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6101.

Question : The function of leghaemoglobin during biological nitrogen fixation in root nodules of legumes is to:

Answer»

CONVERT ATMOSPHERIC `N_2` to `NH_3`
Convert ammonia to nitrite
Transport oxygen for ACTIVITY of NITROGENASE
Protect nitrogenase from oxygen

Answer :D
6102.

Question : The function of IgE is

Answer»

mediate in ALLERGIC response
activation of B-cells
PROTECTION from inhaled and ingested pathogens
stimulation of COMPLEMENT system, PASSIVE immunity to foetus

Answer :A
6103.

Question : The function of insulin hormone is :-

Answer»

To release GLUCOSE from LIVER CELLS and glycogenolysis promotion
To increase CELLULAR glucose uptake and consumption
To increase blood sugar level
To increase lipolysis

Answer :B
6104.

Question : The function of contractile vacuole is_____

Answer»


ANSWER :OSMOREGULATION
6105.

Question : The function of copper ions in copper releasing IUD's is

Answer»

They INHIBIT gametogenesis
They make uterus unsuitable for impantation
They inhibit ovulation
The suppress sperm MOTILITY and fertilising capacity of SPERMS

SOLUTION :The function of copper ions in copper RELEASING IUD's is that they suppress sperm motility and fertilising capacity of sperms.
6106.

Question : The fully processed hnRNA is called as (i)____ and is transproted out of the(ii)____ into the (iii) ____ for translation.

Answer»

`{:("(i)","(II)","(III)"),("MRNA","nucleus","CYTOPLASM"):}`
`{:("(i)","(ii)","(iii)"),("mRNA","cytoplasm","nucleus"):}`
`{:("(i)","(ii)","(iii)"),("tRNA","cytoplasm","nucleus"):}`
`{:("(i)","(ii)","(iii)"),("tRNA","nucleus","cytoplasm"):}`

SOLUTION :(a)
6107.

Question : The function of a selectable maker is

Answer»

ELIMINATING TRANSFORMATION and permitting non-TRANSFORMANTS
Identify ori site
Elimination of non-transformantsand permitting transformants
To destroy recognition SITES

ANSWER :C
6108.

Question : The funcation of the secretion of prostate glands is to

Answer»

inhibit sperm activity
attract sperms
stimulate sperm activity
none of these.

Solution :The Prostate gland is a single large gland that surrounds the URETHRA. It produces a SLIGHTLY alkaline, mailky secretion whichforms `25%` of the volume of semen. It PROCESSES citric acid, enzymes (acid phosphatase, amylase, pepsinogen) and prostaglandins). Secretion of the prostate gland NOURISHES and activites the speratozoa to swim.
6109.

Question : The fruit which breaks into five valves on touching and results in explosive dispersal of seeds is

Answer»

Ecballium
BALSAM
Entada gigas
Ruellia

ANSWER :B
6110.

Question : The fruit of wood apple develop from multicarpellary syncarpous multioculr ovary This is called as

Answer»

ETAERIO of a ACHENE
AMPHISARCA
balausta
hesperidium

ANSWER :B
6111.

Question : The fruit of regma breaks up into single seeded dehisecnt parts called coccl these cocci remain attached onto

Answer»

WINGED APPENDAGES
appapus
STYLOPODIUM
CARPOPHORE

ANSWER :D
6112.

Question : The fruit of coconut is called as

Answer»

Pepo
Etaerio
Pome
Drupe

Answer :D
6113.

Question : The fruit of coconut is :

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Drupe
Hesperidium
Composite FRUIT
Berry

Answer :A
6114.

Question : the frequency of recombination between gene pairs on the same chromosome as measured of the distance between genes was explained by:

Answer»

GREGOR J Mendel
Alfred Sturtevant
Sutton Boveri
T.H. Morgan

Answer :B
6115.

Question : The front end of filament is connected to what?

