The proposed problem book includes tasks 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 (Mendel's III law), inheritance of sex-linked traits. Tasks are sorted by difficulty, with asterisks (*) marking tasks of increased complexity.
The task book contains methodological recommendations, the purpose of which is to help in the independent development of methods for solving problems. The manual provides examples of typical tasks 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, control tasks are offered (Appendix 5), which can be solved both in the classroom and at home (with the subsequent offset of 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 one or another trait or series of traits.
The primary structure of a protein, i.e. the sequence of amino acid residues is encrypted in DNA in the form of a sequence of nitrogenous bases of adenine (A), thymine (T), guanine (G) and cytosine (C). Each amino acid is encoded by one or more sequences of three nucleotides called triplets. Protein synthesis is preceded by the transfer of its code from DNA to messenger RNA (mRNA) - transcription. During transcription, the principle of complementation, or complementarity, is carried out: 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: the amino acids brought to the ribosome by their transfer RNA (tRNA) are combined into a protein polypeptide chain corresponding to mRNA base triplets.
An unambiguous relationship between the sequences of nucleotides in DNA and amino acids in the polypeptide chain of a protein makes it possible to determine the other from one of them. Knowing the changes in DNA, we can say how the primary structure of the protein will change.
Task 1. A fragment of a DNA molecule consists of nucleotides in the following sequence: TAAATGGCAACC. Determine the composition and sequence of amino acids in the polypeptide chain encoded in this region of the gene.
Solution
We write out the DNA nucleotides and, breaking them into triplets, we get the codons of the chain of the DNA molecule:
TAA–ATG–HCA–ACC.
We compose mRNA triplets complementary to DNA codons, and write them down in the line below:
DNA: TAA–ATG–HCA–ACC
mRNA: AUU–UAC–CGU–UTT.
According to the codon table (Appendix 6), we determine which amino acid is encoded by each mRNA triplet:
Ile-Tir-Arg-Trp.
Task 2. A fragment of the molecule contains amino acids: aspartic acid-alanine-methionine-valine. Define:
a) what is the structure of the section of the DNA molecule encoding this amino acid sequence;
b) the number (in%) of different types of nucleotides in this region of the gene (in two chains);
c) the length of this region of the gene.
Solution
a) According to the codon table (Appendix 6), we find mRNA triplets encoding each of the indicated amino acids.
Protein: Asp-Ala-Met-Val
mRNA: GAC–HCA–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 down 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 DNA strand:
2nd DNA strand: GAC-HCA-ATG-GTT
We find the number of nucleotides (ntd): in two chains - 24 ntd, of which A \u003d 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 \u003d 50%, therefore
C + G \u003d 100% - 50% \u003d 50%.
C \u003d G \u003d 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.
Task 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 equal to 100 Da, and one nucleotide - 345 Da.
Solution
Protein X consists of 50,000: 100 = 500 amino acids.
One of the chains of the gene encoding protein X should 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 same length of the gene (double-stranded section of DNA).
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 is the totality of all the genes of an organism.
Phenotype- the totality of all external signs of the body.
Hybrid- an organism formed as a result of crossing individuals that differ in a number of ways.
Alternative signs- contrasting features (white - black, yellow - green).
Locus the location of the gene on the chromosome.
allelic genes- two genes that occupy the same loci in homologous chromosomes and determine alternative traits.
Non-allelic genes are genes that occupy different loci on the chromosomes.
Traits inherited according to Mendel- the most common in solving problems (Appendix 7).
Allelic genes can be in two states: dominant, denoted capital letter Latin alphabet ( BUT, IN, FROM etc.), or recessive, denoted by a lowercase letter ( but, b, from etc.).
Organisms that have the same alleles for the same gene, such as dominant ( AA) or recessive ( aa), are called homozygous. They give one variety of gametes ( BUT) or ( but).
Organisms that have different alleles for the same gene Ah), are called heterozygous. They give two varieties of gametes ( BUT And but).
Symbols:
х – crossing of organisms;
P - parents;
F - children; index means generation number: F 1 , F 2 , F n etc.;
D - gametes of parents or germ cells;
- “shield and spear of Mars”, male;
- "mirror of Venus", female.
Stages of problem solving
1.
Recording the genotypes and phenotypes of the parents.
2.
Write down the possible types of gametes in each parent.
3.
Record possible types of zygotes.
4.
Calculation of the ratio of genotypes and phenotypes of offspring.
