Lethal Genes in the homozygous state can cause the death of offspring even before birth. At the same time, other genotypes survive. As with codominance, in this case three phenotypic classes are formed, but one of the phenotypes does not appear, since individuals carrying lethal genes die. Therefore, the splitting in the offspring differs from Mendelian.
Task 8-1
One of the breeds of chickens is distinguished by shortened legs - a dominant feature (such chickens do not break gardens). This gene also affects the length of the beak. At the same time, in homozygous dominant chickens, the beak is so small that they cannot hatch from the egg and die. In the incubator of a farm that breeds only short-legged hens (long-legged hens are not allowed to breed and are sent for sale), 3,000 chicks were obtained. How many of them were short-legged?
- All hens in the incubator were heterozygous (since homozygous short-legged hens die before birth).
- When heterozygous individuals are crossed with each other, the following offspring are formed:
25% of individuals with the AA genotype - die before birth,
50% of individuals with the Aa genotype are short-legged, 25% of individuals with the aa genotype are long-legged.
That is, short-legged individuals were 2/3 of all surviving offspring - about 2000 pieces.
Task 8-2
When crossing black mice with each other, black offspring are always obtained. When yellow mice are crossed, one third turns out to be black, and two thirds turn out to be yellow. How can these results be explained?
- Black mice are homozygous because all their offspring are uniform.
- Yellow mice are heterozygous, as segregation is observed in their offspring. Since heterozygous individuals carry a dominant trait, the yellow color is dominant.
- Yellow mice, when crossed with each other, never give only yellow offspring. In addition, the segregation in their offspring differs from Mendelian. This suggests that individuals homozygous for the dominant do not survive. Crossing analysis confirms this assumption.
Crossing scheme
Task 8-3
What will happen if we assume that a lethal mutation occurs in an organism, in which only heterozygous individuals die, while homozygous individuals remain viable?
Task 8-4
In mice, the short-tailed gene in the dominant state is lethal, causing the death of the embryo on early stages development. Heterozygotes have shorter tails than normal individuals. Determine the phenotypes and genotypes of offspring arising from crossing long-tailed and short-tailed mice.
Task 8-5
When crossing mirror carps with each other already in first generation splitting was observed: 152 offspring were mirror-like and 78 with normal scales. How to explain these results? What offspring will result from crossing a mirror carp with an ordinary one?
Option 4
In carrots, the orange color of the root crop dominates over the yellow. A homozygous plant with an orange root was crossed with a plant with a yellow root. In the first generation received 15 plants. They were self-pollinated and 120 plants were obtained in the second generation.
1. How various types can a gamete form a parent plant with an orange root?
a) 1;
b) 2;
at 3;
d) 4.
2. How many plants with yellow roots will grow in the second generation?
a) 120;
b) 90;
c) 60;
d) 30.
3. How many heterozygous plants will be in the second generation?
a) 120;
b) 90;
c) 60;
d) 30.
4. How many dominant homozygous plants will be in the second generation?
a) 120;
b) 90;
c) 60;
d) 30.
5. How many plants from the second generation will have an orange root crop?
a) 120;
b) 90;
c) 60;
d) 30.
Answers
Option 1: 1 - b; 2 - in; 3 - a; 4 - b; 5 - c.
Option 2: 1 - b; 2 - a; 3 - in; 4 - in; 5 B.
Option 3: 1 - d; 2 - a; 3 - b; 4 - b; 5 - a.
Option 4: 1 - a; 2 - d; 3 - in; 4 - d; 5 B.
Test control No. 2. Solving problems for dihybrid crossing
Option 1
In peas, high growth dominates over dwarf ones, and the smooth shape of seeds dominates over wrinkled ones. A homozygous tall plant with wrinkled seeds was crossed with a heterozygous plant with smooth seeds and dwarf growth. Received 640 plants.
1. How many will be among the hybrids of tall plants with smooth seeds?
a) 0;
b) 160;
c) 640;
d) 320.
2. How different types can a gamete produce a parent plant with smooth seeds and dwarf growth?
a) 1;
b) 2;
at 3;
d) 4.
