Principles of Inheritance and Variation Test

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Principles of Inheritance and Variation Test - 86

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Question 1
1 / -0

Comparing the genetic maps of homologous genes in human and mouse chromosomes, the genes that make up 40% of human chromosome 1 are present

A
As a single contiguous block in mouse chromosome 1

B
In mouse chromosome 4

C
As a single block in mouse chromosome 3

D
All of the above

Solution

The linkage groups present at adjacent regions of human chromosome 1 are mapped to discrete regions on distal mouse chromosomes 1 and 3. The gene order within the linkage groups is conserved between human and mouse. The 15 genes present between 1q21 and 1q32 on human chromosome 1 are mapped to 29.5 cM on distal mouse chromosome 1 and 6 genes present between human chromosome 1q21 and 1p22 span 15.6 cM on distal mouse chromosome 3. The human CYP2J gene (CYP2J2) that encodes for cytochromes P450 is present on human chromosome 1 and the corresponding mice gene is present at the central region of chromosome 4 distal to the Jun oncogene. The correct answer is D.
Question 2
1 / -0

Mutations can be induced with

A
$$\gamma$$-radiation

B
Infra red radiation

C
IAA

D
Ethylene

Solution

X-rays, $$\gamma$$-rays and fast neutron bombardment techniques can induce point mutations (changes in a single nucleotide) or deletions (loss of a chromosomal segment).
Question 3
1 / -0

What is the inheritance of colour blindness of both parents having a normal vision but mother having a recessive gene for colour blindness?

A
Son - 50% ; Daughter - nil

B
Son - 100% ; Daughter - nil

C
Son - nil ; Daughter - 100%

D
Son - nil ; Daughter - nil

Solution

Colour blindness is a X-linked recessive disorder. One copy of the affected gene in males in each cell is sufficient to cause the disorder (X$$^{c}$$Y). Females with two copies of the affected gene show the disorder (X$$^{c}$$X$$^{c}$$).
Females heterozygous (X$$^{c}$$X) for this trait be normal but serve as a carrier of the disease.
According to the question, the female is carrier for colour blindness (X$$^{c}$$X) and father is normal (XY). The carrier mother for colour blindness will inherit the disease to 50% sons (X$$^{c}$$Y) while the 100% daughter will have normal vision. Correct answer is option A.
Question 4
1 / -0

The correct pathway for the synthesis of skin pigment is

A
Tyrosine - Melanin - Dopaquinone - Dopa

B
Tyrosine - Dopa - Dopaquinone - Melanin

C
Dopa - Tyrosine - Dopaquinone - Melanin

D
Tyrosine - Dopa - Melanin - Dopaquinone

E
Tyrosine - Dopaquinone - Dopa - Melanin

Solution

Melanin is the skin pigment in human. Melanin synthesis starts in the liver with conversion of phenylalanine into tyrosine by the action of phenylalanine hydroxylase. The L-tyrosine is oxidised to L-DOPA catalysed by the action of tyrosinase enzymes within the melanocyte's melanosome. It is followed by oxidation of L-DOPA into Dopaquinone. The Dopaquinone is then channelled to produce either eumelanin or pheomelanin. Correct option is B.
Question 5
1 / -0

The idea of mutations was brought forth by

A
Hugo de Vries, who worked on evening primrose

B
Gregor Mendel, who worked on Pisum sativum

C
Hardy Weinberg, who worked on allele frequencies in a population

D
Charles Darwin, who observed a wide variety of organisms during sea voyage
Solution
A) Correct option A
B) Explanation of correct option
Hugo de Vries did his genetic experiments with Oenothera lamarkiana (evening primrose) and gave the "Mutation theory".
According to his theory mutations are sudden, random, and inheritable and discontinuous variations that serve as raw material for evolution.
C) Explanation of incorrect option
Option B) Gregor Mendel worked on Pisum sativum and proposed the laws of inheritance; he did not explain the variations found in his results.
Option C) Hardy Weinberg studies the allele frequencies in a population that are not evolving.
Option D) According to Charles Darwin, minor heritable variations are the basis of evolution, not the mutations. He proposed the theory of natural selection to explain the process of evolution.
Question 6
1 / -0

A tobacco plant heterozygous for albinism is self pollinated and 1200 seeds are subsequently germinated. How many seedlings would have the parental genotype?

A
900

B
600

C
1200

D
300

Solution

Albinism is a recessive character which means that only homozygous recessive genotype will show it. The heterozygous genotype will be normal but will serve as carrier since the recessive gene is masked by presence of normal dominant gene. According to the question, heterozygous plants were crossed.
Parent generation : Aa x Aa
Gametes
Aa ---->
Aa A a
A AA
normal Aa
normal
a Aa
normal
aa
albino
F$$_1$$ generation genotypic ratio = 1 AA : 2 Aa : 1 aa. Hence, the ratio of parental genotype (Aa)= 2/4 or 1/2. Thus, out of 1200 seedlings, 1200/2=600 seedlings will have parental genotype. Option B is the correct answer.
Question 7
1 / -0

Which one of following symbol and information is correct?

