Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

There is only one possible sequence of amino acids when deduced from a given nucleotide. But multiple nucleotide sequences can be deduced from a single amino acid sequence. Explain this phenomenon.

Answer»

Some amino acids are coded by more than one codon (known as degeneracy of codon), hence on deducing a nucleotide sequence from an amino acid sequence, multiple nucleotide sequences will be obtained. 

For example, isoleucine has three codons AUU, AUC and AUA. Hence a dipeptide Met– Ile can have any of the following nucleotide sequences: 

i. AUG–AUU

ii. AUG–AUC

iii. AUG–AUA

If we deduce amino acid sequences of the above nucleotide sequences, all the three will code for Met–Ile.

2.

What does the lac operon consist of ? How is the operator switch on and off in the expression of gene in this operon? Explain.

Answer»

The lac operon consists of structural genes, promoter, operator, repressor, and inducer. 

When Lactose is Absent 

i. When lactose is absent, i gene regulates and produces repressor mRNA which translate repression. 

ii. The repressor protein binds to the operator region of the operon and as a result prevents RNA polymerase to bind to the operon. 

iii. The operon is switched off. 

When Lactose is Present 

i. Lactose acts as an inducer which binds to the repressor and forms an inactive repressor. 

ii. The repressor fails to bind to the operator region. 

iii. The RNA polymerase binds to the operator and transcribes lac mRNA. 

iv. lac mRNA is polycistronic, i.e., produces all three enzymes, β-galactosidase, permease and transacetylase. 

v. The lac operon is switched on.

3.

Some angiosperm seeds are said to be albuminous, whereas a few others are said to have a perisperm. Explain each with the help of an example.

Answer»

Albuminous seeds retain a part of endosperm as it is not completely used up during embryo development. e.g., wheat, maize, barley, castor, sunflower. 

When remnants of nucellus are persistent it is said to have a perisperm. Example; biack pepper, beat.

4.

Double fertilization is reported in plants of both, castor and groundnut. However, the mature seeds of groundnut are non-albuminous and castor are albuminous. Explain the post fertilization events that are responsible for it.

Answer»

Development of endosperm (preceding the embryo) takes place in both, developing embryo derives nutrition from endosperm Endosperm is retained/persists /not fully consumed in castor, endosperm is consumed in groundnut.

Detailed Answer: 

Endosperm development precedes embryo development. The triploid primary endosperm nucleus (PEN) undergoes repeated mitotic divisions, without cytokinesis, at this stage of development, the endosperm is called free-nuclear endosperm wall formation takes place later on, as a result, the endosperm becomes partly or fully cellular. The cells of the endosperm store food materials, which are later used by the developing embryo.

The endosperm may be completely utilized by the developing embryo before the maturation of seeds as in pea, bean, groundnut etc. in non-albuminous or non-endospermic seeds. In albuminous or endospermic seeds, a portion of endosperm persists in the mature seeds. e.g., castor.

5.

Give one example each of albuminous and non-albuminous seeds.

Answer»

Example : 

Non-albuminous seed  : Pea, bean, mustard. 

Albuminous seed  :  Castor, maize, coconut.

6.

Differentiate between albuminous and non-albminous seeds, giving one example of each.

Answer»

Albuminous - (with residual) endosperm is not completely used up during embryonic development. e.g, wheat / maize / castor / sunflower. 

Non albuminous - (with out residual) endosperm is completely consumed during embryonic development. e.g., pea / groundnut.

7.

You are repeating the Hershey–Chase experiment and are provided with two isotopes: 32P and 15N (in place of 35S in the original experiment). How do you expect your results to be different ?

Answer»

Use of 15N will be inappropriate because method of detection of 35P and 15N is different (32P being a radioactive isotope while 15N is not radioactive but is the heavier isotope of nitrogen). Even if 15N was radioactive then its presence would have been detected both inside the cell (15N incorporated as nitrogenous base in DNA) as well as in the supernatant because 15N would also get incorporated in amino group of amino acids in proteins). Hence, the use of 15N would not give any conclusive results.

8.

Why does endosperm development precede embryo development in angiosperm seeds ? State the  role of endosperm in mature albuminous seeds.

Answer»

As it provides nutrition for the developing embryo. It is an adaptation to provide assured nutrition to the developing embryo. Provides nutrition during and after germination.

9.

Explain the role of 35S and 32P in the experiments conducted by Hershey and Chase.

Answer»

Viruses grown in the medium containing 32P contained radioactive DNA but not radioactive protein because DNA contains phosphorus but proteins do not contain phosphorus. Similarly, viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur.

10.

A youth in his twenties met with an accident and  succumbed to the injuries. His parents agreed to donate his organs. List any two essential clinical steps to be undertaken before any organ transplant. Why is the transplant rejected sometimes? What views would you share with your health club members to promote organ donation?