3. How many of the hybrids will be undersized plants with smooth seeds?
a) 320;
b) 640;
c) 160;
d) 0.
4. How many different genotypes will there be among all hybrid plants?
a) 1;
b) 2;
at 3;
d) 4.
5. How many hybrid plants will be tall?
a) 160;
b) 0;
c) 640;
d) 320.
Option 2
In chickens, feathered legs dominate over non-feathered ones, and a pea-shaped comb dominates over simple ones. Diheterozygous chickens and homozygous roosters with simple crests and feathered legs were crossed. Received 192 chickens.
1. How many types of gametes does a chicken produce?
a) 1;
b) 2;
at 3;
d) 4.
2. How many different genotypes will chickens have?
a) 1;
b) 2;
at 4;
d) 16.
3. How many chickens will have feathered legs?
a) 192;
b) 144;
c) 96;
d) 48.
4. How many chickens will have feathered legs and simple combs?
a) 192;
b) 144;
c) 96;
d) 48.
5. How many different phenotypes will the hybrids have?
a) 1;
b) 2;
at 3;
d) 4.
Option 3
In chickens, shortened legs dominate over normal ones, and a rose-shaped comb dominates over simple ones. As a result of crossing a chicken heterozygous for these traits and a rooster with normal legs and a simple comb, 80 chickens were obtained.
1. How many different types of gametes can a chicken produce?
a) 1;
b) 2;
at 3;
d) 4.
2. How many different types of gametes can a rooster have?
a) 1;
b) 2;
at 3;
d) 4.
3. How many different genotypes will the hybrids have?
a) 4;
b) 8;
at 12;
d) 16.
4. How many chicks will have normal legs and a simple comb?
a) 80;
b) 60;
c) 40;
d) 20.
5. How many chicks will have pink combs?
a) 80;
b) 60;
c) 40;
d) 20.
Option 4
In cows, hornless (hornless) dominates over horn, and black suit over red. A purebred komologo bull of black color was crossed with diheterozygous cows. Received 64 calves.
1. How many different types of gametes does a bull form?
a) 1;
b) 2;
at 3;
d) 4.
2. How many different types of gametes does a cow produce?
a) 1;
b) 2;
at 3;
d) 4.
3. How many different phenotypes are produced by this cross?
a) 1;
b) 4;
at 8;
d) 16.
4. How many different genotypes will the calves have?
a) 1;
b) 2;
at 3;
d) 4.
5. How many polled black diheterozygous calves will there be?
a) 64;
b) 48;
c) 32;
d) 16.
Answers
Option 1: 1 - d; 2 - b; 3 - d; 4 - b; 5 - c.
Option 2: 1 - d; 2 - in; 3 - a; 4 - in; 5 B.
Option 3: 1 - d; 2 - a; 3 - a; 4 - d; 5 - c.
Option 4: 1 - a; 2 - d; 3 - a; 4 - d; 5 - g.
Tasks for monohybrid crossing
Task #1
What pairs are most advantageous to cross to obtain platinum foxes, if platinum dominates over silver, but in the homozygous state, the platinum gene causes the death of the embryo?
Answer: It is most profitable to cross silver and platinum heterozygous foxes.
Task #2
When crossing two white pumpkins in the first generation, 3/4 of the plants were white and 1/4 were yellow. What are the genotypes of the parents if white is dominant over yellow?
Answer: parent plants are heterozygous.
Tasks for dihybrid crossing
Task #3
If a woman with freckles (dominant trait) and wavy hair (dominant trait), whose father had straight hair and no freckles, marries a man with freckles and straight hair (both of his parents with the same traits), what could be do they have children?
Answer: all children in this family will have freckles, and the probability of their birth with straight and wavy hair is 50% each.
Task #4
What are the genotypes of the parent plants if, when crossing red tomatoes (dominant trait) pear-shaped (recessive trait) with yellow spherical ones, it turned out: 25% red spherical, 25% red pear-shaped, 25% yellow spherical, 25% yellow pear-shaped?
Answer: genotypes of parent plants Aabb and aaBb.