A

B

C

D
Question 8
1 / -0

The cause of mutation is

A
Changes in DNA

B
Changes in chromosome

C
Changes in gene

D
All of the above
Solution
A. Correct Option: D

B. Explanation for the correct option:
- Mutation refers to the sudden change in a DNA sequence alerting heritable change in individual phenotype.
- Mutation in genes occurs due to various physical such as mistakes committed by genes while crossing over or chemical reasons such as exposure to harmful radiation.
- It also occurs in the chromosome causing change in the structure or arrangement of the chromosome like abnormal numbers of individual chromosomes or chromosome sets.
Question 9
1 / -0

Sickle cell anaemia is because of

A
More population of house flies.

B
Change in sequences of amino acids in alpha-chain of haemoglobin.

C
Change in sequences of amino acids in beta-chain of haemoglobin.

D
Poor hygienic conditions.

Solution

Sickle cell anaemia is a genetic disorder caused by a mutation that affects the normal development of haemoglobin. The haemoglobin inside blood cells clumps together into solid structures. The clumping distorts the normal shape of the red blood cells, causing them to take on a rigid, sickle shape. The most common haemoglobin of adult humans is known as haemoglobin A (HbA). However, other variant forms of hemoglobin are called hemoglobin S (HbS). Individuals who manufacture HbS exclusively suffer from sickle-cell disease. Individuals with sickle-cell disease have inherited from each parent a gene S encoding the beta chain of haemoglobin. The amino acid sequences of the beta chains of HbA and HbS are identical except for the amino acid at position 6. This position is occupied by glutamic acid in HbA chains, but in HbS beta chains, valine is found there instead.
Question 10
1 / -0

Find out the wrong statement.

A
Mobile genetic elements, transposons were visualized by Barbara McClintock.

B
Udder cell, a somatic cell is used to produce the cloned sheep by nuclear transplantation method.

C
In pedigree analysis, a person immediately affected by an action is called propositions.

D
Dr Ian Wilmut produced a cloned sheep called Dolly.

E
DNA ligases are used to cleave a DNA molecule.
Solution
- Transposons are the genetic elements that can jump from one location to other in the genome. Also called jumping genes, they were first discovered by Barbara McClintok while working on chromosome breakage in maize. Statement A is correct.
Dolly sheep was born on July 5, 1996, through the technique of somatic cell nuclear transfer. It includes the transfer of cell nucleus from an adult udder cell into an unfertilized, enucleated oocyte (developing egg cell). Electric shock is given to the hybrid cell to stimulate it to divide and the resulting blastocyst is implanted in a surrogate mother. Statement B is correct.
Dolly was cloned by Ian Wilmut, Keith Campbell and colleagues at the Roslin Institute. Statement D is correct.
DNA ligase enzyme joins two DNA fragments together via phosphodiester bond between 3’ hydroxyl at the end of one fragment and a 5’ phosphate at the end of another fragment. Statement E is incorrect as the enzyme used to cleave a DNA molecule is called as restriction endonuclease also known as molecular scissors.

In pedigree analysis, a person immediately affected by genetic alteration or any other changes/action is called propositions. Statement C is correct.
So, the correct answer is option E.

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Gametes Aa ----> Aa	A	a
A	AA normal	Aa normal
a	Aa normal	aa albino

Principles of Inheritance and Variation Test - 86

Result

Principles of Inheritance and Variation Test - 86

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Rank

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SHARING IS CARING

Question 1

As a single contiguous block in mouse chromosome 1

In mouse chromosome 4

As a single block in mouse chromosome 3

All of the above

Solution

Question 2

$$\gamma$$-radiation

Infra red radiation

IAA

Ethylene

Solution

Question 3

Son - 50% ; Daughter - nil

Son - 100% ; Daughter - nil

Son - nil ; Daughter - 100%

Son - nil ; Daughter - nil

Solution

Question 4

Tyrosine - Melanin - Dopaquinone - Dopa

Tyrosine - Dopa - Dopaquinone - Melanin

Dopa - Tyrosine - Dopaquinone - Melanin

Tyrosine - Dopa - Melanin - Dopaquinone

Tyrosine - Dopaquinone - Dopa - Melanin

Solution

Question 5

Hugo de Vries, who worked on evening primrose

Gregor Mendel, who worked on Pisum sativum

Hardy Weinberg, who worked on allele frequencies in a population

Charles Darwin, who observed a wide variety of organisms during sea voyage

Solution

Question 6

900

600

1200

300

Solution

Question 7

Question 8

Changes in DNA

Changes in chromosome

Changes in gene

All of the above

Solution

Question 9

More population of house flies.

Change in sequences of amino acids in alpha-chain of haemoglobin.

Change in sequences of amino acids in beta-chain of haemoglobin.

Poor hygienic conditions.

Solution

Question 10

Mobile genetic elements, transposons were visualized by Barbara McClintock.

Udder cell, a somatic cell is used to produce the cloned sheep by nuclear transplantation method.

In pedigree analysis, a person immediately affected by an action is called propositions.

Dr Ian Wilmut produced a cloned sheep called Dolly.

DNA ligases are used to cleave a DNA molecule.

Solution

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