1. Graphical way:
2. Algebraic way:
F 1 ( IN + b) (IN + b)
F 2 = BB + 2bb + bb
3. Punnett lattice:
monohybrid cross
Solving problems for monohybrid crossing includes an analysis of the inheritance of traits determined by only one pair of allelic genes. Mendel determined that when homozygous individuals that differ in one pair of traits are crossed, all offspring are phenotypically uniformly (I Mendel's law).
With complete dominance, hybrids of the first generation have the characteristics of only one of the parents, since in this case the expression of the gene BUT in an allelic pair does not depend on the presence of another gene but(allele but does not appear, therefore it is called recessive), and heterozygotes ( Ah) do not differ phenotypically from homozygotes ( AA).
When monohybrids are crossed in the second generation, the characters are split into the original parental ones in a ratio of 3: 1 in terms of phenotype and 1: 2: 1 in terms of genotype (Mendel's law II): 3/4 of the offspring have signs due to the dominant gene, 1/4 - a sign of recessive gene.
Task 1. Determine the genotypes and phenotypes of the offspring of brown-eyed heterozygous parents.
Given:
BUT- Brown eyes
but- Blue eyes
Define: F 1
Solution
Heterozygous brown-eyed parents Aa
There is a splitting of signs, 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:
BUT- smooth seeds
a- wrinkled seeds
Define: F 1
Solution
According to Mendel's 1st law, all seeds are smooth.
Another entry is also possible.
Homozygotes for this pair of traits form one variety of gametes:
With incomplete dominance, both genes function, so the phenotype of hybrids differs from homozygotes for both alleles ( AA And aa) is an intermediate manifestation of the trait, and in the second generation there is a splitting into three classes in a ratio of 1:2:1 both in terms of genotype and phenotype.
Task 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 offspring will result from crossing heterozygous gooseberry plants with pink fruits
2. What offspring will you get if you pollinate a red-fruited gooseberry with the pollen of a hybrid gooseberry with pink fruits x
Given:
BUT- red fruit color
but- white fruits
F 1-x
Solution
Answer: when crossing hybrid plants with pink fruits in the offspring, splitting occurs in the ratio of phenotype and genotype 1:2:1.
When crossing a red-fruited gooseberry with a pink-fruited one, the offspring will have a splitting according to the phenotype and genotype in a ratio of 1: 1.
Often used in genetics analyzing cross. This is the crossing of a hybrid, whose genotype is unclear, with a homozygous individual for the recessive allele genes. Segregation in offspring according to the trait occurs in a ratio of 1:1.
Dihybrid cross
A dihybrid cross is a cross in which organisms differ in two pairs of alternative traits. Hybrids resulting from such a cross are called diheterozygotes. When two homozygous individuals are crossed, differing 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 of two diheterozygotes (individuals F 1) among themselves in the second generation of hybrids (F 2), splitting of signs according to the phenotype will be observed in the ratio 9: 3: 3: 1 (Mendel's III law). This ratio of phenotypes is the result of the superimposition of two monohybrid cleavages:
, where "n" is the number of feature pairs.
The number of possible gamete variants is 2n, where n is the number of heterozygous pairs of genes in the genome, and 2 is the possible number of gametes in monohybrids.
Examples
The formation of four varieties of gametes is possible, because. In meiosis (prophase I), conjugation and crossing over of chromosomes occur.
Task 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 result will further crossing of such hybrids give?
Given:
BUT- red fruit color
but- yellow fruits
IN- normal growth
b- dwarf growth
Define: 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
Task number 1
Consider the proposed scheme. Write in the answer the missing term, indicated by a question mark in the diagram.
Explanation: consider a cross section of a stem of a woody plant.
In the stem, 4 large parts can be distinguished: the core, wood, cambium and bark. The composition of the bark includes: bast (sieve tubes), peel and cork.
The correct answer is bark.
Task number 2.
Below is a list of research methods. All but two of them are used in genetics. Find two methods "falling out" of 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, genetics does 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 also studied genetic methods) allows me to say with confidence that geneticists precipitate nucleic acids almost every day. acid by centrifugation. The correct answer is 5.
Task number 3.
How many nucleotides make up a tRNA anticodon?
Explanation: An anticodon, like any codon, consists of three nucleotides.
The correct answer is 3.
Task number 4.
All the signs below, except for two, can be used to determine the processes of the light phase of photosynthesis. Identify two signs that "fall out" from the general list, and write down in the table the numbers under which they are indicated.
