monohybrid cross

Task 6 Dihybrid cross

1. In an individual with the aabb genotype, gametes are formed

Task 6 No. 2308 Explanation.

One chromosome from each pair of allelic chromosomes diverge into a gamete. Law of purity of gametes.

Answer: ab

Answer: ab

ab

Source: RESHU USE

2. How many types of gametes are formed in diheterozygous pea plants during dihybrid crossing (genes do not form a linkage group)? Write down a number for your answer.

Task 6 No. 2311 Explanation.

AaBb – AB; Ab; aB; ab.

Answer: 4

Source: RESHU USE

3. In dogs, black coat (A) dominates over brown (a), and short legs (B) dominate over normal leg length (b). Write down the genotype of a black short-legged dog that is heterozygous for leg length only.

Task 6 No. 2401 Explanation.

For a dog to be black, a combination of Aa or AA in the genotype is necessary (variant aa corresponds to the brown coat color). In order for a dog to be short-legged, a combination of BB or BB in the genotype is necessary (the variant BB corresponds to limbs of normal length). Since heterozygosity for only one trait is required, the required genotype is AABb.

Answer: AABb

AABb

Source: RESHU USE

4. When crossing yellow (A) smooth (B) (digomozygote) and green (a) wrinkled (b) peas in F1, all yellow smooth peas were obtained. Determine the genotype of pea seeds in F1 Explanation. According to Mendel’s first law, only when homozygous individuals are crossed, uniformity is obtained in the first generation.

R: AABB x aabb

G: AB ab

F: AaBb – 100%

AABB – yellow smooth

aabb – green wrinkled

Answer: AaBb|AaBv

AaBb|AaBv

Source: RESHU USE

5. When homozygous tomato plants with round red fruits are crossed with pear-shaped yellow fruits (red color – A, yellow – a, round shape – B, pear-shaped – b), offspring with the genotype

Explanation.

According to Mendel’s first law, only when homozygous individuals are crossed, uniformity is obtained in the first generation.

R: AABB x aabb

G: AB ab

F: AaBb – 100% (BbAa)

AABB – red round (BBAA)

aabb – yellow pear-shaped (bbaa)

Answer: AaBb

Answer: AaBb

AaBb

Source: RESHU USE

6. When crossing individuals with AaBb genotypes with AaBb (genes are not linked), the proportion (%) of heterozygotes for both alleles (diheterozygotes) in the offspring will be Explanation.

According to Mendel’s third law, when two diheterozygotes are crossed, the splitting occurs according to the phenotype 9: 3: 3: 1, 16 descendants are formed according to the genotype, and according to the Punnett lattice, the proportion of diheterozygotes is 25%.

Note.

“Heterozygous for both alleles” is the same as DIHETEROZYGOTE.

Answer: 25|25%

25|25%

Source: RESHU USE

7. The rule of uniformity of the first generation will appear if the genotype of one of the parents is aabb, and the other

Task 6 No. 2408 Explanation.

According to the first law of Mendel, homozygous individuals of pure lines are crossed.

Answer: AABB

Answer: AABB|AABB

AABB|AABB

Source: RESHU USE

8. Indicate the genotype of a person if, according to the phenotype, he is fair-haired and blue-eyed (recessive traits)

Task 6 No. 2412 Explanation.

Both traits are recessive, which means that according to the genotype it is a dihomozygote for a recessive trait.

Answer: abb

Answer: abb

aabb

Source: RESHU USE

9. When homozygous tomato plants with red (A) round (B) fruits and plants with yellow (a) pear-shaped (b) fruits in F2 are crossed, phenotypic splitting occurs in the ratio (color genes and fruit shape are located in different pairs of chromosomes). Write your answer as a sequence of numbers showing the ratio of the resulting phenotypes, in descending order

Task 6 No. 2414 Explanation.

In F1, the uniformity of the first generation is manifested, the plant genotype will be heterozygous, in the second generation, when crossing diheterozygotes, splitting occurs 9: 3: 3: 1 according to Mendel’s third law.

