitting occurs according to Mendel’s II law:
by phenotype 3:1
by genotype 1: 2: 1
Task 2. Find the ratio of smooth and wrinkled seeds in peas in the first generation obtained by pollinating plants with wrinkled seeds with pollen from homozygous plants with smooth seeds.
Given:
A– smooth seeds
a– wrinkled seeds
Define: F 1
Solution
According to Mendel's First Law, all seeds are smooth.
Another entry is possible.
Homozygotes for a given pair of traits form one type of gamete:
With incomplete dominance, both genes function, so the phenotype of hybrids differs from homozygotes for both alleles ( AA And ahh) is an intermediate manifestation of the trait, and in the second generation, splitting into three classes occurs in a ratio of 1: 2: 1, both in genotype and phenotype.
Problem 3. Red-fruited gooseberry plants, when crossed with each other, produce offspring with red berries, and white-fruited gooseberry plants produce white ones. As a result of crossing both varieties with each other, pink fruits are obtained.
1. What kind of offspring will be produced by crossing heterozygous gooseberry plants with pink fruits?
2. What kind of offspring will you get if you pollinate a red-fruited gooseberry with pollen from a hybrid gooseberry with pink fruits?
Given:
A– red color of fruits
A– white color of fruits
F 1 – x
Solution
Answer: When hybrid plants with pink fruits are crossed, the offspring undergo cleavage in a ratio of phenotype to genotype of 1:2:1.
When crossing a red-fruited gooseberry with a pink-fruited one, the offspring will be split in phenotype and genotype in a 1:1 ratio.
Often used in genetics test cross. This is the crossing of a hybrid, the genotype of which is unclear, with a homozygous individual for the recessive genes of the allele. Cleavage in the offspring according to the trait occurs in a 1:1 ratio.
Dihybrid cross
A dihybrid cross is a cross in which organisms differ in two pairs of alternative traits. Hybrids obtained from such crossings are called diheterozygotes. When crossing two homozygous individuals that differ from each other in two or more pairs of traits, the genes and their corresponding traits are inherited independently of each other and are combined in all possible combinations. When dihybrid crossing two diheterozygotes (F 1 individuals) with each other in the second generation of hybrids (F 2), a splitting of traits according to phenotype will be observed in the ratio 9: 3: 3: 1 (III Mendel's law). This phenotypic ratio is the result of the superposition of two monohybrid splits:
, where “n” is the number of pairs of features.
The number of possible variants of gametes is 2n, where n is the number of heterozygous gene pairs in the genome, and 2 is the possible number of gametes in monohybrids.
Examples
The formation of four types of gametes is possible, because in meiosis (prophase I) conjugation and crossing over of chromosomes occur.
Problem 4. What characteristics will hybrid apricots obtained as a result of pollination of dihomozygous red-fruited plants of normal growth with pollen of yellow-fruited dwarf plants have? What will be the result of further crossing of such hybrids?
Given:
A– red color of fruits
A– yellow color of fruits
IN– normal growth
b– dwarf stature
Determine: F 1 and F 2
Solution
Answer: when crossing hybrids in F 2, splitting will occur in the ratio:
9/16 – red-fruited, normal growth;
3/16 – red-fruited, dwarf growth;
3/16 – yellow-fruited, normal growth;
1/16 – yellow-fruited, dwarf growth.
To be continued
Test crossbreeding is a type of genetic research carried out to determine the heterozygosity of an organism.
Essence of the method
Test crossbreeding was studied by Gregor Mendel and one of the founders of genetics, William Bateson.
To determine heterozygosity, an analyzer is used - a recessive homozygous organism (aa).
The analyzed individual has a dominant trait in the phenotype. For example, this is a solid coat color in cats.
In the genotype it can be:
- homozygous (AA);
- heterozygous (Aa).
We cross:
R A? (solid colored cat) x aa (Siamese cat)
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G Huh? a a
First option F1: Aa Aa (all kitten colors are solid)
Second option F2: Aa aa (Siamese and solid colors in a 1:1 ratio)
The accuracy of the analysis depends on the number of descendants. The more descendants, the higher the accuracy.
Thus, in the first variant, the analyzed organism produced only A - gametes. This means he is homozygous (pure breed). In the second case of analysis, the cat gave both A - gametes and a - gametes, i.e. it is a heterozygote. Thus, the genotype with complete dominance is determined by the ratio of characteristics during analyzing crossing.
In the case of incomplete dominance in the offspring there are observed intermediate (mixed) signs:
- in the first generation by 100%;
- in the second generation 1:2:1, where 2 is a mixed trait.
Rice. 1. Scheme of analytical crossing.
Determination of a diheterozygote (AaBb) during test crossing occurs when the ratio of phenotypes and genotypes (F) is 1:1:1:1.