1. Photolysis of water
2. Recovery carbon dioxide to glucose
3. Synthesis ATP molecules through 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 in the light and dark phases of photosynthesis.
Light phase processes:
1. Chlorophyll activation
2. Photolysis of water
3. ATP synthesis
4. Creation of NADP H2
5. Formation of free oxygen
Dark phase processes:
1. Carbon dioxide fixation
2. Formation of glucose (Calvin cycle)
The correct answer is 25.
Task number 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 (four haploid cells are formed from a diploid cell in two divisions). Mitosis is the process of division of somatic cells, while two diploid cells are formed from one diploid cell, the process occurs without changing the DNA of the original cell. Therefore, we refer to mitosis - the division of somatic cells, and to everything else - to meiosis. Bivalents - pairs of homologous chromosomes connected by special proteins, this structure is formed during meiosis.
The correct answer is 12222.
Task number 6.
What is the ratio of genotypes obtained by crossing two heterozygotes with complete dominance? Write your 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 variants of genotypes are obtained: AA:2Aa:aa. That is, we write - 211.
The correct answer is 211.
Task number 7.
Below is a list of terms. All of them, except for two, are used to describe genetic processes and phenomena. Find two terms that "fall out" of the general series, and write down the numbers under which they are indicated in the table.
1. Polyploidy
2. Reducer
3. Symbiosis
4. Homozygote
5. Karyotype
Explanation: the terms polyploidy, homozygous, and karyotype refer to genetics. Polyploidy is a multiply increased number of chromosomes. Homozygous is an organism in which both alleles of a given trait are the same. Karyotype - a set of chromosomes, individual for each type of living organisms. A decomposer and symbiosis are concepts related to ecology. The decomposer breaks down organic substances into minerals, and symbiosis is a mutually beneficial relationship between living organisms. The correct answer is 23.
Task number 8.
Establish a correspondence between the method of reproduction and concrete example: for each position given in the first column, match the corresponding position from the second column
Example
A. Fern sporulation
B. Formation of gametes by chlamydomonas
B. Spore formation 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 spawning of fish are related to sexual reproduction.
The correct answer is 12112.
Task number 9.
1. Development of larvae in the body of the host
2. Sexual reproduction
3. The presence of a dense cuticle
4. The presence of bilateral symmetry of the body
5. The presence of a skin-muscular sac
6. Formation of a large number of eggs
Task number 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. The type of nutrition is autotrophic
B. Form organic substances from inorganic
D. The reserve nutrient is starch.
D. V natural systems are decomposers
E. The body is made up of mycelium
Kingdom of organisms
1. Mushrooms
2. Plants
Explanation: fungi are characterized by the following features: the presence of chitin in the cell wall, decomposition organic matter to inorganic, that is, they are decomposers, and also their body consists of mycelium. So we will attribute to plants: autotrophy, starch as a reserve nutrient, the formation of organic substances from inorganic ones. The correct answer is 122211.
Task number 11.
Set the sequence of systematic categories used in the classification of animals, starting with the smallest. Write down the corresponding sequence of numbers in the table.
1. Tiger
2. Feline
3. Predatory
4. Mammals
5. Ussuri tiger
6. Chordates
Explanation:
View - Ussuri tiger
Genus - tiger
Family - cats
Squad - predatory
Class - mammals
Type - chordates
The correct answer is 512346.
Task number 12.
With excitation of the sympathetic nervous system, in contrast to excitation of the parasympathetic nervous system
1. Arteries expand
2. Blood pressure rises
3. Increases intestinal peristalsis
4. The pupil narrows
5. Increases the amount of sugar in the blood
6. Heart contractions become more frequent
Explanation: the sympathetic and parasympathetic divisions of the nervous system are antagonists, if the sympathetic nervous system dilates the pupil, increases pressure, constricts the arteries, increases the amount of sugar in the blood, increases heart rate, reduces intestinal motility, then the parasympathetic system, on the contrary, narrows the pupil, lowers pressure, dilates the arteries, reduces the amount of sugar in the blood, slows down the heartbeat, increases intestinal motility .
The correct answer is 256.
Task number 13.
Establish a correspondence between the value of the reflex and its type: for each position given in the first column, select the corresponding position from the second column.