Answer: 9:3:3:1

Answer: 9:3:3:1|9 : 3 : 3 : 1|9331

9:3:3:1|9 : 3 : 3 : 1|9331

Source: RESHU USE

10. When heterozygous tomato plants with red and round fruits are crossed with individuals recessive in both traits (red A and round B are dominant traits), offspring with genotypes AaBb, aaBb, Aabb, aabb will appear in the ratio. Write down the answer in the form of a sequence of numbers showing the ratio of the resulting phenotypes, in descending order.

Task 6 No. 2416 Explanation.

R: AaBb x aabb

G: AB ab

Ab

aB

ab

F: AaBb; aabb, aabb, aabb

1 : 1 : 1 : 1

Answer: 1:1:1:1|1 : 1 : 1 : 1|1111

1:1:1:1|1 : 1 : 1 : 1|1111

Source: RESHU USE

11. What gametes do individuals with the aaBB genotype have?

Task 6 No. 2419 Explanation.

According to the law of gamete purity, one of each pair of allelic genes goes into the gamete.

Answer: b.

Answer: aB | aB

aB|aB

Source: RESHU USE

12. The genotype of one of the parents will be AaBb if, during analyzing dihybrid crossing and independent inheritance of traits, splitting in the phenotype in the offspring is observed in the ratio. Write down the answer as a sequence of numbers showing the ratio of the resulting phenotypes, in descending order . Explanation.

An analysis cross is a cross with an individual with the aabb genotype.

R: AaBb x aabb

G: AB ab

Ab

aB

ab

F: AaBb; aabb, aabb, aabb

1 : 1 : 1 : 1

Answer: 1 : 1 : 1 : 1|1:1:1:1|1111

1 : 1 : 1 : 1|1:1:1:1|1111

Source: RESHU USE

13. Determine the genotype of an individual of a yellow curly pumpkin, if, during its self-pollination in F1, the splitting of traits according to the phenotype corresponded to 9:3:3:1

Task 6 No. 2422 Explanation.

According to Mendel’s 3rd law, when crossing diheterozygotes AaBb x Aa Bb, 16 offspring are formed, which give a splitting according to the phenotype 9: 3: 3: 1

Answer: AaBb

AaBb

Source: RESHU USE

14. With dihybrid crossing and independent inheritance of traits in parents with AABb and aabb genotypes, splitting in the ratio is observed in the offspring.

Write down the answer in the form of a sequence of numbers showing the ratio of the resulting phenotypes, in descending order.

Task 6 No. 2423 Explanation.

R: AABb x aabb

G: AB ab

Ab

F: AaBb; Aabb

eleven

Answer: 1:1|1 : 1|11

1:1|1 : 1|11

Source: RESHU USE

15. What genotype does the offspring have in F1 when tomato plants are crossed with genotypes AABB and AABB?

Task 6 No. 2426 Explanation.

R: AAvv x aaBB

G: Ab aB

F: AaBb – 100%

Answer: AaBb | AaBv

AaBb|AaBv

Source: RESHU USE

16. In dihybrid crossing (unlinked inheritance) of individuals with dominant AABB and recessive aabb traits in F2, phenotypic splitting occurs in the ratio. Write your answer as a sequence of numbers showing the ratio of the resulting phenotypes, in descending order

Task 6 No. 2427 Explanation. In F1, the uniformity of the first generation is manifested, the plant genotype will be heterozygous, in the second generation, when crossing diheterozygotes, splitting occurs 9: 3: 3: 1 according to Mendel’s third law.

Answer: 9:3:3:1

Answer: 9:3:3:1|9 : 3 : 3 : 1|9331

9:3:3:1|9 : 3 : 3 : 1|9331

Source: RESHU USE

17. What will be the ratio of the splitting of traits by phenotype in the offspring obtained from crossing a diheterozygous black, furry rabbit AaBv with a white, smooth-haired rabbit aavb

Task 6 No. 2428 Explanation.