R AaVv x aavv
G AB Av aV av av
F AaBv Aavv aaBv aavv
Backcrossing
Sometimes backcrossing is carried out - with the parent individual. Its goal is to saturate the hybrid with valuable parental genes.
Sample task
It is known that black color in rabbits is dominant over white. Is it possible to get white rabbits from black parents?
The recessive trait manifests itself in the homozygous state (aa). Such a homozygote can result if each parent has a recessive gene a.
Rice. 2. Drawing of crossing a white and a black rabbit.
Since both parents are black, then gene a will be in both gametes if each of them is heterozygous (Aa).
P Aa x Aa
F AA Aa aA aa
25% of the offspring of heterozygous black rabbits will be white.
Practical use
Test crossbreeding is of great importance in all sectors of animal husbandry. With its help, you can determine the purebred of an individual.
If the dominant trait is economically significant and determines the productivity or other value of animals, early determination of homozygosity helps prevent financial losses.
The fact is that one or more male sires can be the fathers of the entire livestock of the country. Therefore, the requirements for the selection of such animals are very stringent.
The Ukrainian sire of the Podolsk breed, Repp, fathers 50,000 calves every year. The mass of this bull is 1500 kg.
Rice. 3. Bull Repp.
What have we learned?
While studying analytical crossbreeding in 10th grade, we learned that this is an important method of genetic analysis. The ratio of genotypes and phenotypes during test crossing depends on the homozygosity or heterozygosity of the parent individual. If an individual is homozygous, then uniformity in phenotype and genotype is observed in the offspring. If the original organism is heterozygous, then in the offspring there is a divergence in both genotype and phenotype (1:1). Heterozygotes are of less value in livestock production if an economically important trait is associated with the dominant gene.
Task No. 1
Consider the proposed scheme. Write down the missing term in your answer, indicated by a question mark in the diagram.
Explanation: Let's consider a cross section of the stem of a woody plant.
The stem can be divided into 4 large parts: pith, wood, cambium and bark. The bark consists of bast (sieve tubes), peel and cork.
The correct answer is bark.
Task No. 2.
Below is a list of research methods. All of them, except two, are used in genetics. Find two methods that “fall out” from the general series and write down the numbers under which they are indicated.
1. Centrifugation
2. Hybridization
3. Karyotype analysis
4. Crossbreeding
5. Monitoring
Explanation: Among the listed methods, geneticists do not use only monitoring; this method is more related to zoology. The authors of the textbook from which we took this option consider centrifugation to be another non-genetic method, but the education of a biologist (and during my studies I studied methods of genetics, among other things) allows me to say with confidence that geneticists practically every day precipitate nucleic acids. acids by centrifugation. The correct answer is 5.
Task No. 3.
How many nucleotides make up the tRNA anticodon?
Explanation: An anticodon, like any codon, consists of three nucleotides.
The correct answer is 3.
Task No. 4.
All of the following characteristics, except two, can be used to determine the processes of the light phase of photosynthesis. Identify two characteristics that “drop out” from the general list, and write down in the table the numbers under which they are indicated.
1. Photolysis of water
2. Reduction of carbon dioxide to glucose
3. Synthesis of ATP molecules using the energy of sunlight
4. Formation of molecular oxygen
5. Use of the energy of ATP molecules for the synthesis of carbohydrates
Explanation: Here is a list of processes that occur during the light and dark phases of photosynthesis.
Light phase processes:
1. Chlorophyll activation
2. Photolysis of water
3. ATP synthesis
4. Creation of NADPH H2
5. Formation of free oxygen
Dark phase processes:
1. Carbon dioxide fixation
2. Glucose formation (Calvin cycle)
The correct answer is 25.
Task No. 5.
Establish a correspondence between the processes and the method of cell division: for each position given in the first column, select the corresponding position from the second column
Processes
A. Somatic cell division occurs
B. The chromosome set is halved
B. A new combination of genes is formed
D. Conjugation and crossing over occur
D. Bivalents are located along the equator of the cell
Division method
1. Mitosis
2. Meiosis
Explanation: meiosis is the process of formation of germ cells (from a diploid cell, four haploid cells are formed in two divisions). Mitosis is the process of division of somatic cells, in which two diploid cells are formed from one diploid cell, the process occurs without changing the DNA of the original cell. Therefore, we will classify the division of somatic cells as mitosis, and meiosis as the rest. Bivalents are pairs of homologous chromosomes connected by special proteins; this structure is formed during meiosis.
The correct answer is 12222.
Task No. 6.
What ratio of genotypes will be obtained by crossing two heterozygotes with complete dominance? Write the answer as a sequence of numbers in descending order.