Tasks for incomplete dominance
Task number 5
When purebred white chickens are crossed with each other, the offspring turns out to be white, and when black chickens are crossed, they are black. The offspring from white and black individuals turns out to be motley. What kind of plumage will the descendants of a white rooster and a motley hen have?
Answer: half of the chicks will be white and half pied.
Task number 6
Plants of red-fruited strawberries, when crossed with each other, always give offspring with red berries, and plants of white-fruited strawberries - with white ones. As a result of crossing these varieties with each other, pink berries are obtained. What offspring will arise when hybrids with pink berries are crossed?
Answer: half of the offspring will be with pink berries and 25% each with white and red.
Tasks for the inheritance of blood groups
Task number 7
What blood types can children have if both parents have an IV blood group?
Answer: the probability of having children with IV blood group - 50%, with II and III - 25% each.
Task number 8
Is it possible to transfuse blood to a child from a mother if she has a blood type AB, and the father 0 ?
Answer: it is forbidden.
Task number 9
The boy has an IV blood type, and his sister has an I blood type. What are the blood types of their parents?
Answer: II and III.
Task number 10
Two boys (X and Y) were mixed up in the maternity hospital. X has I blood type, Y has II. Parents of one of them with I and IV blood groups, and the other with I and III blood groups. Who is whose son?
Answer: X has parents with I and III groups, Y has parents with I and IV.
Sex-linked inheritance problems
Task number 11
In parrots, the sex-linked dominant gene determines the green color of the plumage, and the recessive gene determines brown. A green heterozygous male is crossed with a brown female. What will the chicks be like?
Answer: half of the males and females will be green, half will be brown.
Task number 12
In Drosophila, the dominant gene for red eyes and the recessive gene for white eyes are located on the X chromosome. What eye color can be expected in first-generation hybrids if a heterozygous red-eyed female and a white-eyed male are crossed?
Answer: the probability of the birth of males and females with red and white eyes - 50% each.
Task number 13
Husbands and wives who are healthy for color blindness have:
– a color-blind son with a healthy daughter;
- a healthy daughter who has 2 sons: one is color blind and the other is healthy;
– a healthy daughter who has five healthy sons.
What are the genotypes of this husband and wife?
Answer: genotypes of parents X D X d, X D Y.
Task number 14
A tortoiseshell cat brought kittens in black, red and tortoiseshell. Is it possible to determine whether a black or red cat was the father of these kittens?
Answer: it is forbidden.
Combined tasks
Task number 15
In cattle, the polled gene dominates the horned gene, and the roan coat color is formed as an intermediate trait when white and red animals are crossed. Determine the probability of the birth of calves similar to the parents from crossing a heterozygous roan bull with a white horned cow.
Answer: the probability of having calves similar to their parents is 25% each.
Task number 16
From crossing two varieties of strawberries (one with a mustache and red berries, the other beardless with white berries), in the first generation all plants were with pink berries and a mustache. Is it possible to breed a beardless variety with pink berries by backcrossing?
Answer: possible, with a probability of 25% when crossing hybrid plants with a beardless parent plant that has white berries.
Task number 17
A man with Rh-negative blood of group IV married a woman with Rh-positive blood of group II (her father has Rh-negative blood of group I). There are 2 children in the family: with Rh-negative blood Group III and with Rh-positive blood of group I. Which child in this family is adopted if the presence of the Rh factor antigen in the erythrocytes is due to the dominant gene?
Answer: adopted child with I blood group.
Task number 18
In one family, four children were born to brown-eyed parents: two blue-eyed with I and IV blood groups, two - brown-eyed with II and IV blood groups. Determine the probability of the birth of the next child brown-eyed with I blood group.
Answer: genotype of a brown-eyed child with I blood group A– I 0 I 0, the probability of having such a child is 3/16, i.e. 18.75%.
Task number 19
A man with blue eyes and normal vision married a woman with brown eyes and normal vision (her relatives all had brown eyes, and her brother was colorblind). What are the children of this marriage?
Answer: all children will be brown-eyed, all daughters with normal vision, and the probability of having sons with color blindness is 50%.