The value of the reflex
A. Provides instinctive behavior
B. Provides adaptation of the body to conditions environment in which many generations of this species lived
C. Allows you to acquire new experience gained during life
D. Determines the behavior of the organism in changing conditions
Type of reflex
1. Unconditional
2. Conditional
Explanation: Let's first understand the terminology. Reflex - the body's response to the action of stimuli or the external environment. Unconditioned reflex - a reflex that is present in a species (or group of species) from birth, developed (fixed from generation to generation (for example, sucking, swallowing, sneezing, etc.). The reflex arc of such reflexes mainly 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 fade away when moving to another territory (or when the reflex is not used). through the brain.Therefore, the correct answer is 1122.
Task number 14.
In what sequence are the parts of the skeleton of the lower limb located in humans, starting from the pelvic girdle? Write down the corresponding sequence of numbers in the table.
1. Phalanges of fingers
2. Metatarsus
3. Thigh
4. Calf
5. Tarsus
Explanation: Consider the skeleton of the human lower limb.
The correct answer is 34521.
Task number 15.
Choose three correct answers from six and write down the numbers under which they are indicated in the table. What embryological evidence of evolution confirms the relationship of man with other vertebrates?
1. Bookmark at the embryo of the gill slits
2. The presence of 46 chromosomes in the cells of the body of a human embryo
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, tail sections
Explanation: biogenetic law Haeckel-Müller says that ontogenesis is a repetition of phylogeny, that is, in the process of embryonic development, we go through all stages of evolution, from one cell to a highly organized multicellular organism, at some stage we have a tail, gills, etc. Therefore, we will refer to the embryonic evidence: the laying of the gill slits, the development of the tail section and the division of the body into three sections (at the stage of the embryo). The correct answer is 136.
Task number 16.
Establish a correspondence between the example of the struggle for existence and the form to which this struggle refers: 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. The use of cattle as a habitat by a bull tapeworm
B. Competition between males for dominance
D. Displacement of the black rat by the gray rat
E. Fox hunting for mice-voles
form of wrestling
1. Intraspecific
2. Interspecies
Task number 17.
Choose three correct answers from 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. Reducing the number of owls, hedgehogs, foxes
2. Large harvest of spruce seeds
4. Cutting trees
5. Deep snow cover in winter
The correct answer is 126.
Task number 18.
Establish a correspondence between the 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. The constancy of the gas composition of the atmosphere
B. Change in the thickness of the ozone layer
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 number 19.
Establish the sequence of processes occurring during the reproduction and development of flowering plants, starting from the moment pollen is formed. 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 the seed from the ovule
5. Formation of generative and vegetative cells
Explanation: put the processes in the correct order. It all starts with the formation of generative and vegetative cells, then the pollen tube germinates, then the sperm enters the embryo sac, the germ cells merge, a triploid cell is formed, and finally the seed is formed from the ovule. The correct answer is 53124.
Task number 20.
Analyze the table. Fill in the blank 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 number 21.
Examine the graph of 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 the enzymatic reaction does not depend on the concentration of the enzyme
2. The rate of the enzymatic reaction significantly depends on the concentration of the enzyme
3. With an increase in the concentration of the enzyme, the reaction rate increases
Explanation: the graph shows that with an increase in the concentration of the enzyme, the rate of the reaction also increases, therefore, answer options 2 and 3 are suitable for describing this dependence. The correct answer is 23.
Task number 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 leguminous, and nodule bacteria live in symbiosis with leguminous plants, which fix molecular nitrogen and process it into a form absorbed by plants.
Task number 23.
Using the picture of the process of sexual reproduction of chlamydomonas, explain what is the essence of sexual reproduction and what is its difference from asexual. As a result of what process are gametes formed, what is their peculiarity? What number in the figure indicates the zygote? How is it different from gametes?
Explanation: during asexual reproduction, the content of a single-celled chlamydomonas is divided into 4 parts by meiosis, resulting in the formation of zoospores that perform the function of settling, then small cells grow to the size of the mother cell and divide by meiosis again. And in adverse conditions sexual reproduction, while biflagellated gametes (3) are formed in the mother cell, they leave the mother cell and merge in pairs with other individuals, the zygote is covered with a dense membrane and hibernates (survives unfavourable conditions), then the zygote divides, resulting 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 diploidy.
Task number 24.
Find errors in the given text. Indicate the numbers of sentences in which errors were made, correct them.
1. The body of the May beetle, covered with skin with a 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 mouth apparatus. 3. Metabolic products are excreted through the green glands. 4. Gas exchange is carried out directly through the walls of the trachea. 5. An open circulatory system consists of the heart and blood vessels.
