Explore topic-wise InterviewSolutions in .

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

1001.

Question : When a green pod containing pea plant is crossed with a yellow pod containing pea plant in the progeny yellow pod containing plants are also produced, so green pod containing plant is

Answer»

only heterozygous
only HOMOZYGOUS
may be homozygous or heterozygous
may be homozygous

Answer :A
1002.

Question : When a hybrid tall pea plant is cross fertilized with a dwarf pea plant , the ratio of tall is to dwarf plants grown from the seeds will be :

Answer»

0.12569444444444

Answer :D
1003.

Question : When a green full pod producing pea plant ils crossed to an yellow constricted pod producing pea plant in the progen gren fullpod and green constricted pod producing pea plants only are formed in 1:1 ratio. The genotype of tested individual is

Answer»

GGFF
GfFf
GGFf
GgFf

Answer :C
1004.

Question : When a hybrid (AaBbCcDd) is selfed then the genotypes AABbCCDd, AaBBCcDd, AaBbCcDd, aabbccdd would be in a proportion of :-

Answer»

`2 : 4 : 8 : 21`
`4 : 8 : 16 : 1`
`4 : 8 : 16 : 27`
`8 : 4 : 16 : 81`

ANSWER :B
1005.

Question : When a gene pair contains two different genes controlling different traits of a character, it is

Answer»

HOMOZYGOUS
HETEROZYGOUS
Monozygous
NONE of the above

ANSWER :B
1006.

Question : When a gerey color seed producing pea plant is crossed to white color seed producing pea plant, in the progeny 164 grey seed producing and 156 white seed producing plants are obtained. This cross is

Answer»

Receprocal cross
Test cross
Incomplete dominance
Codominance

Answer :B
1007.

Question : When a gamete without any fusion develops into a new organism the phenomenon is called SyngamyExternal FertilizationParthenogenesisParthenocarpy

Answer»

Syngamy
External Fertilization
Parthenogenesis
Parthenocarpy

Answer :C
1008.

Question : When a gamete without any fusion develops into a new organism the phenomenon is called:

Answer»

SYNGAMY
EXTERNAL fertilization
Parthenogenesis
PARTHENOCARPY

ANSWER :C
1009.

Question : When a gamete without any fusion develop into a new organisms the phenomenon is called

Answer»

Syngamy
External FERTILIZATION
Parthenogenesis
parthenocarpy

Answer :C
1010.

Question : When a foreign DNA is introduced into an organism, how is it maintained in the host and how is it transferred to the progeny of the organism ?

Answer»

SOLUTION :Foreign gene is USUALLY ligated to a plasmid vector and introduced in the HOST. As plasmid replicates, and makes MULTIPLE copies of itself, so does the foreign gene gets replicated and its copes are made. When the host organism divides, its progeny also receives the plasmid DNA containing the foreign gene.
1011.

Question : When a DNA molecule with one radioactive strand undergoes repeated replications in a medium without radioactivity for five times the number of radio labelled strands in the daughter DNA molecules

Answer»

2
1
16
32

Answer :B
1012.

Question : When a diploid female plant is crossed with a tetraploid male, the ploidy of endosperm cells in the resulting seed

Answer»

tetraploidy
pentaploidy
diploidy
triploidy

Answer :A
1013.

Question : When a dihybrid cross is fit into a Punnet square with 18 boxes, the maximum number of different phenotypes available are

Answer»

8
4
2
16

Answer :B
1014.

Question : When a dihybrid cross is fitted into a pannet square with 16 boxes, the maximum number of different phenotypes available are

Answer»

4
4
16
12

Answer :B
1015.

Question : When a cross is made between tall plant with yellow seeds [TtYy1 and tall plant with seed [Tt yy], What proportion of phenotype in the offspring could be expected to be

Answer»

SOLUTION :
= TALL green - 3/8 or 6/16
= DWARF green - 1/8 or 2/16
=PHENOTYPE RATIO = 3:1
1016.

Question :When a dihybrid cross is fit into a punnett square with 16 boxes , the maximum number of different phenotypes available are :

Answer»

8
4
2
16

Answer :B
1017.

Question : When a dihybrid cross is fit into a Punnet square with 16 boxes, the maximum number of different phenotypes available are

Answer»

8
4
2
16

Answer :B
1018.

Question : When a corss is made pinkflowered and red flowered snapdragon plants, what proportionof phenotypein the offspring could beexpected to be

Answer»

0.25
0.75
0.5
0.125

Answer :C
1019.

