Principles of Inheritance and Variation Test

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Principles of Inheritance and Variation Test - 90

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Weekly Quiz Competition

Question 1
1 / -0

An experiment was performed to test the ability of one wild type and four mutant strains of E.coli bacteria to grow in either minimal media or media that was supplemented with various amino acids.

Supplement None Arginine
Wild-type $$+$$ $$+$$
Strain 1 $$-$$ $$+$$
Strain 2 $$-$$ $$-$$
Strain 3 $$-$$ $$-$$
Strain 4 $$+$$ $$+$$

Supplement Threonine Histidine
Wild-type $$+$$ $$+$$
Strain 1 $$-$$ $$-$$
Strain 2 $$+$$ $$-$$
Strain 3 $$-$$ $$+$$
Strain 4 $$+$$ $$+$$
A $$\left(+\right)$$ sign indicates growth and a $$\left(-\right)$$ sign indicates no growth.
Which of the following is true for mutation in strain 4?

A
Renders it dormant

B
Does not affect its ability to synthesize amino acids

C
Causes it to become haploid

D
Is identical to that in strain 1

E
None of the above
Question 2
1 / -0

An experiment was performed to test the ability of one wild type and four mutant strains of E.coli bacteria to grow in either minimal media or media that was supplemented with various amino acids.

Supplement None Arginine
Wild-type $$+$$ $$+$$
Strain 1 $$-$$ $$+$$
Strain 2 $$-$$ $$-$$
Strain 3 $$-$$ $$-$$
Strain 4 $$+$$ $$+$$

Supplement Threonine Histidine
Wild-type $$+$$ $$+$$
Strain 1 $$-$$ $$-$$
Strain 2 $$+$$ $$-$$
Strain 3 $$-$$ $$+$$
Strain 4 $$+$$ $$+$$
A $$\left(+\right)$$ sign indicates growth and a $$\left(-\right)$$ sign indicates no growth.
Which of the following amino acids are not required by strain 3 to grow?

A
Arginine and histidine

B
Threonine and histidine

C
Histidine only

D
Arginine and threonine

E
Threonine only
Question 3
1 / -0

The given pedigree depicts a rare disorder through three generations of a family.

The number of females in generation II that passed the disorder to their offspring are?

A
0

B
1

C
2

D
3

E
4

Solution

Generation II has total 5 females with 2 affected and 3 normal ones. Out of two affected females, one female is crossed with the normal male. Since the normal parents have normal progeny, the trait exhibit dominant inheritance. The affected female transmits its affected copies of chromosomes to half of its progeny. Thus, the correct answer is option B.
Question 4
1 / -0

A child is born with albinism, an inability to produce or deposit the pigment protein, melanin, in skin cells, as well as other pigment-containing cells of the body. His parents do not display the trait of albinism.
How is this possible?

A
The child has only one allele for skin pigmentation, though his parents each have two. Having fewer genes produces his albino characteristics.

B
A spontaneous mutation, not present in either parent, has resulted in the child producing the albino protein. The albino protein displaces the normal melanin.

C
The normal DNA from both parents has combined to produce a new form of DNA in the child. This abnormal DNA codes for the albino protein.

D
The parents each have one normal DNA allele and one that codes for albinism. The parents produce enough normal melanin protein to allow normal skin color, but the child cannot.

Solution

Albinos lack the dark pigment melanin in the skin, hair and iris. It is an autosomal, recessive genetic disorder. Synthesis of melanin pigment from dihydroxyphenylalanine is absent due to the lack of the enzyme tyrosinase. Only homozygous individuals (aa) are affected b this. It is due to the recessive allele of the long arm of chromosome 11 but may also be caused by another recessive allele of the P gene on the long arm of chromosome 15.

So, the correct option is 'The child has only one allele for skin pigmentation, though his parents each have two. Having fewer genes produces his albino characteristics.'
Question 5
1 / -0

Mendel performed his experiment on garden pea species and gave rules of heridity. Give the scientific name of the species.

A
Zea mays

B
Pisum sativum

C
Phaseolus mungo

D
Rana tigrina

Solution

The scientific name of the garden pea plant on which Mendel performed his experiments is Pisum sativum.