Explanation: Let's take two heterozygotes: Aa and Aa and see what happens when they are crossed. The following genotype options are obtained: AA: 2Aa: aa. That is, let's write down - 211.
The correct answer is 211.
Task No. 7.
Below is a list of terms. All of them, except two, are used to describe genetic processes and phenomena. Find two terms that “fall out” from the general series and write down the numbers under which they are indicated in the table.
1. Polyploidy
2. Decomposer
3. Symbiosis
4. Homozygote
5. Karyotype
Explanation: The terms polyploidy, homozygote and karyotype refer to genetics. Polyploidy is a multiple increase in the number of chromosomes. Homozygote is an organism in which both alleles of a given trait are the same. A karyotype is a set of chromosomes that is individual for each type of living organism. And decomposer and symbiosis are concepts related to ecology. A decomposer breaks down organic substances into mineral ones, and symbiosis is a mutually beneficial relationship between living organisms. The correct answer is 23.
Task No. 8.
Establish a correspondence between the method of reproduction and a specific example: for each position given in the first column, select the corresponding position from the second column
Example
A. Fern sporulation
B. Formation of gametes by Chlamydomonas
B. Formation of spores in sphagnum
D. Yeast budding
D. Fish spawning
Reproduction method
1. Asexual
2. Sexual
Explanation: sporulation, both in ferns and in mosses (sphagnum), refers to the asexual type of reproduction, budding is also asexual reproduction, and the formation of gametes and the spawning of fish refer to sexual reproduction.
The correct answer is 12112.
Task No. 9.
1. Development of larvae in the host body
2. Sexual reproduction
3. The presence of a dense cuticle
4. The presence of bilateral body symmetry
5. Presence of a skin-muscle bag
6. Formation of a large number of eggs
Task No. 10.
Establish a correspondence between the characteristic and the kingdom of organisms: for each position given in the first column, select the corresponding position from the second column.
Characteristic
A. The cell wall contains chitin
B. Autotrophic nutrition type
B. Form organic substances from inorganic substances
D. The reserve nutrient is starch.
D. In natural systems they are decomposers
E. The body consists of mycelium
kingdom organisms
1. Mushrooms
2. Plants
Explanation: fungi are characterized by the following characteristics: the presence of chitin in the cell wall, the decomposition of organic substances to inorganic ones, that is, they are decomposers, and their body consists of mycelium. This means that plants include: autotrophy, starch as a reserve nutrient, the formation of organic substances from inorganic ones. The correct answer is 122211.
Task No. 11.
Establish the sequence of systematic categories used in classifying animals, starting with the smallest. Write down the corresponding sequence of numbers in the table.
1. Tiger
2. Felines
3. Predatory
4. Mammals
5. Ussuri tiger
6. Chordates
Explanation:
Species - Ussuri tiger
Rod - tiger
Family - felines
Squad - predatory
Class - mammals
Type - chordates
The correct answer is 512346.
Task No. 12.
When the sympathetic nervous system is excited, as opposed to when the parasympathetic nervous system is excited
1. Arteries dilate
2. Blood pressure increases
3. Intestinal motility increases
4. The pupil narrows
5. Blood sugar increases
6. Heart contractions become more frequent
Explanation: the sympathetic and parasympathetic sections of the nervous system are antagonists, if the sympathetic nervous system dilates the pupil, increases pressure, narrows the arteries, increases the amount of sugar in the blood, increases heart rate, reduces intestinal motility, then the parasympathetic system, on the contrary, constricts the pupil, lowers pressure, dilates the arteries , reduces the amount of sugar in the blood, slows the heartbeat, enhances intestinal motility.
The correct answer is 256.
Task No. 13.
Establish a correspondence between the meaning of the reflex and its type: for each position given in the first column, select the corresponding position from the second column.
Reflex meaning
A. Provides instinctive behavior
B. Provides adaptation of the organism to the environmental conditions in which many generations of this species lived
B. Allows you to gain new experience gained throughout life
D. Determines the behavior of the organism in changed conditions
Type of reflex
1. Unconditional
2. Conditional
Explanation: First, let's understand the terminology. Reflex is the body’s response to the action of stimuli or the external environment. An unconditioned reflex is a reflex that is present in a species (or group of species) from birth, developed (reinforced from generation to generation (for example, sucking, swallowing, sneezing, etc.). The reflex arc of such reflexes primarily passes through the spinal cord. Conditioned reflex - a reflex that occurs in the human body (or animals) during life in a certain habitat, as an adaptation to this environment. Conditioned reflexes for each organism are individual and when moving to another territory (or when the reflex is not used) the reflex arc of such reflexes disappears. through the brain. Therefore, the correct answer is 1122.
Task No. 14.