R: AaBb x aabb

G: AB ab

Ab

aB

ab

F: AaBb; aabb, aabb, aabb

1 : 1 : 1 : 1

By phenotype: 1 part black hairy : 1 part black smooth : 1 part white hairy : 1 part white smooth.

Genotypic segregation corresponds to phenotypic segregation.

Answer: 1 : 1 : 1 : 1|1:1:1:1|1111

1 : 1 : 1 : 1|1:1:1:1|1111

Source: RESHU USE

18. What is the ratio of genotypes in offspring obtained from crossing individuals with AaBv x AABB genotypes?

Explanation.

The first individual gives 4 types of gametes: AB, Av, aB, av; the second – 1 type of gametes: AB. This means that the ratio of genotypes in the offspring is 1:1:1:1 (AABB, AABv, AaBB, AaBv).

Answer: 1 : 1 : 1 : 1|1:1:1:1|1111

1 : 1 : 1 : 1|1:1:1:1|1111

Source: Yandex: USE training work in biology. Option 2.

19. A woman with blond (a) straight (b) hair married a man with dark curly hair. Write down the genotype of their child who has dark straight hair.

Task 6 No. 12575 Explanation.

Mother’s genotype ♀ aabb; father’s genotype ♂ A_Bb

Curly hair – intermediate inheritance: bb – straight hair; Bb – curly (wavy); BB – shallow.

From the mother, the child receives a gamete – ab; from father – Ab.

Genotype of a child with dark straight hair: Aabb

Answer: Abb

Answer: Abb

Aabb

Source: Unified State Examination in Biology 05/30/2013. main wave. Far East. Option 2.

20. In guinea pigs, black hair dominates over white, and long – over short. Determine the genotype of a diheterozygous individual.

Task 6 No. 12703 Explanation.

Diheterozygote (differs in two pairs of traits) – AaBb

Answer: AaBb

AaBb

Source: Unified State Examination in Biology 05/30/2013. main wave. Far East. Option 5.

21. How many types of gametes does a diheterozygous individual form with full linkage of the studied genes

Task 6 No. 21012 Explanation.

With full linkage, a diheterozygous individual forms two types of gametes:

AB/ab/

Answer: 2

Source: Typical test tasks in biology, edited by G. S. Kalinov, T. V. Mazyarkin. 10 options for tasks. 2017

22. In cattle, black (A) dominates over red (a), polled (B) – over horned (b). When crossing black polled cows with a red horned bull, all the offspring turned out to be black polled. Specify the genotypes of the offspring.

Explanation.

♀A?B? – black polled cows; ♂aabb – red horned bull

This is a task for analyzing crosses, it is necessary to determine the heterozygosity of cows.

Since the offspring of 100% turned out to be black polled, the law of uniformity of hybrids of the first generation (Mendel’s first law) manifests itself – when crossing two homozygous organisms belonging to different pure lines and differing from each other in one pair of alternative manifestations of the trait, the entire first generation of hybrids ( F1) will be uniform and will carry the manifestation of the trait of one of the parents.

Offspring genotype – AaBb

Answer: AaBb

AaBb

Source: RESHU USE

23. Crossed two diheterozygous pumpkin plants with yellow round fruits. Determine the ratio of phenotypes of hybrids of the first generation with complete dominance.

Task 6 No. 21275 Explanation.

With complete dominance, traits are inherited independently, according to Mendel’s III law (the law of independent inheritance): “Splitting for each pair of traits occurs independently of other pairs of traits.”

Splits characteristic of independent inheritance in dihybrid crosses:

There is no splitting (all children are the same) – two homozygotes AABB x aabb (or AAbb x aaBB) were crossed.

Splitting 9:3:3:1 – two diheterozygotes AaBb x AaBb were crossed (Mendel’s third law) .

Answer: 9331

Source: RESHU USE

24. When carrying out an analyzing crossing of a male with a dihetorozygous female, the ratio for the F 1 genotype is (in response, write down only the numbers, without additional signs and symbols)

Explanation.