Explanation: sentence 1 - the body of the May beetle is not covered with skin with a cuticle, but with a chitinous cover, and it is subdivided into the head, chest and abdomen. Proposal 2 - the mouth apparatus of the May language is not piercing, but gnawing. Proposal 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 number 25.
Explain the modification of which organ flowering plant is a head of cabbage.
Explanation: a head of cabbage is a modified bud, since in the second year of life an adult plant is formed from a head of cabbage, which has all the generative organs, that is, a flower and fruits with seeds. Also, a head of cabbage resembles a kidney in structure: a stump is a modified stem, thick leaves and rudimentary buds are located on it.
Task number 26.
Which ecosystem - a potato field or a meadow - has longer and more diverse food chains? Explain the answer.
Explanation: the potato field is an artificial ecosystem while the meadow is a natural ecosystem. Longer and more diverse food chains are in the natural ecosystem, since there are more plants and animals and microorganisms in the meadow and monoculture (potatoes) does not prevail.
Task number 27.
All types of RNA are synthesized on a DNA template. The fragment of the DNA molecule, on which the site of the central loop of tRNA is synthesized, has the following nucleotide sequence: CTTACGGGCATGGCT. Set the nucleotide sequence of the tRNA region that is synthesized on this fragment if the third triplet corresponds to the tRNA anticodon. Explain the answer.
Explanation: according to the given sequence of nucleotides in DNA, we find the sequence of tRNA nucleotides. We will use the principle of complementarity: A=U, G=C.
tRNA: GAAUGCCGUACCA
The third triplet - CCG corresponds to corresponds to mRNA - HGC.
Task number 28.
A man suffering from deafness and color blindness married a healthy woman. They had a son who was deaf and color blind and a daughter with good hearing but color blind. In humans, deafness is an autosomal, recessive trait, color blindness is a recessive trait, sex-linked. Make a scheme for solving the problem. Specify the possible phenotypes and genotypes of children in this family. Determine the probability of the birth of children suffering from both anomalies.
Explanation:
ah - deaf
Aa, AA - not deaf (carries the deafness gene in the second case)
X D X d - healthy mother (carries the gene for color blindness)
X d Y - sick father
AaX D X d x aaX 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)
aaX D X d - deaf, normal vision
AaX d X d - normal hearing, colorblind
aaX d X d - deaf, colorblind
Boys:
AaX D Y - normal hearing, normal vision (healthy child)
AaX 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 Model test tasks. USE 2017. 10 options.
The ratio of phenotypes in analyzing crosses is the same as the ratio of genotypes. This allows you to determine the unknown genotype of one of the parents. The second is a recessive individual according to the studied traits.
As is known, with complete dominance, dominant homozygotes ( AA) and heterozygotes ( aa) have the same phenotype. In other words, it is impossible to draw an unambiguous conclusion about the genotype based on the trait shown. In this case, analyzing cross comes to the rescue. Depending on which offspring are obtained, a conclusion is made about the unknown genotype of one of the parents, since against the background of the recessive alleles of the second parent, all the alleles of the first are manifested.
So heterozygous aa forms two types of gametes: A And a. The second recessive parent produces only gametes. a. As a result of their crossing, half of the offspring will have the genotype aa, second half - aa. That is, a splitting of 1:1 will be observed. The phenotypes will also be different, and their ratio will also be 1:1.
If the parent being tested was homozygous AA, it forms only one type of gametes - A. In this case, the result of analyzing crossing will be the uniformity of all descendants both in terms of genotype and phenotype. All of them will be heterozygotes aa. Ratio 1:0.
Thus, depending on the obtained phenotypes of the offspring, a conclusion is made about the genotype of the studied specimen.
More complex example This is a dihybrid cross. If the individual under study is dominant in two traits, then its genotype can be as AABB, and AaBb, as well as AABb or AaBB. All four variants with complete dominance have the same phenotypic manifestation. However, when analyzing crosses, each of these genotypes gives its own unique splitting.
1. If the genotype was AABB, then when crossing with a recessive individual aabb, all descendants will be uniform. Their genotype will AaBb, and the phenotype is identical to the studied parent.
2. In case AaBb and with an independent distribution of genes, gametes of four types are formed: AB, Ab, aB, ab. When crossed with gamete ab, four different genotypes will be obtained: AaBb, Aabb, aaBb, aabb. Their ratio will be 1: 1: 1: 1. The ratio of phenotypes will be the same, since in individuals AaBb two dominant traits will appear, in individuals Aabb- dominant trait only for the first gene, in individuals aaBb- dominant trait only for the second gene, individuals aabb are recessive for both genes.