In what order are the parts of the lower limb skeleton located in humans, starting with the pelvic girdle? Write down the corresponding sequence of numbers in the table.
1. Fingers
2. Metatarsus
3. Thigh
4. Shin
5. Tarsus
Explanation: Consider the skeleton of a human lower limb.
The correct answer is 34521.
Task No. 15.
Choose three correct answer options out of six and write down the numbers under which they are indicated in the table. What embryological evidence of evolution supports human kinship with other vertebrates?
1. Anlage of gill slits at the embryo
2. The presence of 46 chromosomes in the cells of the human embryo body
3. Development of the caudal region in the embryo
4. Presence of homologous organs
5. Development of vestigial organs
6. Division of the body into the head, trunk, and caudal sections
Explanation: The Haeckel-Müller biogenetic law says that ontogeny is a repetition of phylogeny, that is, in the process of embryonic development we go through all stages of evolution, from a single cell to a multicellular highly organized organism, at some stage we have a tail, gills, etc. . Therefore, embryonic evidence includes: the formation of gill slits, the development of the tail, and the division of the body into three sections (at the embryonic stage). The correct answer is 136.
Task No. 16.
Establish a correspondence between an example of the struggle for existence and the form to which this struggle relates: for each position given in the first column, select the corresponding position from the second column.
Example
A. Determination of nesting sites in the forest by crossbills
B. Use of cattle as a habitat by the bovine tapeworm
B. Competition between males for dominance
D. Replacement of a black rat by a gray rat
D. Fox hunting for field mice
Fight form
1. Intraspecific
2. Interspecific
Task No. 17.
Choose three correct answers out of six and write down the numbers under which they are indicated in the table.
What biotic factors can lead to an increase in the number of mouse-like rodents in a spruce forest?
1. Reduction in the number of owls, hedgehogs, foxes
2. Large harvest of spruce seeds
4. Tree cutting
5. Deep snow cover in winter
The correct answer is 126.
Task No. 18.
Establish a correspondence between a characteristic of the environment and its factor: for each position given in the first column, select the corresponding position from the second column.
Characteristic
A. Constancy of the gas composition of the atmosphere
B. Changing the thickness of the ozone screen
B. Change in air humidity
D. Change in the number of consumers
D. Change in the number of producers
Environmental factors
1. Biotic
2. Abiotic
The correct answer is 222111.
Task No. 19.
Establish the sequence of processes occurring during the reproduction and development of flowering plants, starting from the moment of pollen formation. Write down the corresponding sequence of numbers in the table.
1. Penetration of sperm into the embryo sac
2. Formation of a triploid cell
3. Pollen tube germination
4. Formation of a seed from an ovule
5. Formation of generative and vegetative cells
Explanation: Let's arrange the processes in the correct order. It all starts with the formation of generative and vegetative cells, then the pollen tube grows, then the sperm penetrates the embryo sac, the sex cells fuse, a triploid cell is formed and, finally, the seed is formed from the ovule. The correct answer is 53124.
Task No. 20.
Analyze the table. Fill in the empty cells of the table using the concepts and terms, examples given in the list. For each lettered cell, select the appropriate term from the list provided.
List of terms and concepts:
1. Biological progress
2. The presence of webbed limbs in waterfowl
3. Presence of warm-bloodedness in chordates
4. Aromorphosis
5. Divergence
6. Biological regression
Correct answer: 142.
Task No. 21.
Study the graph of the reaction rate versus enzyme concentration. Select statements that can be formulated based on the analysis of the proposed schedule. Write down the numbers of the selected statements in your answer.
1. The rate of an enzymatic reaction does not depend on the concentration of the enzyme
2. The rate of an enzymatic reaction depends significantly on the concentration of the enzyme
3. As the enzyme concentration increases, the reaction rate increases
Explanation: The graph shows that as the concentration of the enzyme increases, the reaction rate also increases, so answer options 2 and 3 are suitable for describing this dependence. The correct answer is 23.
Task No. 22.
It is known that when growing clover, soybeans, and beans, fertilizing with nitrogen fertilizers is not required. Explain why.
Explanation: All of these plants are legumes, and nodule bacteria live in symbiosis with leguminous plants, which fix molecular nitrogen and process it into a form digestible by plants.
Task No. 23.
Using a drawing of the process of sexual reproduction of Chlamydomonas, explain the essence of sexual reproduction and how it differs from asexual reproduction. As a result of what process are gametes formed, what are their features? What number in the picture indicates the zygote? How is it different from gametes?