Answer»

Blood group matching, and tissue matching should be done prior to the organ transplant, the body is able to identify the'non-self' triggers the cell-mediated immune response, this rejects the graft.

Views:

The cornea can be transplanted to anyone and a  blind can see the world. HeartflungA(deny cal be transplanted and a Person is gifted -with life the mindset to volunteer to register for organ donation - (particularly eye donation). Even the live donors can donate some of their body parts e.g. kidney, liver, bone or its parts. Members of youth club should be motivated to propagate awareness in the society with regard to organ donation so that precious life in emergency may be saved.

11.

Differentiate between Monocot & dicot stem.

Answer»
Monocot stemDicot stem
(1) Multicellular hairs absent (1) Multicellular epidermal hair present
(2) Hypodermis sclerenchymatous(2) Hypodermis Collenchymatous
(3) Vascular bundles numerous and scattered. (Atactostele)(3) Vascular bundles limited, arranged as a broken ring (Eustele)
(4) Vascular bundles collateral, conjoint, closed, endarch(4) Vascular bundles collateral, conjoint, open, endarch
(5) Endodermis, pericycle absent(5) Endodermis, pericycle preseent
(6) Stem shows ground tissue (6) Stem show cortex and pith
12.

Government has decided to ban the sale of gutka. Give your views.

Answer»

Many states of India have banned the sale, manufacture, distribution and storage of gutka and all its variants. Gutka is banned under the provision to ban any food product containing harmful adulterants in the centrally enacted Food Safety and Regulation Act. This ban is enforced by the state Public Health Ministry, the State Food and Drug Administration and the local police. Gutka should be banned because cheating of gutka may cause a number of diseases. It may cause throat and lung cancer, discolours teeth and damages the digestive and nervous system.

13.

Given alongside is an enlarged view of one microsporangium of a mature anther. (i) Name ‘a’, ‘b’ and ‘c’ wall layers. (ii) Mention the characteristics and function of the cells forming wall layer ‘c’.

Answer»

(i) 

= Endothecium, 

= Middle layers, 

= Tapetum 

(ii) Tapetum provides nourishment to the developing pollen grains. The tapetal cells also secrete Ubisch granules that provide sporopollenin and other materials for exine formation.

14.

Why are the tumour cells dangerous ?

Answer»

The tumour cells are dangerous because- 

-They invade and damage the surrounding normal cells because of their rapid growth. 

- The normal surrounding cells are starved because of their competing for vital nutrients. 

- Some of the tumour cells are sloughed off and migrate to different sites of the body where they develop into new and secondary tumour by metastasis.

15.

Name the cellular genes which when activated can cause cancer.

Answer»

Cellular oncogenes or proto-oncogenes.

16.

Difference between Dicot Stem and Monocot Stem.

Answer»
Dicot StemMonocot Stem
1. The ground tissue is differentiated into cortex, endodermis, pericy and pith. 1. The ground tissue is made up of similar cells.
2. The vascular bundles are arranged in a ring.2. The vascular bundles are scattered throughout the ground tissue.
3. Vascular bundles are open, without bundle sheath and wedge-shaped outline.3. Vascular bundles are closed, surrounded by sclerenchymatous bundle sheath, oval or rounded in shape.
4. The stem shows secondary growth  due to presence of cambium between xylem and phloem.4. Secondary growth is absent.
5. Stomata have kidney-shaped guard cells.5. Stomata have dumb bell-shaped guard cells.

17.

All human beings have cellular oncogenes but only a few suffer from cancer disease. Give reasons.

Answer»

All humans have cellular oncogenes or proto-oncogenes, but only a few suffer from cancer because cancer only occurs on activation of oncogenes. This activation is induced by carcinogens which can be physical, chemical or biological. The chemical carcinogens present in tobacco smoke have been identified as a major cause of lung cancer.

18.

From the reappearance of recessive trait in F2 generation, Mendel concluded that(A) factors do not mix with each other in F1 generation.(B) factors remain together in F1 generation.(C) factor mix with each other in F1 generation.(D) both (A) and (B)

Answer»

Correct answer is

(D) both (A) and (B)

19.

Explain the following in context of cancer:(i) Benign tumour(ii) Malignant tumour(iii) Oncogens/Carcinogens(iv) Oncogenes(v) Contact inhibition

Answer»

(i) Benign tumours are the masses of cells which remain confined to their original location and do not spread to other parts of the body and cause little damage.

(ii) Malignant tumours are the masses of proliferating cells called neoplastic or tumour cells. These grow very rapidly, invading and damaging the surrounding normal tissues.

(iii) Transformation of normal cells into cancerous, neoplastic cells may be induced by physical, chemical or biological agents. These agents are called carcinogens. For example X-rays, gamma rays, UV radiations and some chemicals like EtBr.

