Genetics And Evolution Test

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Genetics And Evolution Test - 1

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Weekly Quiz Competition

Question 1
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Which of the following pairs of traits were not selected by Mendel for his hybridisation experiments on pea plants?

A
Stem height (tall/dwarf) and Flower colour (violet/white)

B
Flower position (axial/terminal) and Pod shape (inflated/constricted)

C
Pod colour (green/yellow) and Seed shape (round/wrinkled)

D
Pod colour (violet/yellow) and Seed shape (inflated/constricted)

Solution

The seven contrasting traits studied by Mendel were:
1. Stem height (tall/dwarf)
2. Flower colour (violet/white)
3. Flower position (axial/terminal)
4. Pod shape (inflated/constricted)
5. Pod colour (green/yellow)
6. Seed shape (round/wrinkled)
7. Seed colour (yellow/green)
Question 2
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Which of the following statements about dihybrid cross, involving yellow round seeded plant and green wrinkled seeded plants, is incorrect?

A
In the dihybrid cross the phenotypes round, yellow; wrinkled, yellow; round, green and wrinkled, green appeared in the ratio 9 : 3 : 3 : 1.

B
F1 heterozygous (RrYy ) plant can produce four types of gametes namely RY, Ry, rY and ry.

C
In the dihybrid test cross the phenotypes round, yellow; wrinkled, yellow; round, green and wrinkled, green appeared in the ratio 9 : 3 : 3 : 1.

D
Based upon dihybrid crosses Mendel proposed the Law of Independent Assortment.

Solution

In the dihybrid test cross the phenotypes round, yellow;wrinkled, yellow; round, green and wrinkled, green appeared in the ratio 1 : 1 : 1 : 1 and is not 9 : 3 : 3 : 1.
Question 3
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Which of the following statements about sickle cell anemia is incorrect?

A
Sickle-cell anaemia is an autosome linked recessive trait that can be transmitted from parents to the offspring when both the partners are heterozygous (carrier for the gene).

B
The disease is controlled by a single pair of allele, Hb^A and Hb^S. Out of the three possible genotypes only homozygous individuals for HbS (Hb^SHb^S) show the diseased phenotype.

C
The defect is caused by the substitution of Glutamic acid (Glu) by Valine (Val) at the sixth position of the beta globin chain of the haemoglobin molecule.

D
The substitution of amino acid in the globin protein results due to the single base substitution at the sixth codon of the beta globin gene from GUG to GAG.

Solution

The substitution of amino acid in the globin protein results due to the single base substitution at the sixth codon of the beta globin gene from GUG to GAG. This statement is incorrect. The substitution of amino acid in the globin protein results due to the single base substitution at the sixth codon of the beta globin gene from GAG to GUG.
Question 4
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Which of the following statements about genetic disease is true?

A
Turner's syndrome is caused due to the absence of one of the Y chromosomes, i.e. 45 with X0.

B
Klinefelter's syndrome is caused due to the presence of an additional copy of X-chromosome resulting into a karyotype of 47, XXY.

C
Klinefelter's syndrome is caused due to the absence of one of the X chromosomes, i.e. 45 with X0.

D
The cause of Down's syndrome is the presence of an additional copy of the chromosome number 20 (trisomy of 20).

Solution

Klinefelter's syndrome is caused due to the presence of an additional copy of X-chromosome resulting into a karyotype of 47, XXY.
Question 5
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Henking (1891) could trace a specific nuclear structure all through spermatogenesis in a few insects, and it was also observed by him that 50 per cent of the sperm received this structure after spermatogenesis, whereas the other 50 percent sperm did not receive it. Which of the following name was given to this structure by him?

A
X-body

B
Barr body

C
Autosomes

D
Y-body

Solution

Henking (1891) could trace a specific nuclear structure all through spermatogenesis in a few insects, and it was also observed by him that 50 per cent of the sperm received this structure after spermatogenesis, whereas the other 50 percent sperm did not receive it. Henking gave name to this structure as the X body.
Question 6
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T.H Morgan hybridised yellow-bodied, white-eyed females drosophila to brown-bodied, red-eyed males and intercrossed their F1 progeny. Which of the following represents the percentage of recombinants and parental type in F2?

A
98.5% recombinants and 1.5% parantel

B
98.7% recombinants and 1.3% parantel

C
68.8% recombinants and 37.2% parantel

D
1.3% recombinants and 98.7% parantel

Solution

It is correct. When yellow-bodied, white-eyed females drosophila to brown-bodied, red-eyed males and intercrossed their F1 progeny. The percentage of recombinants and parental type in F2 is 1.3% recombinants and 98.7% parantel.
Question 7
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Which of the following types of disorder is transferred from a phenotypically normal, but carrier female to only some of the male progeny?

A
Autosomal recessive

B
Sex-linked recessive

C
Sex-limited dominant

D
Sex-linked dominant

Solution

Sex-linked recessive is the type of disorder transferred from a phenotypically normal, but carrier female to only some of the male progeny.
Question 8
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The condition of a karyotype 2n + 1, 2n - 1, 2n + 2 and 2n - 2 is called ___________.

A
euploidy

B
polyploidy

C
aneuploidy

D
hexaplody

Solution

The condition of a karyotype 2n + 1, 2n - 1, 2n + 2 and 2n - 2 is called aneuplody.
Question 9
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A person having genotype IAIB, represents the the blood group as AB. Which of the following explains this?

A
Co dominance

B
Incomplete dominance

C
Pleiotropy

D
Epistasis

Solution

In co dominanace both genes in heterozygous condition are equally expressed. AB blood group is due to co dominance.
Question 10
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Which of the following groups of organisms show Z Z/ZW type of sex determination?