Analyzing cross between a male and a heterozygous female:

P ♀ AaBb x ♂ aabb

G ♀ab; Ab; aB; AB ♂ab

F 1 1 aabb : 1Aabb : 1 aaBb : 1 AaBb

Answer: 1111

Source: RESHU USE

25. How many types of gametes does the CsVv zygote form if genes C (c) and B (c) are linked and inherited together?

Write only the number as your answer.

Explanation.

Genes are located on the same chromosome and are inherited together. There is no crossing over, so this zygote produces only two types of gametes: CB and St.

Answer: 2

Answer: 2

Source: RESHU USE

26. How many phenotypic groups are formed when two diheterozygotes are crossed?

Dihybrid cross

1. In an individual with the aabb genotype, gametes are formed

Task 6 No. 2308

Explanation.

One chromosome from each pair of allelic chromosomes diverge into a gamete. Law of purity of gametes.

Answer: ab

Answer: ab

ab

Source: RESHU USE

2. How many types of gametes are formed in diheterozygous pea plants during dihybrid crossing (genes do not form a linkage group)? Write down a number for your answer.

Task 6 No. 2311

Explanation.

AaBb – AB; Ab; aB; ab.

Answer: 4

Source: RESHU USE

3. In dogs, black coat (A) dominates over brown (a), and short legs (B) dominate over normal leg length (b). Write down the genotype of a black short-legged dog that is heterozygous for leg length only.

Task 6 No. 2401

Explanation.

For a dog to be black, a combination of Aa or AA in the genotype is necessary (variant aa corresponds to the brown coat color). In order for a dog to be short-legged, a combination of BB or BB in the genotype is necessary (the variant BB corresponds to limbs of normal length). Since heterozygosity for only one trait is required, the required genotype is AABb.

Answer: AABb

AABb

Source: RESHU USE

4. When crossing yellow (A) smooth (B) (digomozygote) and green (a) wrinkled (b) peas in F1, all yellow smooth peas were obtained. Determine the genotype of pea seeds in F1

Task 6 No. 2402

Explanation.

According to Mendel’s first law, only when homozygous individuals are crossed, uniformity is obtained in the first generation.

R: AABB x aabb

G: AB ab

F: AaBb – 100%

AABB – yellow smooth

aabb – green wrinkled

Answer: AaBb|AaBv

AaBb|AaBv

Source: RESHU USE

5. When homozygous tomato plants with round red fruits are crossed with pear-shaped yellow fruits (red color – A, yellow – a, round shape – B, pear-shaped – b), offspring with the genotype

Task 6 No. 2404

Explanation.

According to Mendel’s first law, only when homozygous individuals are crossed, uniformity is obtained in the first generation.

R: AABB x aabb

G: AB ab

F: AaBb – 100% (BbAa)

AABB – red round (BBAA)

aabb – yellow pear-shaped (bbaa)

Answer: AaBb

Answer: AaBb

AaBb

Source: RESHU USE

6. When crossing individuals with AaBb genotypes with AaBb (genes are not linked), the proportion (%) of heterozygotes for both alleles (diheterozygotes) in the offspring will be

Task 6 No. 2406

Explanation.

According to Mendel’s third law, when two diheterozygotes are crossed, the splitting occurs according to the phenotype 9: 3: 3: 1, 16 descendants are formed according to the genotype, and according to the Punnett lattice, the proportion of diheterozygotes is 25%.

Note.

“Heterozygous for both alleles” is the same as DIHETEROZYGOTE.

Answer: 25|25%

25|25%

Source: RESHU USE

7. The rule of uniformity of the first generation will appear if the genotype of one of the parents is aabb, and the other

Task 6 No. 2408

Explanation.

According to the first law of Mendel, homozygous individuals of pure lines are crossed.

Answer: AABB

Answer: AABB|AABB

AABB|AABB

Source: RESHU USE

8. Indicate the genotype of a person if, according to the phenotype, he is fair-haired and blue-eyed (recessive traits)

Task 6 No. 2412

Explanation.

Both traits are recessive, which means that according to the genotype it is a dihomozygote for a recessive trait.