Question : When a cross is conducted between black feathered hen and a white feathered cock , blue feathered fowls are formed . When these fowls are allowed for interbreeding , in F_1 -generation , there are 20 blue fowls . What would be the number of black and white fowls .

Answer»

BLACK 20 , WHITE 10
Black 20 , white 20
Black 10 , white 10
Black 10 , white 20

Answer :C
1020.

Question : When a connective is prolonged into a feathery appendix beyond the anthers it is called ………….andis found inn………….

Answer»

distractile NERIUM
appendicualte SALVIA
DIVARICATE salvia
APPENDICULATE nerium

ANSWER :D
1021.

Question : When a cell is plasmolysed

Answer»

TURGOR PRESSURE BECOMES negative
Pressure POTENTIAL is negative
Osmotic potential is negative
All the above

Answer :D
1022.

Question : "When a cell is placed in isotonic solution water will come out of it by exosmosis" . Is it true or false .

Answer»


ANSWER :HYPERTONIC
1023.

Question : When a cell is fully turgid which of the following will be zero

Answer»

TURGOR PRESSURE/pressure potential
Wall pressure
Suction pressure/DPD/water potential
Osmotic pressure (SOLUTE pressure)

ANSWER :C
1024.

Question : When 12 CO_2 molecueles are utilized in C_2 cycle, number of troise phosphates exported out from the chloroplast into the cytosol for the synthesis of hexose will be

Answer»

2
4
6
12

Answer :B
1025.

Question : When 54 molecules of CO_2 fixed by RuBisCO in a C_3 plant , number of G_3 -P participate in regeneration phase respectively

Answer»

90,18
54,54
60 , 48
18, 90

Answer :D
1026.

Question : wheat variety having high protein content is

Answer»

Kalyan SONA
Himgiri
Sharbati SONORA
ATLAS 66

Answer :D
1027.

Question : wheat root cells have 42 chromosomes. the number of chromosomes in a cell of pollen grain is

Answer»

`14`
`21`
`28`
`42`

ANSWER :B
1028.

Question : Wheat variety, Atlas-66 is improved for …………. .

Answer»

HIGH proteins
high carbohydrates
high fats
high vitamins

Answer :A
1029.

Question : Wheat is which of the following types of fruit

Answer»

Berry
Nut
Caryopsis
Legume(POD)

ANSWER :C
1030.

Question : Whch one of the following is an important breed of fowl?

Answer»

Nageswari
Jersey
Rhode ISLAND Red
Khaki campbell

Answer :C
1031.

Question : Whatever be the life span, death of every individual organism is a certainty, i.e., no individual is immortal, except

Answer»

HUMAN beings
Amoeba and Paramoecium
Single-celled organisms
Both B and C.

Answer :D
1032.

Question : What you meant by resource partitioning in an ecosystem.

Answer»

SOLUTION : If TWO SPECIES foe the same resource, they avoid competition by choosing different times for feeding or different foraging patterens.This is called resource PARTITIONING.
1033.

Question : What .X. indicates in the given figure?

Answer»

Geitonogamy
Homogamy
Autogamy
Cleistogamy

Answer :C
1034.

Question : What would happen when one grows a recombinant bacterium in a bioreactor but forget to add antibiotic to the medium in which the recombinant is growing ?

Answer»

Solution :In the absence of ANTIBIOTIC, there will be no pressure on RECOMBINANTS to retain the PLASMID (containing the gene of your interest). Since, maintaining a high copy number of plasmids is a METABOLIC burden to the microbial cells, will thus tend to lose the plasmid.
1035.

Question : What would happen when one grows a recombinant bacterium in a bioreactor but forget to add antibiotic to the medium in which the recombinant is growing?

Answer»

SOLUTION :In the absence of antibiotic, all bacteria will grow successfully. There will be no selection pressure. So there will be no pressure on transformed bacteria to RETAIN the recombinant PLASMID and they MAY TEND to lose these plasmids.
1036.

Question : What would happen to immune system, if thymus gland is removed from the body of a person?

Answer»

Solution :Thymus is the primary lymphoid organ.
In thymus GLAND, immature lymphocytes DIFFERENTIATE into antigen-sensitive lymphocytes.
If thymus gland is removed from the BODY of a person, the immune system BECOMES WEAK.
As a result the person.s body becomes prone to infectious diseases.
1037.

Question :What would happen if oxygen availability to activated sludge flocs is reduced ?(A) It will slow down the rate of degradation of organic matter.(B) The centre of flocs will become anoxic, which would cause death of bacteria and eventually breakage of flocs.(C) Flocs would increase in size as anaerobic bacteria would grow around flocs.(D) Protozoa would grow in large numbers.