(iv) The genes which may lead to oncogenic transformations of the cells are called oncogenes.

(v) Contact inhibition—Whenever normal cells come in contact with each other, after a definite time they inhibit each other’s excess growth and multiplication. This property of normal cells is called contact inhibition which maintains the normal shape and size of the body. But cancer cells appear to have lost this property which results in their uncontrolled growth and multiplication.

20.

Why do some adolescents start taking drugs? How can the situation be avoided?

Answer»

Reasons for alcohol abuse in adolescents: 

i. Social pressure. 

ii. Curiosity and need for adventure, excitement and experiment. 

iii. To escape from stress, depression and frustration. 

iv. To overcome hardships of daily life. 

v. Unstable or unsupportive family structure. 

For measures to avoid taking drug are as follows: 

i. Avoid undue peer pressure. 

ii. Educating and counselling the problems and stresses to avoid disappointments and failures in life. 

iii. Seeking help from parents and peers. 

iv. Looking for danger signs to take appropriate measures on time. 

v. Seeking professional and medical help whenever required.

21.

Reappearance of recessive trait in F2 generation is due to(A) Law of independent assortment(B) Law of dominance(C) Law of codominance(D) Law of purity of gametes

Answer»

Correct answer is

(D) Law of purity of gametes

22.

Mendel’s principle of segregation is based on separation of alleles during(A) gamete formation(B) seed formation(C) pollination(D) embryonic development

Answer»

(A) gamete formation

The law of segregation states that when a pair of allelomorphs are brought together in the F1 hybrid they co-exist or remain together in the hybrid without blending or in any way contaminating each other and they separate completely and remain pure during the formation of gametes.

23.

According to Mendel, plants of F1 generation show(A) law of dominance(B) purity of gametes(C) independent assortment of genes(D) all of these

Answer»

(A) law of dominance

The character which is expressed in F1 generation is dominant and the recessive character is suppressed in F1 generation.

24.

Mendel formulated the law of dominance and law of purity of gametes on the basis of(A) test cross(B) back cross(C) monohybrid cross(D) dihybrid cross

Answer»

Correct answer is

(C) monohybrid cross

25.

The factors which represent the contrasting pairs of characters are called(A) dominant and recessive(B) alleles(C) homologous pairs(D) determinants

Answer»

Correct option is (B) alleles

26.

Which of the following Mendel’s laws has not been proved to be true in all cases?(A) Law of segregation(B) Mendel’s second law of inheritance(C) Law of dominance(D) Law of purity of gametes

Answer»

Correct answer is C.

In some cases, there is incomplete dominance or no dominance. Law of dominance could not support such cases. Hence, it is not universally acceptable.

27.

ABO blood grouping is controlled by gene I which has three alleles and show co-dominance. There are six genotypes. How many phenotypes in all are possible?(A) six(B) Three(C) Four(D) Five

Answer»

(C) Four

The six genotypes are - IAIA or IAi, IBIB or IBi, IAIB, ii.

The four phenotypes are – A, B, AB, O.

28.

Genes located on same locus but show more than two different phenotypes are called(A) polygenes(B) multiple alleles(C) co-dominants(D) pleiotropic genes

Answer»

 Correct Option is (B) multiple alleles

29.

Innermost wall layer of microsporangium which nourishes the developing pollen grain is called ......

Answer»

Innermost wall layer of microsporangium which nourishes the developing pollen grain is called Tapetum 

30.

The second law of inheritance proposed by Mendel deals with(A) dominance(B) independent assortment(C) segregation(D) epistasis

Answer»

Correct answer is C.

Mendel’s first law is the law of dominance. Law of independent assortment is the third law. Epistasis is a drawback in Mendel’s studies, where intergenic suppression of characters is observed.

31.

State Mendel’s second law of inheritance or law of segregation or law of purity of gametes.

Answer»

Law of segregation states that “when the two alleles for a contrasting character are brought together in a hybrid, they do not mix or contaminate but segregate or separate out from each other during gamete formation”. Law of segregation is also known as law of purity of gametes, as gametes have only one allele.

32.

Blood grouping in humans is controlled by(A) 4 alleles in which A is dominant.(B) 3 alleles in which AB is co-dominant.(C) 3 alleles in which none is dominant.(D) 3 alleles in which A is dominant.

Answer»

Correct Option is (B) 3 alleles in which AB is co-dominant.

33.

The outermost and innermost wall layers of microsporangium in an anther are respectively: a. Endothecium and tapetum b. Epidermis and endodermis c. Epidermis and middle layer d. Epidermis and tapetum

Answer» d. Epidermis and tapetum
34.

“Gametes are never hybrid”. It is a statement of law of(A) dominance(B) segregation(C) independent assortment(D) unit character

Answer»

 Correct Option is (B) segregation

35.