A
Pigeon, Peacock and Drosophila

B
Pigeon, Peacock and Hen

C
Grasshopper, Housfly and Drosophila

D
Man, Housefly and Drosophila

Solution

Pigeon, peacock and Hen show Z Z/ZW type of sex determination.
Question 11
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Occasionally, a single gene may express more than one effect. The phenomenon is called pleiotropy. Which of the following is an example of pleiotropy?

A
Klinefelter's Syndrome and Turner's Syndrome

B
Downs syndrome and Klinefelter's Syndrome

C
Phenyl ketonurea and Downs syndrome

D
Sickle cell anaemia and Phenyl ketonurea

Solution

Sickle cell anaemia and Phenyl ketonurea are examples for pleiotropy.
Question 12
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Mother and father of a person with 'O' blood group have 'A' blood group. What would be the genotype of both mother and father?

A
Both mother and father are heterozygous for 'A' blood group.

B
Mother is heterozygous for 'A' and father is homozygous for 'A' blood group.

C
Both mother and father are homozygous for 'A' blood group.

D
Mother is homozygous for 'A' blood group and father is heterozygous for 'A'.

Solution

Heterozygous A blood group parents have O group child.
Question 13
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In peas, tallness is dominant over dwarfness and violet colour of flowers is dominant over the white colour. When a tall plant bearing violet flowers was pollinated with a dwarf plant bearing white flowers, the different phenotypic groups were obtained in the progeny as given below.

Tall, Violet = 136 Tall, White = 134 Dwarf, Violet = 136 Dwarf, White = 132. __________ is the genotypes of the two parents.

A
TTVV and ttvv

B
TTVV and TTvv

C
TTVV and TTVV

D
TtVv and ttvv

Solution

The given result is obtained only in a test cross. This involves cross between heterozygous plant and homozygous recessive plants, i.e, TtVv and ttvv.
Question 14
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A diploid organism is heterozygous for 3 loci, how many types of gametes can be produced?

A
2 types

B
4 types

C
8 types

D
16 types

Solution

A diploid organism is heterozygous for 3 lociie AaBbCc. Then types of gamete produced by these organism is 2n, i.e, 2³= 8. Therefore 8 type is posssible.

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Genetics And Evolution Test - 1

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Question 1

Stem height (tall/dwarf) and Flower colour (violet/white)

Flower position (axial/terminal) and Pod shape (inflated/constricted)

Pod colour (green/yellow) and Seed shape (round/wrinkled)

Pod colour (violet/yellow) and Seed shape (inflated/constricted)

Solution

Question 2

In the dihybrid cross the phenotypes round, yellow; wrinkled, yellow; round, green and wrinkled, green appeared in the ratio 9 : 3 : 3 : 1.

F1 heterozygous (RrYy ) plant can produce four types of gametes namely RY, Ry, rY and ry.

In the dihybrid test cross the phenotypes round, yellow; wrinkled, yellow; round, green and wrinkled, green appeared in the ratio 9 : 3 : 3 : 1.

Based upon dihybrid crosses Mendel proposed the Law of Independent Assortment.

Solution

Question 3

Sickle-cell anaemia is an autosome linked recessive trait that can be transmitted from parents to the offspring when both the partners are heterozygous (carrier for the gene).

The disease is controlled by a single pair of allele, HbA and HbS. Out of the three possible genotypes only homozygous individuals for HbS (HbSHbS) show the diseased phenotype.

The defect is caused by the substitution of Glutamic acid (Glu) by Valine (Val) at the sixth position of the beta globin chain of the haemoglobin molecule.

The substitution of amino acid in the globin protein results due to the single base substitution at the sixth codon of the beta globin gene from GUG to GAG.

Solution

Question 4

Turner's syndrome is caused due to the absence of one of the Y chromosomes, i.e. 45 with X0.

Klinefelter's syndrome is caused due to the presence of an additional copy of X-chromosome resulting into a karyotype of 47, XXY.

Klinefelter's syndrome is caused due to the absence of one of the X chromosomes, i.e. 45 with X0.

The cause of Down's syndrome is the presence of an additional copy of the chromosome number 20 (trisomy of 20).

Solution

Question 5

X-body

Barr body

Autosomes

Y-body

Solution

Question 6

98.5% recombinants and 1.5% parantel

98.7% recombinants and 1.3% parantel

68.8% recombinants and 37.2% parantel

1.3% recombinants and 98.7% parantel

Solution

Question 7

Autosomal recessive

Sex-linked recessive

Sex-limited dominant

Sex-linked dominant

Solution

Question 8

euploidy

polyploidy

aneuploidy

hexaplody

Solution

Question 9

Co dominance

Incomplete dominance

Pleiotropy

Epistasis

Solution

Question 10

Pigeon, Peacock and Drosophila

Pigeon, Peacock and Hen

Grasshopper, Housfly and Drosophila

Man, Housefly and Drosophila

Solution

Question 11

Klinefelter's Syndrome and Turner's Syndrome

Downs syndrome and Klinefelter's Syndrome

Phenyl ketonurea and Downs syndrome

Sickle cell anaemia and Phenyl ketonurea

Solution

Question 12

Both mother and father are heterozygous for 'A' blood group.

Mother is heterozygous for 'A' and father is homozygous for 'A' blood group.

Both mother and father are homozygous for 'A' blood group.

Mother is homozygous for 'A' blood group and father is heterozygous for 'A'.

Solution

The disease is controlled by a single pair of allele, Hb^A and Hb^S. Out of the three possible genotypes only homozygous individuals for HbS (Hb^SHb^S) show the diseased phenotype.