Explanation: during asexual reproduction, the contents of a single-celled Chlamydomonas are divided into 4 parts using meiosis, resulting in the formation of zoospores, which perform the function of dispersal, then small cells grow to the size of the mother cell and divide again by meiosis. And in unfavorable conditions, sexual reproduction occurs, and biflagellate gametes are formed in the mother cell (3), they leave the mother cell and merge in pairs with other individuals, the zygote is covered with a dense shell and overwinters (survives unfavorable conditions), then the zygote divides, as a result which results in the formation of 4 diploid chlamydomonas, which grow to maternal size.
The zygote in the figure is indicated by the number 5. It differs from haploid gametes in its diploidity.
Task No. 24.
Find errors in the given text. Indicate the numbers of the sentences in which errors were made and correct them.
1. The body of the cockchafer, covered with skin and cuticle, is divided into the head, trunk and abdomen. 2. The digestive system of beetles begins on the head with a mouth opening with a piercing mouthpart. 3. Metabolic products are secreted through green glands. 4. Gas exchange occurs directly through the walls of the trachea. 5. The open circulatory system consists of the heart and blood vessels.
Explanation: sentence 1 - the body of the cockchafer is not covered with skin with a cuticle, but with a chitinous covering, and it is divided into the head, chest and abdomen. Sentence 2 - the mouthparts of the May tongue are not piercing, but gnawing. Proposition 3 - green glands are not organs of the excretory system, since the metabolic products of the cockchafer are excreted through the Malpighian vessels and the fat body.
Task No. 25.
Explain what organ of a flowering plant a head of cabbage is a modification of.
Explanation: a head of cabbage is a modified bud, since in the second year of life an adult plant is formed from the head of cabbage, which has all the generative organs, that is, a flower and fruits with seeds. Also, a head of cabbage resembles a bud in structure: the stump is a modified stem, with thick leaves and rudimentary buds located on it.
Task No. 26.
Which ecosystem, a potato field or a meadow, has longer and more diverse food chains? Explain your answer.
Explanation: a potato field is an artificial ecosystem, and a meadow is a natural ecosystem. Longer and more diverse food chains are found in a natural ecosystem, since there are more plants, animals and microorganisms in the meadow and monoculture (potatoes) does not predominate.
Task No. 27.
All types of RNA are synthesized on a DNA template. The fragment of the DNA molecule on which the region of the central loop of tRNA is synthesized has the following nucleotide sequence: CTTACGGGGCATGGCT. Establish the nucleotide sequence of the tRNA region that is synthesized on this fragment if the third triplet corresponds to the tRNA anticodon. Explain your answer.
Explanation: Based on the given sequence of nucleotides in DNA, we will find the sequence of tRNA nucleotides. We will use the principle of complementarity: A=U, G=C.
tRNA: GAAUGCCCCGUACCCGA
The third triplet - CCG corresponds to the mRNA - GGC.
Task No. 28.
A man suffering from deafness and color blindness married a healthy woman. They had a son who was deaf and colorblind and a daughter who had good hearing but was colorblind. In humans, deafness is an autosomal, recessive trait, while color blindness is a recessive trait, linked to sex. Make a diagram for solving the problem. Indicate the possible phenotypes and genotypes of children in this family. Determine the probability of having children suffering from both anomalies.
Explanation:
aa - deaf
Aa, AA - not deaf (carries the deafness gene in the second case)
X D X d - healthy mother (carries the color blindness gene)
X d Y - sick father
АаХ D X d x ааX d Y
Gametes: AX D, aX D, AX d, aX d x aX d, aY
Possible gentypes of children:
Girls:
AaX D X d - normal hearing, normal vision (healthy child)
ааX D X d - deaf, normal vision
AaX d X d - normal hearing, colorblind
aaX d X d - deaf, color blind
Boys:
AaX D Y - normal hearing, normal vision (healthy child)
АаX D Y - deaf, normal vision
AaX d Y - normal hearing, colorblind
AaX d Y - deaf, colorblind
Among boys, the probability of having both anomalies is 1/4, and among girls too, that is, 25%.
G.S. Kalinova, T. V. Mazyarkina Biology Typical test tasks. Unified State Exam 2017. 10 options.
In the study of the inheritance of traits, geneticists proceed from the idea that the development of each trait is determined by a separate gene.
Therefore, in a dihybrid cross, development is to study the inheritance of two genes.
In monohybrid crosses, it was found that a number of pairs of pea traits: smooth - wrinkled, yellow - green seeds, tall - short plant growth, purple - white flowers, etc. - reveal cleavage in the offspring of the hybrid (in F 2) according to phenotype in a ratio of 3: 1. Of each such pair of characters, one turns out to be dominant, the other - recessive. For dihybrid crossing, Mendel took homozygous pea plants that differed simultaneously in two pairs of traits. The mother “seed” plant had smooth seeds (the gene that determines this trait will be designated B) and a yellow color of the seeds, the gene of which will be designated A; both of these traits are dominant. The paternal "pollen" plant had recessive characteristics: wrinkled and green seeds. The parental forms were homozygous for two pairs of traits or for two genes determining them; The genotype of the mother plant can be designated AABB, and the paternal plant - aabb. However, the distribution of characteristics in the parent forms does not matter in this case. Mendel also crossed plants with smooth and green seeds with plants with wrinkled and yellow seeds, i.e. aaBB and AAbb.