3. If the genotype of the parent under study was AABb, then only two types of gametes are formed: AB And Ab. Test cross hybrids will have two genotypes AaBb And Aabb in a ratio of 1: 1. At the same time, according to the first trait, all individuals are uniform, and according to the second trait, splitting is observed, that is, half with a dominant trait, the second with a recessive one.
4. If the parent had a genotype AaBB, then the splitting will be observed only for the first gene. offspring genotypes - AaBb And aaBb.
Thus, depending on which of the four ratios of offspring is observed in the analyzing cross, a conclusion is made about the genotype of the individual under study.
The ratios obtained by linking genes are different. Linked genes are localized on the same chromosome and are inherited together during the formation of gametes. Often it is possible to determine not only the genotype, but also the linkage groups, including their presence or absence.
Suppose, as a result of a dihybrid analyzing cross, a ratio of phenotypes of 5: 2: 2: 5 was obtained. Let it be that for 5 specimens dominant in two traits there are 5 recessive in both traits and 2 each, which are dominant only in one of the traits. That is, the splitting by genotype will be as follows: 5 ( AaBb) : 2 (Aabb) : 2 (aaBb) : 5 (aabb).
The predominance in hybrids from analyzing crossing of some genotypes over others, and not the complete exclusion of the latter, indicates that incomplete linkage of genes is observed. There are always more gametes with the original linkage of genes than with a new one, which is formed as a result of crossing over.
Here, the initial linkage groups are genes A And B, while the genes a And b are on a different chromosome. Therefore gametes AB And ab more has been formed. Accordingly, as a result of the analyzing crossing of hybrids AaBb And aabb turned out to be more. If the coupling were complete, then only such hybrids would be obtained. However, due to crossing over, recombinant chromosomes with new linkage groups appeared in some cells - Ab And aB, which led to the emergence of a number of hybrids Aabb And aaBb.
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 dihybrid crossing, 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, high - low plant growth, purple - white flowers, etc. - show splitting in the offspring of the hybrid (in F 2) according to the phenotype in relation to 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 differ simultaneously in two pairs of traits. The mother "seed" plant had smooth seeds (the gene that determines this feature, we will conditionally denote B) and yellow seeds, the gene of which will be denoted A; Both of these traits are dominant. The paternal "pollen" plant had recessive traits: wrinkled and green seeds. Parental forms were homozygous for two pairs of traits or for two genes that determine them; the genotype of the mother plant can be designated as AABB, and that of the paternal plant as aabb. However, the distribution of features in parental 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 of the genes is located on a separate chromosome, then we should expect that mature eggs and sperm with a haploid set of chromosomes will have only one allele of each gene. Then the gametes of the mother plant should carry the alleles A and B (or a and B), and the paternal - a and b (or A and b). Fertilization of the AB egg with ab sperm will lead to the formation of a dihybrid zygote F 1 in the somatic cells of the hybrid embryo, a double set of chromosomes will be restored, and the hybrid will be heterozygous for two allelic pairs, i.e., diheterozygous AaBv. The same genotype is also formed in the case of the combination of gametes Ab and aB.
The hybrid pea seeds in our example, having the hereditary structure of AaBb in phenotype, as one would expect with complete dominance, will turn out to 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 in meiosis, four varieties of gametes are formed: AB, aB, Ab, ab. The form aabb gives only one sort of gametes - ab. With the equiprobable implementation of all combinations of gametes, four types of zygotes are formed in an equal ratio l AaBb: 1aaBb: 1Aabb: 1aabb. Analyzing crossing allows us to most quickly investigate the genotype of a hybrid organism according to the genes of interest to us.
Mendel also made a test cross 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 forms four varieties of gametes in equal proportions and, therefore, is heterozygous for both allelic pairs.
Cleavage by phenotype. Mendel's third law
In the self-pollination progeny of fifteen F1 dihybrid plants, Mendel produced 556 seeds, of which there were 315 smooth yellows, 101 wrinkled yellows, 108 smooth greens, and 32 wrinkled greens.
As we already know, in monohybrid crossing with complete dominance in F 2, splitting in the phenotype in a ratio of 3: 1, in the genotype 1: 2: 1 is observed. Imagine that each individual pair of Aa and Bb behaves in inheritance in the same way as in a monohybrid cross. There are grounds for such an assumption; Recall the known mechanism of 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 during 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, one can again apply the method of accounting for each pair of traits separately. To do this, all 556 seeds of the second generation must be divided into two classes: 1) in shape: 315 + 108 = 423 smooth and 101 + 32 = 133 wrinkled; 2) by color: 315 + 101 = 416 yellow and 108 + 32 = 140 green.