Answer: abb

Answer: abb

aabb

Source: RESHU USE

9. When homozygous tomato plants with red (A) round (B) fruits and plants with yellow (a) pear-shaped (b) fruits in F2 are crossed, phenotypic splitting occurs in the ratio (color genes and fruit shape are located in different pairs of chromosomes). Write your answer as a sequence of numbers showing the ratio of the resulting phenotypes, in descending order

Task 6 No. 2414

Explanation.

In F1, the uniformity of the first generation is manifested, the plant genotype will be heterozygous, in the second generation, when crossing diheterozygotes, splitting occurs 9: 3: 3: 1 according to Mendel’s third law.

Answer: 9:3:3:1

Answer: 9:3:3:1|9 : 3 : 3 : 1|9331

9:3:3:1|9 : 3 : 3 : 1|9331

Source: RESHU USE

10. When heterozygous tomato plants with red and round fruits are crossed with individuals recessive in both traits (red A and round B are dominant traits), offspring with genotypes AaBb, aaBb, Aabb, aabb will appear in the ratio. Write down the answer in the form of a sequence of numbers showing the ratio of the resulting phenotypes, in descending order.

Task 6 No. 2416

Explanation.

R: AaBb x aabb

G: AB ab

Ab

aB

ab

F: AaBb; aabb, aabb, aabb

1 : 1 : 1 : 1

Answer: 1:1:1:1|1 : 1 : 1 : 1|1111

1:1:1:1|1 : 1 : 1 : 1|1111

Source: RESHU USE

11. What gametes do individuals with the aaBB genotype have?

Task 6 No. 2419

Explanation.

According to the law of gamete purity, one of each pair of allelic genes goes into the gamete.

Answer: b.

Answer: aB | aB

aB|aB

Source: RESHU USE

12. The genotype of one of the parents will be AaBb if, during analyzing dihybrid crossing and independent inheritance of traits, splitting in the phenotype in the offspring is observed in the ratio. Write your answer as a sequence of numbers showing the ratio of the resulting phenotypes, in descending order

Task 6 No. 2420

Explanation.

An analysis cross is a cross with an individual with the aabb genotype.

R: AaBb x aabb

G: AB ab

Ab

aB

ab

F: AaBb; aabb, aabb, aabb

1 : 1 : 1 : 1

Answer: 1 : 1 : 1 : 1|1:1:1:1|1111

1 : 1 : 1 : 1|1:1:1:1|1111

Source: RESHU USE

13. Determine the genotype of an individual of a yellow curly pumpkin, if, during its self-pollination in F1, the splitting of traits according to the phenotype corresponded to 9:3:3:1

Task 6 No. 2422

Explanation.

According to Mendel’s 3rd law, when crossing diheterozygotes AaBb x Aa Bb, 16 offspring are formed, which give a splitting according to the phenotype 9: 3: 3: 1

Answer: AaBb

AaBb

Source: RESHU USE

14. With dihybrid crossing and independent inheritance of traits in parents with AABb and aabb genotypes, splitting in the ratio is observed in the offspring.

Write down the answer in the form of a sequence of numbers showing the ratio of the resulting phenotypes, in descending order.

Task 6 No. 2423

Explanation.

R: AABb x aabb

G: AB ab

Ab

F: AaBb; Aabb

eleven

Answer: 1:1|1 : 1|11

1:1|1 : 1|11

Source: RESHU USE

15. What genotype does the offspring have in F1 when tomato plants are crossed with genotypes AABB and AABB?

Task 6 No. 2426

Explanation.

R: AAvv x aaBB

G: Ab aB

F: AaBb – 100%

Answer: AaBb | AaBv

AaBb|AaBv

Source: RESHU USE

16. In dihybrid crossing (unlinked inheritance) of individuals with dominant AABB and recessive aabb traits in F2, phenotypic splitting occurs in the ratio. Write your answer as a sequence of numbers showing the ratio of the resulting phenotypes, in descending order

Task 6 No. 2427

Explanation.