If we assume that each gene is located on a separate chromosome, then we should expect that mature eggs and sperm, which have a haploid set of chromosomes, will have only one allele of each gene. Then the gametes of the mother plant should carry alleles A and B (or a and B), and the paternal plant - a and b (or A and b). Fertilization of the egg AB by the sperm ab will lead to the formation of a dihybrid zygote F 1 in the somatic cells of the hybrid embryo, the double set of chromosomes will be restored, and the hybrid will be heterozygous for two allelic pairs, i.e. diheterozygous AaBb. The same genotype is formed in the case of the combination of gametes Ab and aB.
Hybrid pea seeds in our example, having the hereditary structure AaBb in phenotype, as would be expected with complete dominance, will be smooth and yellow.
To make sure that the F 1 hybrid is heterozygous for two AaBb genes, we can use the already known method of analyzing crossing. To do this, the F 1 hybrid should be crossed with a form that is homozygous for both recessive traits - aabb. In a hybrid, four types of gametes are formed in meiosis: AB, aB, Ab, ab. The aabb form produces only one type of gamete - ab. When all combinations of gametes are equally probable, four types of zygotes are formed in an equal ratio l AaBb: 1aaBb: 1Aabb: 1aabb. Analyzing crossing allows us to most quickly study the genotype of a hybrid organism for genes of interest to us.
Mendel also made test crosses of F 1 hybrid plants (smooth and yellow seeds) with plants homozygous for two recessive genes (wrinkled and green seeds). In the offspring he obtained four classes of seeds in numerical ratios very close to the expected splitting of 1:1:1:1, namely: smooth yellow - 55 (AaBb), smooth green - 51 (aaBb), wrinkled yellow - 49 (Aabb) , wrinkled green - 53 (aabb).
Thus, genetic methods have shown that a dihybrid organism produces four types of gametes in equal proportions and, therefore, is heterozygous for both allelic pairs.
Cleavage according to fepotype. Mendel's third law
In the progeny of fifteen dihybrid F 1 plants, Mendel obtained 556 seeds, of which 315 were smooth yellow, 101 wrinkled yellow, 108 smooth green and 32 wrinkled green.
As we already know, in a monohybrid cross with complete dominance in F 2, there is a cleavage by phenotype in the ratio of 3: 1, by genotype 1: 2: 1. Let us imagine that each individual pair of Aa and Bb behaves in inheritance in the same way as in a monohybrid cross. There are reasons for such an assumption; remember the mechanism we know about chromosome segregation in meiosis. In this case, in a dihybrid plant, both female and male, containing both allelic pairs, four varieties of gametes (AB, Ab, aB, ab) will be formed in meiosis, which, during fertilization, can freely combine with each other and give 16 types of zygotes.
To find out how each pair of alleles behaves in the offspring of a dihybrid, you can again apply the method of taking into account each pair of traits separately. To do this, all 556 seeds of the second generation must be divided into two classes: 1) according to shape: 315 + 108 = 423 smooth and 101 + 32 = 133 wrinkled; 2) by color: 315 + 101 = 416 yellow and 108 + 32 = 140 green.
Knowing that splitting for each pair of characteristics occurs in a ratio of 3: 1, we can say that of the total number of seeds, 3/4 should be smooth and 1/4 wrinkled. By making the appropriate calculations (556x 3 / 4 - 417 and 556x 1 / 4 = 139), we obtain the theoretically expected numerical ratios of seeds in F 2 for each pair of traits 417: 139. From the above calculations it is clear that in a dihybrid cross for each pair of alleles There is a natural splitting in a ratio of 3:1.
To imagine how the combination of two pairs of alleles Aa and Bb is carried out simultaneously, and also to establish the nature of segregation in F 2 while simultaneously taking into account both characteristics, you can go in two ways. The first way is to construct a Punnett lattice. The Punnett grid allows you to establish all possible combinations of male and female gametes during fertilization, as well as determine the phenotypes and genotypes of F2 individuals.
The second way is purely mathematical, based on the law of combination of two or more independent phenomena. This law states that if two events are independent, then the probability that they will occur simultaneously is equal to the product of the probabilities of each of them.
