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Test: Genetics and Evolution (December 23) - NEET MCQ


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15 Questions MCQ Test Daily Test for NEET Preparation - Test: Genetics and Evolution (December 23)

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Test: Genetics and Evolution (December 23) - Question 1

The two polynucleotide chains in DNA are

[2007]

Detailed Solution for Test: Genetics and Evolution (December 23) - Question 1

The two chains are antiparallel, one aligned in 5’→3’  direction the other in 3’ → 5’ direction.

Test: Genetics and Evolution (December 23) - Question 2

In pea plants, yellow seeds are dominant to green. If a heterozygous yellow seeded plant is crossed with a green seeded plant, what ratio of yellow and green seeded plants would you expect in F1 generation ?

[2007]

Detailed Solution for Test: Genetics and Evolution (December 23) - Question 2

Let GG  ⇒ homozygous yellow seed plant.
Gg ⇒ heterozygous green
gg ⇒ homozygous green According to the question : 

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Test: Genetics and Evolution (December 23) - Question 3

DNA fingerprinting refers to

[2004]

Detailed Solution for Test: Genetics and Evolution (December 23) - Question 3

DNA finger printing is the technique of determining nucleotide sequences of certain areas of DNA which are unique to each individual. DNA contains noncistronic hypervariable repeat sequences called VNTR. DNA fingerprinting involves the identification of these VNTRs.

Test: Genetics and Evolution (December 23) - Question 4

Genetic map is one that

[2003]

Detailed Solution for Test: Genetics and Evolution (December 23) - Question 4

Genetic mapping help in studying the architecture of the chromosomes and of entire genome.

Test: Genetics and Evolution (December 23) - Question 5

Industrial melanism is an example of

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Detailed Solution for Test: Genetics and Evolution (December 23) - Question 5

The replacement of grey coloured moth by dark coloured melanic species due to industrial smoke is called industrial melanism. Initially on the lichen covered tree trunks, the mutant moths was more conspicuous due to its black colouration and was therefore more susceptible to predation by birds. The large scale burning of coal during industrial revolution resulted in the deposition of sooty particulate matter on tree trunks. Hence the grey coloured moths became more conspicuous to predatory birds. Thus the population of black moths  increased considerably. 

Test: Genetics and Evolution (December 23) - Question 6

In a random mating population in equilibrium, which of the following brings about a change in gene frequency in a non-directional manner?                                                                                    [2003]

Detailed Solution for Test: Genetics and Evolution (December 23) - Question 6

Migration refers to the movement of individuals into and out of population. Mutation refers to random and sudden heritable variations or changes arising in the genetic constitution.

Test: Genetics and Evolution (December 23) - Question 7

Sickle cell anaemia has not been eliminated from the African population because

[2006]

Detailed Solution for Test: Genetics and Evolution (December 23) - Question 7

In sickle cell anaemia RBCs become sickle shaped which are not supportive for the growth of malarial parasite Plasmodium so it provides immunity against malaria disease.

Test: Genetics and Evolution (December 23) - Question 8

Both sickle cell anemia and Huntington's chorea are

[2006]

Detailed Solution for Test: Genetics and Evolution (December 23) - Question 8

Sickle cell anaemia is a biochemical disorder in which shape of RBCs become sickle-shaped due to the defective haemoglobin. Haemoglobin becomes useless for oxygen transport.
Huntington Chorea in a disease in which atrophy of brain occurs resulting to respiratory irragulations, articulation of speech and irregular limbs movements take place. They both are genetic disease present in any person since birth hence congenital diseases.

Test: Genetics and Evolution (December 23) - Question 9

During transcription, the DNA site at which RNA polymerase binds is called

[2003]

Detailed Solution for Test: Genetics and Evolution (December 23) - Question 9

Regulator is a gene which forms a biochemical for suppressing the activity of operator gene. Promoter is the gene which provides the point of attachment to RNA polymerase required for transcription of structural genes.

Test: Genetics and Evolution (December 23) - Question 10

Two different species can not live for long duration in the same niche or habitat. This law is

[2002]

Detailed Solution for Test: Genetics and Evolution (December 23) - Question 10

Gause's hypothesis states that two different competing species cannot coexist together for a long in the same niche, one has to get eliminated by the other competing species, if they do coexist, then they do so as a result of niche differentiation, i.e. differentiation of then realized niches.

Test: Genetics and Evolution (December 23) - Question 11

Which one of the following describes correctly the homologous structures ?

[2003]

Detailed Solution for Test: Genetics and Evolution (December 23) - Question 11

Those organs which have a common origin and are built on the same anatomical pattern, but perform different functions and are modified accordingly. 

Test: Genetics and Evolution (December 23) - Question 12

Darwin in his 'Natural Selection Theory' did not believe in any role of which one of the following in organic evolution ?

[2003]

Detailed Solution for Test: Genetics and Evolution (December 23) - Question 12

In any population , there is always competition for space and food resources. 

Test: Genetics and Evolution (December 23) - Question 13

Two genes R and Y are located very close on the chromosomal linkage map of maize plant. When RRYY and rryy genotypes are hybridized, the F2 segregation will show

[2007]

Detailed Solution for Test: Genetics and Evolution (December 23) - Question 13

When the linked genes are situated quite close, the chances of crossing over are highly reduced. Due to this, large number of parental gametes are formed and only few recombinant gametes are formed. This results in higher number of parental types in F2 generation as compared to recombinants.

Test: Genetics and Evolution (December 23) - Question 14

A common test to find the genotype of a hybrid is by

[2007]

Detailed Solution for Test: Genetics and Evolution (December 23) - Question 14

crossing of one F1 progeny with male parent. e.g

To find the genotype of  hybrid test cross is done.

Test: Genetics and Evolution (December 23) - Question 15

In recent years, DNA sequences (nucleotide sequence) of mt-DNA and Y chromosomes were considered for the study of human evolution, because

[2003]

Detailed Solution for Test: Genetics and Evolution (December 23) - Question 15

In human cells, chromosomes are present in both nucleus as well as mitochondria. Most of the proteins in humans are encoded by the nuclear genome but mitochondrial DNA plays a significant role too. Mitochondria contain their own DNA that is separate from the nuclear DNA. mtDNA shows maternal inheritance. The nuclear DNA which is equally inherited from both the mother and the father, mitochondrial DNA (mtDNA) is inherited only from the mother. This makes mitochondrial DNA particularly interesting and useful for human evolutionary studies. Normally, mtDNA is identical to that of the mother. But like all DNA molecules, mtDNA molecules also undergo mutations that change the sequence slightly. Thus, mtDNA of a daughter cell would be different from the mtDNA of its mother and this difference is proportional to the amount of time that has passed and therefore proves to be very useful for studying human evolution. Similar is the case with Y chromosome that is inherited from the father and, therefore, follows a similar evolutionary pattern like mtDNA.

So, the correct answer is option C.

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