In F1, the uniformity of the first generation is manifested, the plant genotype will be heterozygous, in the second generation, when crossing diheterozygotes, splitting occurs 9: 3: 3: 1 according to Mendel’s third law.

Answer: 9:3:3:1

Answer: 9:3:3:1|9 : 3 : 3 : 1|9331

9:3:3:1|9 : 3 : 3 : 1|9331

Source: RESHU USE

17. What will be the ratio of the splitting of traits by phenotype in the offspring obtained from crossing a diheterozygous black, furry rabbit AaBv with a white, smooth-haired rabbit aavb

Task 6 No. 2428

Explanation.

R: AaBb x aabb

G: AB ab

Ab

aB

ab

F: AaBb; aabb, aabb, aabb

1 : 1 : 1 : 1

By phenotype: 1 part black hairy : 1 part black smooth : 1 part white hairy : 1 part white smooth.

Genotypic segregation corresponds to phenotypic segregation.

Answer: 1 : 1 : 1 : 1|1:1:1:1|1111

1 : 1 : 1 : 1|1:1:1:1|1111

Source: RESHU USE

18. What is the ratio of genotypes in offspring obtained from crossing individuals with AaBv x AABB genotypes?

Task 6 No. 11559

Explanation.

The first individual gives 4 types of gametes: AB, Av, aB, av; the second – 1 type of gametes: AB. This means that the ratio of genotypes in the offspring is 1:1:1:1 (AABB, AABv, AaBB, AaBv).

Answer: 1 : 1 : 1 : 1|1:1:1:1|1111

1 : 1 : 1 : 1|1:1:1:1|1111

Source: Yandex: USE training work in biology. Option 2.

19. A woman with blond (a) straight (b) hair married a man with dark curly hair. Write down the genotype of their child who has dark straight hair.

Task 6 No. 12575

Explanation.

Mother’s genotype ♀ aabb; father’s genotype ♂ A_Bb

Curly hair – intermediate inheritance: bb – straight hair; Bb – curly (wavy); BB – shallow.

From the mother, the child receives a gamete – ab; from father – Ab.

Genotype of a child with dark straight hair: Aabb

Answer: Abb

Answer: Abb

Aabb

Source: Unified State Examination in Biology 05/30/2013. main wave. Far East. Option 2.

20. In guinea pigs, black hair dominates over white, and long – over short. Determine the genotype of a diheterozygous individual.

Task 6 No. 12703

Explanation.

Diheterozygote (differs in two pairs of traits) – AaBb

Answer: AaBb

AaBb

Source: Unified State Examination in Biology 05/30/2013. main wave. Far East. Option 5.

21. How many types of gametes does a diheterozygous individual form with full linkage of the studied genes

Task 6 No. 21012

Explanation.

With full linkage, a diheterozygous individual forms two types of gametes:

AB/ab/

Answer: 2

Source: Typical test tasks in biology, edited by G. S. Kalinov, T. V. Mazyarkin. 10 options for tasks. 2017

22. In cattle, black (A) dominates over red (a), polled (B) – over horned (b). When crossing black polled cows with a red horned bull, all the offspring turned out to be black polled. Specify the genotypes of the offspring.

Task 6 No. 21126

Explanation.

♀A?B? – black polled cows; ♂aabb – red horned bull

This is a task for analyzing crosses, it is necessary to determine the heterozygosity of cows.

Since the offspring of 100% turned out to be black polled, the law of uniformity of hybrids of the first generation (Mendel’s first law) manifests itself – when crossing two homozygous organisms belonging to different pure lines and differing from each other in one pair of alternative manifestations of the trait, the entire first generation of hybrids ( F1) will be uniform and will carry the manifestation of the trait of one of the parents.

Offspring genotype – AaBb

Answer: AaBb

AaBb

Source: RESHU USE

23. Crossed two diheterozygous pumpkin plants with yellow round fruits. Determine the ratio of phenotypes of hybrids of the first generation with complete dominance.

Task 6 No. 21275

Explanation.

With complete dominance, traits are inherited independently, according to Mendel’s III law (the law of independent inheritance): “Splitting for each pair of traits occurs independently of other pairs of traits.”