As has been shown, splitting for each pair of alleles during dihybrid crossing occurs as two independent phenomena. The appearance of individuals with a dominant trait during monohybrid crossing occurs in 3/4 of all cases, and with recessive ones - 1/4. Therefore, the probability that the characteristics smooth shape and yellow color of seeds will appear simultaneously is equal to the product 3/4 X 3/4 = 9/16, smooth shape and green color - 3/4 X 1/4 = 3/16, wrinkled shape and yellow color - 1/4 X 3/4 = 3/16 and wrinkled shape and green color - 1/4 X 1/4 = 1/16. In other words, the product of individual probabilities gives the ratio of phenotypic segregation classes 9 / 16: 3 / 16: 3 / 16: 1 / 16, or 9:3:3:1.
Let's return to the example of segregation by characteristics obtained from the analysis of 556 F 2 seeds in Mendel's experiment. It is easy to verify that the seeds he received were distributed among the classes of combinations of characteristics in a ratio close to the expected one. In order to calculate the theoretically expected numbers by class, 556 seeds should be multiplied by 9/16, 3/16, 3/16 and 1/16, respectively. Consequently, the ratio of phenotypic segregation classes in an F 2 dihybrid cross with complete dominance fits into the formula 9: 3: 3: 1.
Now it should be clear why, when counting each pair of alternative characters separately, the ratio of the number of smooth seeds to the number of wrinkled ones was 12: 4, or in empirical numbers 423: 133, and yellow to green -12: 4, or 416: 140, i.e. For each pair the ratio was 3:1. The same results can be obtained using a Punnett lattice, in which the 16 genotypes described above are divided into four classes by phenotype in the same ratio of 9:3:3:1.
Thus, in a dihybrid crossing, each pair of characters, when split in the offspring, behaves in the same way as in a monohybrid crossing, i.e., independently of the other pair of primates.
Based on the simultaneous analysis of the inheritance of several pairs of alternative characters, Mendel established a pattern of independent distribution of factors, or genes, which is known as Mendel's third law. Menlel wrote: “There is no doubt that for all tested characters the following statement is equally valid: the descendants of hybrids that combine several significantly different characters are members of a combination series in which the development series of each pair of differing characters are combined. This simultaneously proves that the behavior of each pair of differing characters in a hybrid combination is independent of other differences in both original plants.”
And then Mendel formulates the actual principle of independence of the combination of hereditary factors: “Constant characters that are found in various forms of a related plant group can enter through repeated artificial fertilization into all compounds that are possible according to the rules of combination.”
Segregation by genotype
The formula 9:3:3:1 expresses the ratio of phenotypic segregation in F2 in a dihybrid cross.
It is necessary to analyze the same segregation by genotype. Obviously, in the case of complete dominance, this can only be done by crossing individuals of all 16 genotypes, which can be obtained by combining four varieties of female and male gametes with a homozygous recessive form of aabb. Since, when splitting by phenotype, each pair of alleles behaves independently, then splitting by genotype will manifest itself in accordance with the same pattern, but in different ratios.
By analyzing the F2 genotypes using the Punnett grid, we can determine the frequency of different genotypes, which will give us the splitting formula 1: 2: 2: 4: 1: 2: 1: 2: 1. Knowing that in a monohybrid cross, the genotype splitting corresponds to 1AA: 2Aa: 1aa for one pair of alleles and 1BB: 2Bb: 1bb for another, you can calculate the probability of the appearance of genotypes of different classes during dihybrid crossing.
The probability of occurrence of the AA genotype is 1/4. Accordingly, for Aa - 1/2 and for aa - 1/4. The same will happen for another allelic pair: BB - 1/4, Bb - 1/2, bb - 1/4. By multiplying the two probabilities, one can obtain all segregation classes by genotype. As a result of this calculation, the same 9 classes of segregation by genotype 1: 2: 2: 4: 1: 2: 1: 2: 1 are obtained, which could be established using the Punnett lattice.
As we have seen, with a monohybrid crossing, the number of classes of segregation by phenotype is 2 (3: 1), and by genotype - 3 (1: 2: 1); in dihybrid crossing, the number of phenotypic classes of segregation is 4, and genotypic - 9. Therefore, in the case of two genes, the number of classes corresponds to 2 2 for the phenotype, and 3 2 for the genotype. In the future, when analyzing the splitting of several genes in polyhybrid crosses, we will make sure that the derived formulas are also valid for these crosses.
It should be said about the rules for writing formulas for various genotypes and phenotypes. With complete dominance, homozygous forms are phenotypically indistinguishable from heterozygous ones; Thus, AABB is indistinguishable from AaBb, AABb, AaBB. For the purpose of shorthand writing, similar phenotypes of homozygotes and heterozygotes are sometimes denoted by the phenotypic radical A-B-. By substituting different alleles into such a radical in place of the dash, one can obtain similar phenotypes (for example, for the radical A-bb, genotypes AAbb and Aabb will have similar phenotypes).