Task number 20
In canaries, the sex-linked dominant gene determines the green color of the plumage, and the recessive gene determines brown. The presence of a crest depends on an autosomal dominant gene, its absence depends on an autosomal recessive gene. Both parents are green with tufts. They had 2 chicks: a green male with a crest and a brown female without a crest. Determine the genotypes of the parents.
Answer: P: ♀ X Z Y Ah; ♂ X Z X K Ah.
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 grant can be used for in-depth study 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 by a capital letter of the Latin alphabet ( BUT, AT, FROM etc.), or recessive, denoted by a lowercase letter ( a, b, With 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 ( a).
Organisms that have different alleles for the same gene Ah), are called heterozygous. They give two varieties of gametes ( BUT and a).
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 ( AT + b) (AT + 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 a(allele a 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
a- 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
a- 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 combined in all possible combinations. In the case of a 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 a ratio of 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
a- yellow fruits
AT- 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
No. 4. When crossing with each other, red-fruited strawberry plants always give offspring with red fruits, and white-fruited plants with white ones. By crossing these varieties, plants are obtained that produce pink fruits. What offspring will arise when strawberry plants with pink fruits are crossed with each other, assuming monogenic control of the fruit color trait? What offspring will be obtained in backcrosses of pink-fruited plants with the original parental forms?
Solution:
No. 5. When crossing rye plants with purple (due to the presence of anthocyanin) and green (lack of pigment) seedlings in F2, 4584 plants with purple and 1501 plants with green seedlings were obtained. Explain splitting. Determine the genotypes of the original plants. What phenotype did the F1 plants have?
Solution:
No. 35. From crossing a silver-sable male mink with normal dark females, 345 silver-sable and 325 dark minks were obtained in the offspring. Litter size averaged 5.11 puppies. When crossing ……………..
Solution:
No. 38. One breed of chickens is distinguished by shortened legs; such chickens do not tear gardens. This feature is dominant. The gene controlling it also causes shortening of the beak at the same time. At the same time, in homozygous chickens, the beak is so small that they are not able to …………………..
Solution:
No. 43. When breeding black and white hens in the offspring, consisting of 42 chickens, there were 20 black and white, 12 black and 10 pure white. How can this be explained? How is black and white plumage inherited? What cross should be made to get only black and white chickens?
Solution:
No. 62. In the maternity hospital, four babies were born in one night, who had blood types 0, A, B and AB. The blood groups of the four parental pairs were: I pair - 0 and 0; II pair - AB and 0; III pair - A and B; IV pair - B and B. Four babies can be distributed with complete certainty among parental pairs. How to do it? What are the genotypes of all parents and children?
Solution:
No. 1. When crossing two varieties of tomatoes, one of which had yellow and the other red fruits, F1 hybrids had red fruits, and in the second generation - 58 red and yellow fruits. Explain splitting. What are the genotypes of …………….
Solution:
No. 304. In common wheat, the grain color is determined by the interaction of several genes according to the type of non-cumulative polymer. In this case, a different number of genes can be involved. In one of the crosses of the red and white grain lines in F2, a splitting of 63/64 red grains was observed: 1/64 white grains. How many genes………
Solution:
No. 183. In shepherd's purse, the shape of the fetus depends on two pairs of polymeric genes. A plant with triangular fruits is crossed with a plant with ovoid fruits. In the offspring, ¾ of the plants had triangular fruits and ¼ - ovoid. Determine the genotypes of the parents. What happens from self-pollination of a parent plant with triangular fruits?
Solution:
No. 245. From crossing yellow parakeets with blue in the first generation, all the descendants turned out to be green, and in the second - 56 green, 18 blue, 20 yellow and 6 white. Explain splitting, determine the genotypes of birds of all colors.
Solution:
No. 249 Color variations in the color of horses are determined various combinations alleles of three genes: aBE - bay, ABE - savrasai, Abe - nightingale, aBe - brown, abe - red, AbE - bulano-savrasai, …………..