Splits characteristic of independent inheritance in dihybrid crosses:

There is no splitting (all children are the same) – two homozygotes AABB x aabb (or AAbb x aaBB) were crossed.

Splitting 9:3:3:1 – two diheterozygotes AaBb x AaBb were crossed (Mendel’s third law) .

Answer: 9331

Source: RESHU USE

24. When carrying out an analyzing crossing of a male with a dihetorozygous female, the ratio for the F 1 genotype is (in response, write down only the numbers, without additional signs and symbols)

Task 6 No. 21278

Explanation.

Analyzing cross between a male and a heterozygous female:

P ♀ AaBb x ♂ aabb

G ♀ab; Ab; aB; AB ♂ab

F 1 1 aabb : 1Aabb : 1 aaBb : 1 AaBb

Answer: 1111

Source: RESHU USE

25. How many types of gametes does the CsVv zygote form if genes C (c) and B (c) are linked and inherited together?

Write only the number as your answer.

Task 6 No. 21282

Explanation.

Genes are located on the same chromosome and are inherited together. There is no crossing over, so this zygote produces only two types of gametes: CB and St.

Answer: 2

Answer: 2

Source: RESHU USE

26. How many phenotypic groups are formed when two diheterozygotes are crossed?

Task 6 No. 21285

Explanation.

Digeterozygotes when crossing form 4 phenotypic groups

Answer: 4

monohybrid cross

1. What percentage of individuals of roan suit can be obtained by crossing cattle of red (AA) and white (aa) suit with incomplete dominance

Task 6 No. 2301

Explanation.

R: AA x aa

G: A a

F: Aa – 100%

With incomplete dominance, heterozygotes have an intermediate trait.

Correct answer – 100%

Answer: 100|100%

100|100%

Source: RESHU USE

2. What percentage of night beauty plants with pink flowers can be expected from crossing plants with red (A) and white (a) flowers (incomplete dominance)

Task 6 No. 2302

Explanation.

R: AA x aa

G: A a

F: Aa – 100%

With incomplete dominance, heterozygotes have an intermediate trait.

In this case, it does not matter at all which of the signs is dominant white or red. The main thing is that when they are crossed with each other in the offspring, we will get 100% of plants with an intermediate trait Aa – pink.

Answer: 100|100%

100|100%

Source: RESHU USE

3. Determine the ratio of genotypes in the offspring when crossing heterozygous plants of the night beauty

Task 6 No. 2303

Explanation.

R: Aa x Aa

G: A A

a a

F: AA 2Aa aa

According to Mendel’s second law, when heterozygous plants are crossed, the offspring are split according to the genotype 1:2:1

Answer: 1:2:1|1 : 2 : 1|121|211|2:1:1

1:2:1|1 : 2 : 1|121|211|2:1:1

Source: RESHU USE

4. Homozygous dominant gray sheep die when switching to roughage, while heterozygous ones survive. Determine the genotype of a gray viable individual

Task 6 No. 2316

Explanation.

The task describes one trait, heterozygotes survive, which means the Aa genotype

Answer: Aa

Aa|Aa

Source: RESHU USE

5. During self-pollination of a heterozygous tall pea plant (high stem – A), the proportion of dwarf forms is (%)

Task 6 No. 2319

Explanation.

R: Aa x Aa

G: A A

a a

F 2: AA 2Aa aa

According to Mendel’s second law, when heterozygous organisms are crossed, the offspring splits according to the genotype 1:2:1, according to the phenotype 1:3, i.e., 25% of recessive individuals appear in the offspring.

Answer: 25|25%

25|25%

Source: RESHU USE

6. What is the probability (%) of having tall children in heterozygous parents with short stature (short stature dominates tall stature)

Task 6 No. 2324

Explanation.

R: Aa x Aa

G: A A

a a

F: AA 2Aa aa

According to Mendel’s second law, when heterozygous organisms are crossed, the offspring splits according to the genotype 1:2:1, according to the phenotype 1:3, i.e., 25% of recessive individuals appear in the offspring.

