Test: Heredity And Variation 3 - From Past 28 Years Questions


25 Questions MCQ Test Biology 28 Years NEET/AIPMT Question Papers of Class 12 | Test: Heredity And Variation 3 - From Past 28 Years Questions


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QUESTION: 1

A self-fertilizing trihybrid plant forms

[2004]

Solution:

A self-fertilizing trihybrid plant forms 8 different gametes and 64 different zygotes. The genotype of trihybrid plant is AaBbCc. Number of gametes formed = 2n = 23 = 8 Here n = no. of heterozygosity = 3 in present case.
No. of zygotes formed = 82 = 64.

QUESTION: 2

After a mutation at a geneticlocus the character of an organism changes due to the change in

[2004]

Solution:

Change in genetic locus, changes the gene and in turn the amino acid it codes for. This alters the nature of protein synthesized which produces change in the organism. DNA replication is not affected neither the method of protein synthesis. Synthesis of  RNA over DNA template is called transcription.

QUESTION: 3

In order to find out the different types of gametes produced by a pea plant having the genotype AaBb, it should be crossed to a plant with the genotype:

[2005]

Solution:

In order to find out the gamete or the genotype of an unknown individual, Scientists perform a test cross. In test cross the individual in question is crossed with the  homozygous recessive parent. Hence the answer is aabb.

QUESTION: 4

At a particular locus, frequency of ‘A’ allele is 0.6 and that of ‘a’ is 0.4. What would be the frequency of heterozygotes in a random mating population at equilibrium?

[2005]

Solution:

The mathematical expression to calculate allele frequency can be given as p + q = 1. Where p represents the frequency of dominant allele and q represents the frequency of recessive allele. The bionomial expression of Hardy Weinberg’s law to calculate genotype frequency is p2 + 2pq + q2 = 1 where p2 = homozygous dominant 2pq = heterozygous dominant  q2 = homozygous recessive.
Frequency of heterozygous individuals = 2 × 0.6 × 0.4 = 0.48.

QUESTION: 5

The salivary gland chromosomes in the dipteran larvae, are useful in gene mapping because:

[2005]

Solution:

Polytene chromosomes were first reported by EG Balbiani in 1881. They are quite common in salivary glands of insects and are therefore popularly called salivary gland  chromosomes. Polytene chromosomes also occur in other organs of insects antipodal cells (of embryo sac), endosperm cells and suspensor cell of embryo. The chromosomes can reach length of 200 µm and contain 1000 to 16000 times DNA as compared to the ordinary somatic chromosomes. The giant chromosomes are formed by somatic pairing between homologous chromosomes, and repeated replication (endomitosis) of their chromatids.

QUESTION: 6

G-6-P dehydrogenase deficiency is associated with haemolysis of:

[2005]

Solution:

Glucose-6-P dehydrogenase is the first enzyme of glucose oxidation during Pentose Phosphate Pathway. RBC contain haemoglobin which combines with oxygen to form oxyhaemoglobin which gives its oxygen for oxidation of food. In haemolysis there is destruction of RBCs with release of haemoglobin into plasma resulting in juandice. So, now RBCs cannot provide oxygen for oxidation of food thereby causing deficiency of G-6-P dehydrogenase.

QUESTION: 7

A woman with normal vision, but whose father was colour blind, marries a colour blind man. Suppose that the fourth child of this couple was a boy. This boy

[2005]

Solution:

Since the woman’s father was colour blind. She would be a carrier of the colour blindness gene. When she marries a colour blind man. Their progeny could be 

QUESTION: 8

Which of the following is not a hereditary disease?

[2005]

Solution:

Cystic fibrosis : It is a common disorder of caucasian race in which thick and more salty mucus blocks the respiratory tract. The homozygous recessive condition produces the defective protein which regulates chloride transport channel.
Cretinism : In this disorder the physical growth, mental growth and sexual growth in children is retarded. Such a dwarf and sterile child is called a cretin. It is due to hyposecretion of thyroid hormones.
Thalassemia : Due to defective production of α or β chains of haemoglobin, autosomal recessive.
Haemophilia : Sex linked disorder due to defective recessive gene.