So, the correct answer is 'Pisum sativum'.
Question 6
1 / -0

Based on the given pedigree, the individual III-5 will have _______ genotype.

A
Carrier

B
Unaffected

C
Homozygous recessive

D
Homozygous dominant

E
Heterozygous

Solution

The shaded symbol in pedigree analysis represents affected individuals making. A cross between normal II-5 and II-6 gives affected progeny (III-5) showing that both parents are the carrier for the trait and the trait is recessive. Since a recessive trait is exhibited by homozygous genotype only, III-5 is homozygous recessive for the trait. The trait is homozygous recessive.
Thus, the correct answer is option C.
Question 7
1 / -0

The tumor suppressor gene APC, when mutant, causes a precancerous condition called familial adenomatous polyposis (FAP). FAP has autosomal dominant inheritance.
What is the probability that two individuals who are heterozygous for the APC mutation would have four children, all without the APC mutation?

A
1/64

B
0.25

C
1/16

D
0.125

E
0.5

Solution

Since FAP (familial adenomatous polyposis) has autosomal dominant inheritance, so individuals with AA or Aa genotype will be APC mutant and those with aa genotype will be normal or without APC mutation.
The cross can be represented as shown in the figure.
Hence probability for the children that will be born without APC mutation is:
No. of children with aa genotype/ Total no. of children with all possible genotypes
= 1/4
= 0.25
So, the correct answer is '0.25'.
Question 8
1 / -0

According to the chromosomal theory of inheritance, which of the following type of cell division contributes maximum for recombination and genetic variation?

A
Meiosis

B
Mitosis

C
Amitosis

D
Both A and B

Solution

The higher organisms reproduce through sexual reproduction in which the zygote is formed by the fusion of gametes. The gametes are formed by meiotic division. Genetic recombination takes place during meiosis. There is an exchange of the genetic material between two homologous chromosomes. This recombination introduces genetic variation in the organism. The chromosomal theory states that the recombination event results in variation.
Thus, the correct answer is option A.
Question 9
1 / -0

In a pea plant, the yellow colour of the seed dominates over the green colour. The genetic make up of the green colour of the seed can be shown as _______

A
GG

B
Gg

C
Yy

D
yy
Question 10
1 / -0

Name the scientist who converted Mendel's conclusions into principles of heredity

A
De Vries

B
Tschermak-seysenegg

C
Carl Correns

D
T. H. Morgan

Solution

Carl Erich Correns was a German botanist and geneticist, who is notable primarily for his independent discovery of the principles of heredity, and for his rediscovery of Gregor Mendel's earlier paper on that subject, which he achieved simultaneously but independently of the botanists Erich Tschermak von Seysenegg. He also converted Mendel’s conclusions into principles of heredity.
So, the conclusion is ‘Carl correns’.

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Principles of Inheritance and Variation Test - 90

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Principles of Inheritance and Variation Test - 90

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SHARING IS CARING

Question 1

Renders it dormant

Does not affect its ability to synthesize amino acids

Causes it to become haploid

Is identical to that in strain 1

None of the above

Question 2

Arginine and histidine

Threonine and histidine

Histidine only

Arginine and threonine

Threonine only

Question 3

0

1

2

3

4

Solution

Question 4

The child has only one allele for skin pigmentation, though his parents each have two. Having fewer genes produces his albino characteristics.

A spontaneous mutation, not present in either parent, has resulted in the child producing the albino protein. The albino protein displaces the normal melanin.

The normal DNA from both parents has combined to produce a new form of DNA in the child. This abnormal DNA codes for the albino protein.

The parents each have one normal DNA allele and one that codes for albinism. The parents produce enough normal melanin protein to allow normal skin color, but the child cannot.

Solution

Question 5

Zea mays

Pisum sativum

Phaseolus mungo

Rana tigrina

Solution

Question 6

Carrier

Unaffected

Homozygous recessive

Homozygous dominant

Heterozygous

Solution

Question 7

1/64

0.25

1/16

0.125

0.5

Solution

Question 8

Meiosis

Mitosis

Amitosis

Both A and B

Solution

Question 9

GG

Gg

Yy

yy

Question 10

De Vries

Tschermak-seysenegg

Carl Correns

T. H. Morgan

Solution

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