Answer»

Connective
Placenta
Thalamus or petal
Anther

Answer :C
6116.

Question : The frist clinical gene therapy was done of the treatment of:

Answer»

SCID
Toxoid
Toxin
Antibiotic

Answer :D
6117.

Question : The frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes was explained by :

Answer»

T.H. Morgan
GREGOR J. MENDEL
ALFRED Sturtevant
Sutton Boveri

Solution :Alfred Sturtevant
6118.

Question : The frequency of an autosomal recessive gene is 0.4. Then what will be the frequency of heterozygotes in progeny among 4000 individual ?

Answer»

2000
1920
2400
1600

Answer :B
6119.

Question : The frequency of crossing over would be higher if

Answer»

TWO genes are located closely
Two genes are FAR apart on a chromosome
Two genes are not located in the same chromosome
None of the above

Answer :B
6120.

Question : The frequencies of alleles in a Mendelian population are subjected to influence of

Answer»

GENETIC DRIFT
Mutation
Natural selection
All the above

SOLUTION :All the above
6121.

Question : The freely movable joint is called______.

Answer»


ANSWER :SYNOVIAL JOINT
6122.

Question :The free-living fungus Trichoderma can be used for

Answer»

Killing insects
BIOLOGICAL control of plant diseases
Controlling butterfly caterpillars
Producing antibiotics

SOLUTION :The free-living fungus Trichoderma can be USED for biological control of plant diseases. This is commonly found in soil.
6123.

Question : The fragmentation of a forest and deforestation lead to the disappearance of species and drastic climatic variation. a. Give some adverse effects of indiscriminate cutting down of forests. b. Suggest some measures to check deforestation.

Answer»

Solution :The indiscriminate cutting down of forest is also KNOWN as deforestation. Due to deforestation the main problem is the soil erosion. The rain drops DIRECTLY fall on the ground causing soil erosion. The soil will be carried to the RIVER and gets deposited in the river bed. The FORESTS are also having a significant role in the climatic regulation. The measures for checking deforestation can be only done by public awareness and STRICT legislations.
6124.

Question : The free - living fungus Trichoderma can be used for :

Answer»

KILLING insects
biological CONTROL of PLANT diseases
controlling BUTTERFLY caterpillars
producing antibiotics

Answer :B
6125.

Question : The free genes started absorbing the organic compounds from the prebiotic soup and evolved into

Answer»

ANAEROBIC autotrophs
AEROBIC HETEROTROPHS
Aerobic protobionts
Anaerobic heterotrophs

Answer :D
6126.

Question : The free energy per mole -of a substance in a chemical system.

Answer»


ANSWER :CHEMICAL POTENTIAL
6127.

Question : The fraction of water held by particles of soil surfaces.

Answer»


ANSWER :HYGROSCOPIC WATER
6128.

Question : The fraction of single heterozygous individuals in the F_(2) generation of a typical dihyrbid cross

Answer»

`8//16`
`10//16`
`9//16`
`6//16`

ANSWER :A
6129.

Question : The fraction of single heterozygous double dominant individuals in the F_(2) of a Mendelian dihybrid cross

Answer»

`4//16`
`1//16`
`8//16`
`9//16`

ANSWER :A
6130.

Question : The fraction of double homozygous recombinants in the F_(2)of a Mendelian dihybrid cross is

Answer»

`6//16`
`2//16`
`4//16`
`3//16`

ANSWER :B
6131.

Question : The fraction of single homozygous individuals in the F_(2) of a typical dihybrid cross is

Answer»

`9//16`
`4//16`
`2//16`
`1//16`

ANSWER :D
6132.

Question : The four elements that make up 99% of all elements found in a living system are

Answer»

H,O, C, N
C, H, O, S
C, H, O , P
C, N, O, P

Answer :A
6133.

Question : The four pyramids (A, B, C and D) given below represent four different type of ecosystem ecological pyramids, which one of these is correctly identified in the options given, along with its correct ecosystem & type of pyramid:-

Answer»


ANSWER :C
6134.