QUESTION: 9

A women with 47 chromosomes due to three copies of chromosome 21 is characterized by:

[2005]

Solution:

Down's syndrome is caused by the presence of an extra chromosome number 21 and the off spring has 47 chromosomes.

QUESTION: 10

Haemophilia is more commonly seen in human males than in human females because:    

[2005]

Solution:

This disease is due to an x - linked recessive mutation. Males suffer this disorder since they have only one X chromosome and hence express any trait on this chromosome.

QUESTION: 11

If a colour blind woman marries a normal visioned man, their sons will be

[2006]

Solution:

Colour blindness in a X-chromosome linked character. So they’ll be having all colour blind sons and carrier daughters.

QUESTION: 12

Which one of the following is an example of polygenic inheritance ?

[2006]

Solution:

Polygenic inheritance is the inheritance of traits which are dependent on the no. of genes such as the skin colour of human beings. eg. AABB is black AaBB in neither dark nor black. AaBb is again wheatish Aabb is light and aabb is white colour.

QUESTION: 13

Cri-du-chat syndrome in humans is caused by the

[2006]

Solution:

Cri-du-chat syndrome in humans caused by the partial deletion of the short arm of chromosome no. 5. In this child has abnormalities in his/her facial expressions, mental retardation, speak like and they cry like that of cat, heart does not work properly.

QUESTION: 14

Both sickle cell anemia and Huntington's chorea are

[2006]

Solution:

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.

QUESTION: 15

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

[2006]

Solution:

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.

QUESTION: 16

In Mendel’s experiment with garden pea, round seed shape (RR) was dominant over wrinkled seeds (rr), yellow cotyledon (YY) was dominant over green cotyledon (yy). What are the expected phenotypes in the F2 generation of the cross RRYY × rryy?

[2006]

Solution:


So the ratio of round seeded with yellow cotyledons : Round seeded green cotyledons : Wrinkled seeded yellow cotyledons : Wrinkle seeded green cotyledons in 9 : 3 : 3 : 1. So in the F2 generation the round seeded character remains with yellow cotyledons & wrinkled seeds with yellow cotyledons.

QUESTION: 17

Test cross involves

[2006]

Solution:

A cross of F1 hybrid with its recessive homozygous parent is called the test cross. It is done to determine the genotype of a given plant. If the given plant has homozygous dominant traits then on test cross it gives all dominant trait plants but if it is heterozygous dominant than it gives dominant and recessive phenotypes in 1 : 1 ratio.

QUESTION: 18

Phenotype of an organism is the result of

[2006]

Solution:

Phenotype is the appearance one organism shows while genotype is the gene complement it has from its ancestors. These genes only show their effect in phenotype but environment also plays an important role in this hence phenotype is a result of genotype and environmental interaction.

QUESTION: 19

How many different kinds of gametes will be produced by a plant having the genotype AABbCC ?

[2006]

Solution:

It would make only two types of gametes, these are ABC & AbC.

QUESTION: 20

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

[2007]

Solution:

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

To find the genotype of  hybrid test cross is done.

QUESTION: 21

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]

Solution:

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.

QUESTION: 22

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]

Solution:

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

QUESTION: 23

The two polynucleotide chains in DNA are

[2007]

Solution:

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

QUESTION: 24

Inheritances of skin colour in humans is an example of

[2007]

Solution:

Inheritance of skin colour in human is controlled by three genes, A B and C which is polygenic inheritance.

QUESTION: 25

A human male produces sperms with the genotypes AB, Ab, aB, and ab pertaining to two diallelic characters in equal proportions. What is the corresponding genotype of this person?

[2007]

Solution:


So, the corresponding genotype will be AaBb.