Solution:
No. 267. From crossing white and blue rabbits, 28 black rabbits were obtained in F1, and 67 black, 27 blue and 34 white in F2. How are black, blue, and white coat colors inherited in rabbits? Explain splitting. Determine the genotypes of parents and offspring.
Solution:
No. 283. Two varieties of flax are crossed, one of which had a pink flower color and normal petals, and the other had a white flower color and normal petals. In F1, the color of the flowers is pink, the petals are normal. Splitting occurred in F2: 42 pink normals, ……….
Solution:
You can be sure of the quality of this work. Part of the control is shown below:
At incomplete dominance heterozygotes do not show any of the traits of the parents. At intermediate inheritance hybrids carry an average expression of traits.
At co-dominance heterozygotes exhibit both parental traits. An example of intermediate inheritance is the inheritance of the color of the fruits of strawberries or flowers of the night beauty, codominance - the inheritance of the roan suit in cattle.
Task 3-1
When crossing plants of red-fruited strawberries with each other, plants with red berries are always obtained, and white-fruited ones with white ones. As a result of crossing both varieties, pink berries are obtained. What offspring will result from the pollination of red-fruited strawberries with the pollen of a plant with pink berries?
- Plants with red and white fruits, when crossed with each other, did not give splitting in the offspring. This indicates that they are homozygous.
- Crossing homozygous individuals that differ in phenotype leads to the formation of a new phenotype in heterozygotes (pink color of fruits). This indicates that in this case the phenomenon of intermediate inheritance is observed.
- Thus, plants with pink fruits are heterozygous, while those with white and red fruits are homozygous.
Crossing scheme
AA - red fruits, aa - white fruits, Aa - pink fruits.
50% of the plants will have red and 50% pink fruits.
Task 3-2
In the "night beauty" plant, the inheritance of flower color is carried out according to an intermediate type. Homozygous organisms have red or white flowers, while heterozygous organisms have pink flowers. When two plants were crossed, half of the hybrids had pink and half had white flowers. Determine the genotypes and phenotypes of the parents.
Task 3-3
The calyx shape of strawberries can be normal and leaf-shaped. In heterozygotes, the cups are intermediate in shape between normal and leaf-shaped. Determine possible genotypes and phenotypes of offspring from crossing two plants with an intermediate calyx shape.
Task 3-4
Kohinoor minks (light, with a black cross on the back) are obtained by crossing white minks with dark ones. Crossing white minks with each other always gives white offspring, and crossing dark minks always gives dark ones. What offspring will be obtained from crossing cochinur minks with each other? What offspring will result from crossing kohinoor minks with whites?
Task 3-5
They crossed a motley rooster and a chicken. The result was 26 variegated, 12 black and 13 white chickens. What trait is dominant? How is the color of plumage inherited in this breed of chickens?
Task 3-6
In one Japanese variety of beans, when self-pollinated by a plant grown from a light spotted seed, it was obtained: 1/4 - dark spotted seeds, 1/2 - light spotted and 1/4 - seeds without spots. What offspring will result from crossing a plant with dark spotted seeds with a plant that has seeds without spots?
- The presence of segregation in the offspring indicates that the original plant was heterozygous.
- The presence of three classes of phenotypes in the offspring suggests that in this case codominance takes place. Segregation by phenotype in a ratio of 1:2:1 confirms this assumption.
When a plant with dark spotted seeds is crossed with a plant without spots (both forms are homozygous), all offspring will be uniform and will have light spotted seeds.
Task 3-7
In cows, the genes for red (R) and white (r) color are codominant to each other. Heterozygous individuals (Rr) - roans. The farmer bought a herd of roan cows and decided to keep only them and sell the red and white ones. What color bull should he buy in order to sell as many calves as possible?
Task 3-8
By crossing radish plants with oval roots, 68 plants with round, 138 with oval and 71 with long roots were obtained. How is the root shape inherited in radishes? What offspring will be obtained from crossing plants with oval and round roots?
Task 3-9
When strawberries with pink fruits were crossed with each other, 25% of individuals producing white fruits and 25% of plants with red fruits turned out to be in the offspring. The rest of the plants had pink fruits. Explain your results. What is the genotype of the examined individuals?