Question :The four nitrogen base sequences which form the code words for DNA language are :

Answer»

UTAC
ACTU
AGGU
ATCG

Answer :D
6135.

Question : The four nitrogen base sequence which form the code words for DNA language are

Answer»

UTAC
ACTU
AGCU
ATCG

ANSWER :D
6136.

Question : The fossilised microbial mats which consist of filamentous prokaryotes and trapped sediments are

Answer»

STROMATOLITES
Sulphur bacteria
Saprophytes
NONE of the above

Answer :A
6137.

Question : The four carbon compound formed during the regeneration of RUBP in Calvin cycle

Answer»

Sedoheptulose PHOSPHATE
Xylulose phosphate
ERYTHROSE phosphate
Ribose phosphate

ANSWER :C
6138.

Question : The formula of population density is ……….(A) N_(t+1) = N_(t) + [(B-I)-(D+E)](B) N_(t+1) = N_(t) + [(B+I)-(D+E)](C) N_(t+1) = N_(t) + [(D+E)-(B+I)](D) N_(t+1) = N_(t) + [(B+I)-(D-E)]

Answer»

`N_(t+1) = N_(t) + [(B-I)-(D+E)]`
`N_(t+1) = N_(t) + [(B+I)-(D+E)]`
`N_(t+1) = N_(t) + [(D+E)-(B+I)]`
`N_(t+1) = N_(t) + [(B+I)-(D-E)]`

ANSWER :B
6139.

Question : The formula of growth rate for population in given time is :

Answer»

`DT//DN = RN`
`dt//rN = dN`
`rN//dN = dt`
`dN//dt = rN`

ANSWER :D
6140.

Question : The formula of growth rate for population in a given time is

Answer»

dt/dN= rN
dt/rN = dN
rn/dN=dt
dN/dt=rN

Answer :D
6141.

Question : The formula for exponential population growth

Answer»

`dN//RN = rN`
`DT//dN = rN`
`dN//rnN= r//N`
`rN//dN = dt`

ANSWER :A
6142.

Question : The formula for exponential population growth is :

Answer»

RN / DN = DT
dN / dt = rN
dt.dN = rN
dN / rN = dt

Answer :B
6143.

Question : The formula, dN/dt = g represents

Answer»

environmental RESISTANCE
vital index
carrying CAPACITY
EXPONENTIAL growth

Answer :D
6144.

Question : The formation of primary endosperm nucleus is called as triple fusion . Why is it called so ?

Answer»

Solution :The primary endosperm nucleus is formed by the fusion of secondary nucleus and a MALE gamete and the secondary NUCLEOUS is formed by the fusion of two haploid POLAR nuclei . As three fusions are taking place in the formation of primary endosperm nucleus it is CALLED TRIPLE fusion .
6145.

Question : The formation of pericarp takes place from :

Answer»

OVARY wall
Outer INTEGUMENT
Inner integument
Placenta

Answer :A
6146.

Question : The formation of living and non-living world simultaneously is the main feature of

Answer»

Theory of PANSPERMIA
Theory of cosmozoic ORIGIN
Theory of catastrophism
Theory of ABIOGENESIS

Answer :B
6147.

Question : The formation of microspores from pollen mother cells through meiosis is known as?

Answer»

SOLUTION :Microsorogenesis
6148.

Question : The formation of erythrocyte in foetus takes place in

Answer»

LIVER and spleen
Red BONE marrow
Blood plasma
Sarcoplasm

Answer :A
6149.

Question : The formation of chromatid takes place in,

Answer»

`G_1` PHASE
`G_2` phase
S phase
M phase

Answer :C
6150.

Question : The form of carbon used for the carboxylation of phosphoenolpyruvate in C_4 plants is

Answer»

`CH_4`
`HCO_3^(-)`
`H_2 CO_3`
`C_2H_4`

ANSWER :B