Genetics- a science that studies genes, mechanisms of inheritance of traits and variability of organisms. During the process of reproduction, a number of traits are passed on to the offspring. It was noticed back in the nineteenth century that living organisms inherit the characteristics of their parents. The first to describe these patterns was G. Mendel.
Heredity– the property of individual individuals to transmit their characteristics to their offspring through reproduction (through germ and somatic cells). This is how the characteristics of organisms are preserved over a number of generations. When transmitting hereditary information, its exact copying does not occur, but variability is always present.
Variability– the acquisition by individuals of new properties or the loss of old ones. This is an important link in the process of evolution and adaptation of living beings. The fact that there are no identical individuals in the world is due to variability.
Inheritance of characteristics is carried out using elementary units of inheritance - genes. The set of genes determines the genotype of an organism. Each gene carries encoded information and is located in a specific place in the DNA.
Genes have a number of specific properties:
- Different traits are encoded by different genes;
- Constancy - in the absence of a mutating effect, the hereditary material is transmitted unchanged;
- Lability – the ability to succumb to mutations;
- Specificity - a gene carries special information;
- Pleiotropy – one gene encodes several traits;
Under the influence of environmental conditions, the genotype gives different phenotypes. The phenotype determines the degree to which the organism is influenced by environmental conditions.
Allelic genes
The cells of our body have a diploid set of chromosomes; they, in turn, consist of a pair of chromatids, divided into sections (genes). Different forms of the same genes (for example, brown/blue eyes), located in the same loci of homologous chromosomes, are called allelic genes. In diploid cells, genes are represented by two alleles, one from the father and one from the mother.
Alleles are divided into dominant and recessive. The dominant allele determines which trait will be expressed in the phenotype, and the recessive allele is inherited, but does not manifest itself in a heterozygous organism.
There are alleles with partial dominance, such a condition is called codominance, in which case both traits will appear in the phenotype. For example, flowers with red and white inflorescences were crossed, resulting in red, pink and white flowers in the next generation (pink inflorescences are a manifestation of codominance). All alleles are designated by letters of the Latin alphabet: large - dominant (AA, BB), small - recessive (aa, bb).
Homozygotes and heterozygotes
Homozygote is an organism in which alleles are represented only by dominant or recessive genes.
Homozygosity means the presence of the same alleles on both chromosomes (AA, bb). In homozygous organisms, they code for the same traits (for example, the white color of rose petals), in which case all offspring will receive the same genotype and phenotypic manifestations.
Heterozygote is an organism in which alleles have both dominant and recessive genes.
Heterozygosity is the presence of different allelic genes in homologous regions of chromosomes (Aa, Bb). The phenotype of heterozygous organisms will always be the same and is determined by the dominant gene.
For example, A – brown eyes, and – blue eyes, an individual with genotype Aa will have brown eyes.
Heterozygous forms are characterized by splitting, when when crossing two heterozygous organisms in the first generation we get the following result: by phenotype 3:1, by genotype 1:2:1.
An example would be the inheritance of dark and light hair if both parents have dark hair. A is a dominant allele for dark hair, and is recessive (blond hair).
R: Aa x Aa
G: A, a, a, a
F: AA:2Aa:aa
*Where P – parents, G – gametes, F – offspring.
According to this diagram, you can see that the probability of inheriting a dominant trait (dark hair) from parents is three times higher than a recessive one.
Diheterozygote- a heterozygous individual that carries two pairs of alternative characteristics. For example, Mendel's study of the inheritance of traits using pea seeds. The dominant characteristics were yellow color and smooth seed surface, while the recessive characteristics were green color and rough surface. As a result of the crossing, nine different genotypes and four phenotypes were obtained.
Hemizygote- this is an organism with one allelic gene, even if it is recessive, it will always manifest itself phenotypically. Normally they are present on sex chromosomes.
Difference between homozygote and heterozygote (table)
| Differences between homozygous and heterozygous organisms | ||
|---|---|---|
| Characteristic | Homozygote | Heterozygote |
| Alleles of homologous chromosomes | Identical | Different |
| Genotype | AA, aa | Aa |
| The phenotype is determined by the trait | By recessive or dominant | By dominant |
| First generation monotony | + | + |
| Split | Doesn't happen | From the second generation |
| Manifestation of a recessive gene | Characteristic | Suppressed |
Reproduction and crossing of homozygotes and heterozygotes leads to the formation of new characteristics that are necessary for living organisms to adapt to changing environmental conditions. Their properties are necessary when breeding crops and breeds with high quality indicators.