Knowing that the splitting for each pair of characters occurs in a ratio of 3: 1, we can say that out of the total number of seeds there should be 3/4 smooth and 1/4 wrinkled. Making the corresponding 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 dihybrid crossing for each pair of alleles there is a regular splitting in the ratio of 3:1.
To imagine how the combination of two pairs of alleles Aa and Bb is carried out simultaneously, as well as to establish the nature of splitting in F 2 while simultaneously taking into account both traits, one can go in two ways. The first way is to construct a Punnett lattice. The Punnett lattice makes it possible to establish all possible combinations of male and female gametes during fertilization, as well as to determine the phenotypes and genotypes of F 2 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 phenomena are independent, then the probability that they will occur at the same time 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 - 1/4. Therefore, the probability that the features of a smooth shape and yellow color of seeds will appear simultaneously is equal to the product of 3/4 X 3/4 \u003d 9/16, smooth shape and green color - 3/4 X 1/4 \u003d 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 the individual probabilities gives a ratio of cleavage classes by phenotype of 9/16:3/16:3/16:1/16, or 9:3:3:1.
Let us return to the example of splitting according to traits obtained by analyzing 556 F 2 seeds in Mendel's experiment. It is easy to verify that the seeds obtained by him were distributed among the classes of the combination of traits in a ratio close to expected. In order to calculate the theoretically expected numbers by class, the 556 seeds should be multiplied by 9/16, 3/16, 3/16, and 1/16, respectively. Therefore, the ratio of splitting classes by phenotype in F 2 dihybrid crossing with complete dominance fits into the formula 9: 3: 3: 1.
Now it should be clear why, when counting each pair of alternative traits 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 the Punnett grid, in which the 16 genotypes described above are divided by phenotype into four classes in the same ratio of 9:3:3:1.
Thus, in dihybrid crossing, each pair of characters, when splitting in the offspring, behaves in the same way as in monohybrid crossing, i.e., independently of the other pair of primates.
Based on the simultaneous analysis of the inheritance of several pairs of alternative traits, 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 the characters subjected to experiments, the following statement is equally valid: the descendants of hybrids that combine several essentially different characters are members of a combination series in which the series of development of each pair of differing characters are connected. This simultaneously proves that the behavior in the hybrid combination of each pair of differing characters is independent of other differences in both parent plants.
And then Mendel formulates the actual principle of the 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.”
Cleavage by genotype
The formula 9:3:3:1 expresses the ratio of phenotypic segregation in F 2 in a dihybrid cross.
It is necessary to analyze the same cleavage by genotype. Obviously, in the case of complete dominance, this can only be done by crossing individuals of all 16 genotypes, which can result from a combination of four varieties of female and male gametes with a homozygous recessive form aabb. Since each pair of alleles behaves independently during splitting according to the phenotype, the splitting according to the genotype will also manifest itself in accordance with the same pattern, but in different proportions.
By analyzing the F 2 genotypes on 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 crossing, the splitting according to the genotype corresponds to 1AA: 2Aa: 1aa for one pair of alleles and 1BB: 2Bb: 1bb for the other, one can calculate the probability of the appearance of genotypes of different classes in a dihybrid cross.
The probability of the appearance of the AA genotype is 1/4. Accordingly, for Aa - 1/2 and for aa - 1/4. The same will be for another allelic pair: BB - 1/4, Bb - 1/2, bb - 1/4. By multiplying two probabilities, you can get all the classes of splitting by genotype. As a result of this calculation, the same 9 classes of splitting according to the 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 monohybrid crossing, the number of splitting classes by phenotype is 2 (3: 1), and by genotype - 3 (1: 2: 1); in dihybrid crossing, the number of phenotypic splitting classes is 4, and genotypic - 9. Therefore, in the case of two genes, the number of classes corresponds to the phenotype 2 2, and the genotype - 3 2. 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 of various genotypes and phenotypes. With complete dominance, homozygous forms are indistinguishable by phenotype from heterozygous ones; so, AABB is indistinguishable from AaBb, AABb, AaBB. For purposes of brevity, when writing, similar phenotypes of homozygotes and heterozygotes are sometimes denoted by the phenotypic radical A-B-. By substituting different alleles in place of the dash in such a radical, similar phenotypes can be obtained (for example, for the A-bb radical, similar phenotypes will be in the AAbb and Aabb genotypes).