Answer: 25|25%

25|25%

Source: RESHU USE

7. When a plant heterozygous for one pair of traits is crossed with a homozygous one, the proportion of homozygotes in the offspring will be

Task 6 No. 2332

Explanation.

R: aa x aa

G: a A

a a

F: Aa aa

50% 50%

With such a crossing, 50% heterozygotes and 50% homozygous recessive individuals are obtained.

OR

R: AA x Aa

G: A A a

F: AA Aa

50% 50%

With such a crossing, 50% heterozygotes and 50% homozygous dominant individuals are obtained.

Answer: 50|50%

50|50%

Source: RESHU USE

8. What number of phenotypes is formed in the offspring when crossing Aa x Aa in case of complete dominance?

Write down a number for your answer.

Task 6 No. 2333

Explanation.

R: Aa x Aa

G: A a A a

F: AA 2Aa aa

According to Mendel’s second law (the law of splitting), when heterozygous organisms are crossed in the offspring, splitting occurs according to the genotype 1:2:1, according to the phenotype 1:3, which means that 2 varieties of phenotypes are formed.

Answer: 2

Source: RESHU USE

9. What is the probability (in %) of dark-haired parents (Aa) having children with blond hair (dark color dominates over light)?

Task 6 No. 2334

Explanation.

R: Aa x Aa

G: A A

a a

F: AA 2Aa aa

According to the second law of Mendel, when heterozygous organisms are crossed, the offspring splits according to the genotype 1:2:1, according to the phenotype 1:3, i.e. 25% of recessive individuals (fair-haired) appear in the offspring

Answer: 25|25%

25|25%

Source: RESHU USE

10. Determine the ratio of phenotypes in offspring in a monohybrid crossing of two heterozygous organisms with complete dominance. Write down the answer in the form of a sequence of numbers showing the ratio of the resulting phenotypes, in descending order.

Task 6 No. 20468

Explanation.

When monohybrid crossing of two heterozygous organisms with complete dominance, the ratio of phenotypes is 3: 1 (Mendel’s splitting rule)

Answer: 31

Answer: 31

Source: Demo version of the USE-2017 in biology.

11. How many alleles of one gene does the egg of a flowering plant contain?

Write down only the appropriate number in your answer.

Task 6 No. 20781

Explanation.

each gamete contains only one allele of each gene.

Answer: 1

Source: RESHU USE

12. How many different phenotypes are obtained by self-pollination of plants with pink corolla petals (heterozygote) in case of incomplete dominance?

Task 6 No. 20956

Explanation.

When self-pollinating plants with pink corolla petals in case of incomplete dominance

Aa x Aa we get 3 phenotypes: AA red; Aa pink; aa white

Note.

The manifestation of the trait (color) is not important, you need to know that when heterozygotes self-pollinate with incomplete dominance, 3 phenotypes are obtained

Answer: 3

Source: Typical test tasks in biology, edited by G. S. Kalinov, T. V. Mazyarkin. 10 options for tasks. 2017

13. The round shape of the fruit of the tomato dominates over the pear-shaped, the red color of the fruit – over the yellow. Determine the genotype of a tomato with pear-shaped yellow fruits.

Task 6 No. 21127

Explanation.

pear-shaped and yellow fruits are recessive traits, so the genotype is aabb

Answer: aabb

aabb|aabb

Source: RESHU USE

14. What is the probability of birth (in%) of healthy boys in a family where the mother is healthy and the father is sick with hypertrichosis, a disease caused by the presence of a gene linked to the Y chromosome?

Task 6 No. 21264

Explanation.

because this gene is linked to the Y chromosome, then all sons will be sick with hypertrichosis. This means that the probability of the birth of healthy boys in this family is 0%

Answer: 0

Source: RESHU USE

15. In humans, protruding ears (B) dominate over the gene for normally closed ears (b). What is the father’s genotype if the mother’s genotype is bb, and among their children there were 50% lop-eared and 50% with normally flattened ears?

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