Answer»

It will slow down the rate of degradation of ORGANIC matter.
The centre of flocs will become anoxic, which would cause death of bacteria and eventually breakage of flocs.
Flocs would INCREASE in size as anaerobic bacteria would grow around flocs.
PROTOZOA would grow in large numbers.

Solution : If oxygen availability to activated sludge FLOES is reduced the centre of flocks will become anoxic, which would cause death of bacteria as they require oxygen and eventually breakage of flocks. As a result foul smelling products are produced in anaerobic DECOMPOSITION.
1038.

Question : What would happen if oxygen availability to activated sludges flocs is reduced ?

Answer»

It will slow down the rate of degradation of organic matter
The CENTER of floes will become anoxic . which would cause death of bacteria and eventually breakage of floes
Flocs would increase in SIZE as anaerobic bacteria would GROW AROUND flocs
Protozoa would grow in large numbers

Answer :B
1039.

Question : What would happen if histones were to be mutated and made rich in acidic amino acids such as aspartic acid and glutamic acid in place of basic amino acids such as lysine and arginine?

Answer»

Solution :If HISTONE proteins were rich in acidic amino acids INSTEAD of basic amino acids then they may not have any role in DNA packaging in eukaryotes as DNA is ALSO negatively charged molecule. The packaging of DNA around the nucleosome would not happen. Consequently, the chromatin FIBRE would not be formed.
1040.

Question :What would happen if our intestine harbours microbial flora exactly similar to that found in the rumen of cattle ?

Answer»

SOLUTION :If our intestine HARBOURS microbial flora EXACTLY similar to that found in the RUMEN of cattle then we would be able to digest the cellulose PRESENT in our food.
1041.

Question : Whate would happen if in a gene encoding a polypeptide of 50 amino acids , 25^(th) codon (UAU) is mutated to UAA

Answer»

A POLYPEPTIDE of 24 amino acids will be FORMED
Two polypeptides of 24 and 25 amino acids will be formed
A polypeptide of 49 amino acids will be formed
A polypeptide of 25 amino acids will be formed

Answer :a
1042.

Question :What would happen if a large volume of untreated sewage is discharged into a river ?

Answer»

SOLUTION : Due to increasing urbanisation, sewage is being produced in much larger QUANTITIES than ever before. However, the number of sewage treatment PLANTS has not increased enough to treat such large quantities. So, the untreated sewage is often DISCHARGED directly into rivers leading to their POLLUTION and increase in waterborne diseases.
1043.

Question : What would Earth be like without the greenhouse effect?

Answer»

Solution :GREENHOUSE EFFECT is vital for the sustenance of life. Greenhouse gases like CO2, water vapour etc absorb some of the reflected sun.s radiation and radiate BACK it to the Earth surface, thus maintaining the Earth.s warm condition. Without this effect, life on Earth would be difficult or rather impossible for existence or become hostile to most LIVING organisms.
1044.

Question : How many full turns of the Calvin cycle are required to make one molecule of glucose

Answer»

5
4
12
6

Answer :D
1045.

Question : What would be the ploidy of the cell of the tetrad ?

Answer»


ANSWER :HAPLOID;
1046.

Question : What would be the phenotypic ratio in F_(2) generation in a dihybrid cross if both the genes show lethality in homozygous dominant state -

Answer»

`2 : 2 : 2 : 2`
`4 : 2 : 2 : 1`
`6 : 3 : 2 : 1`
`9 : 3 : 3 : 1`

Answer :B
1047.

Question : What would be the per cent growth or birth rate per indiviual per hour for the same population mentional in the previous question?

Answer»

100
200
50
150

Answer :B
1048.

Question : What would be the percent growth or birth rate per individual per hour for the same population mentioned in the previous question (Question 10)?

Answer»

100
200
50
150

Answer :A
1049.

Question : What would be the per cent growth or birth rate per individual per hour for the same population mentioned in the previous ques- tion (Question 10 )?

Answer»

100
200
50
150

Solution :Initial number of PARAMECIUM = 50
Number of Paramecium after 1 HOUR = 150
Birth rate = `("Number of NEW Paramecium")/("Initial number of Paramecium")xx100`
`=(100)/(50)xx100=200%`
1050.

Question : What would be the per cent growth or birth rate per individual per hour for the same population mentioned in the previous question ?

Answer»

100 
200 
50 
150 

SOLUTION :The per cent growth or birth rate per individual per HOUR
= `("FINAL population - Initial population ")/("Initial population") XX 100`
`= (150 - 50)/(50) xx 100 = 200`