Numerous experiments have confirmed the correctness of the regularities established by Mendel. At the same time, facts appeared showing that the numerical ratios obtained by Mendel during the splitting of the hybrid generation were not always observed. This indicated that the relationships between genes and traits are more complex. It turned out: the same gene can influence the development of several traits; the same trait can develop under the influence of many genes.
It should be noted that the interaction of genes is of a biochemical nature, that is, it is not genes that interact with each other, but their products. The product of a eukaryotic gene may be either a polypeptide, or tRNA, or rRNA.
TYPES OF INTERACTION OF ALLELIC GENES
There are complete dominance, incomplete dominance, codominance, allelic exclusion.
allelic genes called genes located in identical loci of homologous chromosomes. A gene can have one, two or more molecular forms. The appearance of the second and subsequent molecular forms is a consequence of gene mutation. If a gene has three or more molecular forms, it is called multiple allelism. Of the entire set of molecular forms, only two can be present in one organism, which is explained by the pairing of chromosomes.
Complete dominance
Complete dominance- this is a type of interaction of allelic genes, in which the phenotype of heterozygotes does not differ from the phenotype of homozygotes according to the dominant, that is, the product of the dominant gene is present in the phenotype of heterozygotes. Complete dominance is widespread in nature, taking place during inheritance, for example, the color and shape of pea seeds, eye color and hair color in humans, the Rh antigen, and many others. others
The presence of the Rh antigen (Rh factor) of erythrocytes is determined by the dominant Rh gene. That is, the genotype of a Rh-positive person can be of two types: either RhRh, or Rhrh; the genotype of a Rh-negative person is rhrh. If, for example, the mother is Rh-negative, and the father is Rh-positive and heterozygous for this trait, then with this type of marriage, both a Rh-positive and a Rh-negative child can be born with the same probability.
Rh-positive fetus and Rh-negative mother may have an Rh conflict.
This is the name of the type of interaction of allelic genes, in which the phenotype of heterozygotes differs both from the phenotype of homozygotes for the dominant and from the phenotype of homozygotes for the recessive and has an average (intermediate) value between them. It takes place during the inheritance of the color of the perianth of the night beauty, snapdragon, the color of the coat of guinea pigs, etc.
Mendel himself encountered incomplete dominance when he crossed a large-leaved pea variety with a small-leaved one. Hybrids of the first generation did not repeat the trait of any of the parental plants; they had leaves of medium size.
When crossing homozygous red-fruited and white-fruited varieties of strawberries, the entire first generation of hybrids has pink fruits. When these hybrids are crossed with each other, we get: by phenotype - 1/4 red-fruited, 2/4 pink-fruited and 1/4 white-fruited plants, by genotype - 1/4 AA, 1/2 Aa, 1/4 aa (and by phenotype, and genotype ratio 1:2:1). Correspondence of splitting by genotype to splitting by phenotype is characteristic of incomplete dominance, since heterozygotes differ phenotypically from homozygotes.
Codominance
Codominance- a type of interaction of allelic genes, in which the phenotype of heterozygotes differs both from the phenotype of homozygotes for the dominant and from the phenotype of homozygotes for the recessive, and the products of both genes are present in the phenotype of heterozygotes. It takes place during the formation, for example, of the IV blood group of the system (AB0) in humans.
In order to imagine how the inheritance of blood groups occurs in a person, you can look at the birth of children with which blood type is possible for parents who have one - the second, the other - the third blood group and are heterozygous for this trait.
| R | ♀I A i 0 II(A) |
× | ♂I B i 0 III(B) |
||
| Types of gametes | I A | i 0 | I B | i 0 | |
| F | i 0 i 0 I(0) 25% |
I A i 0 II(A) 25% |
I B i 0 III(B) 25% |
I A I B IV (AB) 25% |
|
Allelic exclusion
Allelic exclusion the absence or inactivation of one of a pair of genes is called; in this case, the product of another gene is present in the phenotype (hemizygosity, deletion, heterochromatization of the chromosome region in which the desired gene is located).
TYPES OF INTERACTION OF NON-ALLELIC GENES
Complementarity, epistasis, polymerism.
Non-allelic genes- genes located either in non-identical loci of homologous chromosomes, or in different pairs of homologous chromosomes.
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