Inheritance of skin colour in humans is an example of(A) Point mutation(B) Polygenic inheritance(C) Co-dominance(D) Chromosomal aberration

Answer»

 Correct Option is (B) Polygenic inheritance

36.

In the absence of an agreement to the contrary, the partners areA. entitled for 6% interest on their capitals, only when there are profitsB. entitled for 9% interest on their capitals, only when there are profitsC. entitled for interest on their capitals at the bank rate, only when there are profitD. not entitled for interest on their capitals

Answer» Correct Answer - D
37.

Rohit, a partner is to carry out dissolution and he gets Rs 50,000 as remuneration. Realisation Expenses were Rs 25,000. What will be the amount debited to Realisation Account ?A. Rs 50,000B. Rs 75,000C. Rs 25,000D. None of these

Answer» Correct Answer - B
38.

A dicotyledonous plant bears flowers but never produces fruits and seeds. The most probable cause for the above situation is: a. Plant is dioecious and bears only pistillate flowers b. Plant is dioecious and bears both pistillate and staminate flowers c. Plant is monoecious d. Plant is dioecious and bears only staminate flowers.

Answer» d. Plant is dioecious and bears only staminate flowers.
39.

In case of fixed capital, partners will haveA. credit balances in their Capital AccountsB. debits balances in their Capital AccountsC. may have credit or debit balances in their Capital AccountsD. None of the above

Answer» Correct Answer - A
40.

Appearance of new combinations in F2 generation in a dihybrid cross proves the law of _______. (A) dominance(B) segregation(C) independent assortment(D) purity of gametes

Answer»

 Correct Option is (C) independent assortment

41.

New character combinations appear in F2 generation of a dihybrid cross mainly because of(A) dominance(B) recessiveness(C) principle of unit character(D) independent assortment

Answer»

Correct answer is

(D) independent assortment

42.

During dihybrid cross, the ratio of yellow : green and round : wrinkled in F2 generation is(A) 1 : 3(B) 3 : 1(C) 9 : 3(D) 3 : 9

Answer»

(B) 3 : 1

Yellow round = 9

Yellow wrinkled = 3

Green round = 3

Green wrinkled = 1

From above,

i. Yellow coloured seeds = 9 + 3 = 12

Green coloured seeds = 3 + 1 = 4

∴ Yellow : Green = 12:4 = 3:1

ii. Similarly,

Round seeds = 9 + 3 = 12

Wrinkled seeds = 3 + 1 = 4

∴ Round wrinkled = 12:4 = 3:1

43.

In the absence of Partnership Deed, partners are paid remunerationA. `"@ "Rs 10,000` per partner per monthB. `"@ "Rs 20,000` per partner per monthC. `"@ "Rs 30,000` per partner per monthD. None of the above

Answer» Correct Answer - D
44.

In an embryo sac, the cells that degenerate after fertilisation are: a. Synergids and primary endosperm cell b. Synergids and antipodals c. Antipodals and primary endosperm cell d. Egg and antipodals.

Answer» b. Synergids and antipodals
45.

In case of fixed capital, Partners Current Accounts will haveA. credit balancesB. debit balancesC. credit or debit balancesD. None of these

Answer» Correct Answer - C
46.

In a typical complete, bisexual and hypogynous flower the arrangement of floral whorls on the thalamus from the outermost to the innermost is: a. Calyx, corolla, androecium and gynoecium b. Calyx, corolla, gynoecium and androecium c. Gynoecium, androecium, corolla and calyx d. Androecium, gynoecium, corolla and calyx

Answer» a. Calyx, corolla, androecium and gynoecium
47.

Current Accounts of partners are maintaned ifA. capitals are fixedB. capitals are fluctuatingC. Both (a) and (b)D. None of these

Answer» Correct Answer - A
48.

Which is the triploid tissue in a fertilised ovule? How is the triploid condition achieved?

Answer»

The triploid tissue in the ovule is the endosperm. Its triploid condition is achieved by the fusion of two polar nuclei and one nucleus of male gamete, referred to as triple fusion.

49.

Natural selection theory of Darwin is objected, because it(A) stresses upon slow and small variations.(B) stresses upon inter-specific struggle.(C) explains natural calamities with heavy toll.(D) none of these

Answer»

 Correct Option is (A) stresses upon slow and small variations

50.

Introduction of exotic species is one of the major threats of biodiversity.1. Cite any two examples of these exotic species in your locality. 2. Recent illegal introduction of a fish for aquaculture poses a threat to indigenous cat fishes in our rivers. Name it. 3. How co-extinctions affects biodiversity?

Answer»

1. Eichornia - Lantana Camera 

2. African cat fish (clarius gariepinus) 

3. When a species becomes extinct, the species associated with it in an obligatory way also become extinct. This is called co-extinction.