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3601.

Question : Two hormones………..(a)…………and (b)………..synthesize in hypothalamus and transported in pituitary gland through ……….(C)……….. And ………..(d)……….respectively.

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a=oxytocin `to` C=portal CIRCULATION
B=ADH`IMPLIES` d=direct release
a=ADH`implies` C=AXONAL TRANSPORT
b=TSHRF `implies~d=portal circulation
a=ACTH `implies` =axonal transport
b=MSH`implies`c=portal circulation
a=TSHRF `implies` c=axonal transport
b=ADH `implies` d=portal circulation

Answer :B
3602.

Question : Two heterozygous parents are crossed. If the two loci are linked, what would be the distribution of phenotypic features in F_1generation for a dihybrid cross?

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Solution : If the two loci are linked, there will be two possibilities:
(a) If the genes are completely linked: In the CASE of COMPLETE linkage, all the offspring will show parental phenotypes. (B) If the genes are incompletely linked: In incomplete linkage, there is occurrence of crossing over and new combinations are FORMED but the parental combinations are more NUMEROUS than the new or nonparental combinations.
3603.

Question : Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in F_1, generation for a dihybrid cross?

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Solution :If the TWO loci in parents are completely linked, there would be no segregation.
=The `F_1` generation will EXHIBIT parental characters only.
= But, if the two loci are incompletely linked then segregation would occur PARTLY and the `F_1` generation will exhibit both parental and recombinant characteristics, but the recombinant will be in a very small proportion.

= Only two types of GAMETES will be produced from each parent as the two loci are linked.
3604.

Question : Two groupsof isolated thylakoids are placed in an acidic bathing solutinso that H^(+)diffuses into the thylakoids they are then transferred to a basicbathing solution and one groups is in the dark select belwothe choice that descibes what youexpect each group of thylakoids to produce {:("In Light","In Dark"),("ATP only","Nothing"),("ATP" O_(2),"ATP only"),("ATP" O_(2) "glucose","ATP" O_(2)),("ATP" O_(2),O_(2)):}

Answer»


ANSWER :B
3605.

Question : Two heterozygous parents are crossed. If the three loci are linked completely what would be the distribution of phenotypic features in resulting generation for a trihybrid cross ?

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`1:2:1`
`27:9:9:3:9:3:3:1`
`9:3:3:`
`3:1`

ANSWER :D
3606.

Question : Two halves of pectoral girdle fuse in

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Lateral side
Mid-DORSAL SIDES
Mid-VENTRAL LINE
Both dorsal and ventral lines.

Answer :C
3607.

Question : Two genes .A. and .B. are linked. In a dihybrid cross involving these two genes, the F_1 heterozygote is crossed with homozygous recessive parental type (aa bb). What would be the ratio of offspring in the next generation ?

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1:1:1:1
9:3: 3:1
3:1
`1:1`

Solution :Two genes .A. and .B. are linked. In a dihybrid CROSS involving these two genes, the F heterozygousis crossed with homozygous RECESSIVE parental type (AA bb). The RATIO of offspring in the next generation will be 1: 1. This type of cross is called as Test cross.
3608.

Question : Two genes R and Y are located very close on the chromosomal linkage map of maize plant. When RRY and rryy genotypes are hybridized, then F_(2) segregation will show

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HIGHER NUMBER of there combinant typcs
segregation in the EXPECTED 9:3:3:1 ratio
segregation in 3:1 ratio
higher number of TE paretal types

Answer :D
3609.

Question : Two genes A and B are linked in a dihybrid cross involving these two genes, the F_(1) heterozygote is crossed with homozygous recessive parental type (aa bb). What would be the ratio of offspring in the next generation?

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`1:1:1:1`
`9:3:3:1`
`3:1`
`1:1`

ANSWER :D
3610.

Question : Two genes a and b are linked and show 30% recombination. If ++ / ++ individual married with ab / ab then the types and proportion of gametes in F1 will be -

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`++ 35% : AB : 35% : + a 15% : +B 15%`
`++ 15% : ab : 15% : + a 35% : +b 35%`
`++ 35% : ab : 15% : + a 35% : +b 15%`
`++ 15% : ab : 35% : + a 15% : +b 35%`

ANSWER :A
3611.

Question : Two friends are eating together on a table. One of them suddenly starts coughing while swallowing some food. This coughing would have been due to improper movement of

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Neck
Tongue
EPIGLOTTIS
DIAPHRAGM

ANSWER :C
3612.

Question : Two gametes are similar in appearance they are called(A) isogametes(B) homogasmetes(C) isogametes or homogametes(D) none of the above

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ISOGAMETES
homogasmetes
isogametes or HOMOGAMETES
NONE of the above

ANSWER :C
3613.

Question : Two-dimensional polyacrylamide gel electrophoresis was developed by

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Maizel
Raymon
Martin and synge
patrick O' farrel

Answer :D
3614.

Question : Two different species cannot live for long duration in the same niche or habitat. This law is:

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ALLEN's law
Gause's law
Weismann's theory
Competitive EXCLUSION principle

Answer :B
3615.

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

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ALLEN's law 
Bergmen's law 
COMPETITIVE EXCLUSION principle
Weisman's theory 

ANSWER :B
3616.

Question : Two different plants associated with discovery of two different phytohormones having commone biosynthetic precursor exhibit one of the following characters each I. Versatile anthers II. Compound spadix III. Pentalocular ovary IV. Trifoliate compound leaves

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II and III
I and III
II andIV
I and IV.

Solution :Two hormones with COMMON biosynthetic precursor are GIBBERELLIN and abscisic acid. Gibberellin was discovered in CONNECTION with bakane DISEASE of Rice (having versatile anther) while abscisic acid was discovered in connection with cotton (having 3-5 locular OVARY).
3617.

Question : Two crosses between the same pair of genotypes or phenotypes in which the sources of the gametes are reversed in one cross, is known as

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TEST CROSS
RECIPROCAL cross
DIHYBRID cross
reverse cross

Solution :reciprocal cross
3618.

Question : Two core techniques that enabled birth of modern biotechnology. Identify and explain these two core techniques.

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Solution :(i) Genetic engineering - Techniques to alter the CHEMISTRY of genetic material (RNA and DNA) to introduce these into host organisms and thus change the phenotype of the host organism.
Maintenance of sterile ambiance in chemical engineering processes to enable growth of only the DESIRED MICROBE in large quantities for the manufacture of iotechnological products LKE aritibiotics, vaccines, enzymes, etc.
3619.

Question : Two chambered heart in the development of human being indicate the fundamental

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CHORDATE CHARACTER
VERTEBRATE character
Pisces character
Amphibian character

ANSWER :C
3620.

Question : Two cells A & B are contiguous. Cell A has osmotic pressure 10 atm, turgor pressure 7 atm and diffusion pressure deficit 3 atm. Cell B has osmotic pressure 8 atm, turgor pressure deficit 3 atm and diffusion pressure deficit 5 atm. The result will be

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MOVEMENT of WATER from CELL A to B
Movement of water from cell B to A
No movement of water
Equilibrium between the two

Answer :A
3621.

Question : Two cells A and B are continuous . Cell A has osmotic pressure 10 atm, turgor pressure 7 atm and diffusion pressure deficit 3 atm. Cell B has osmotic pressure 8 atm, turgor pressure 3 atm and diffusion pressure deficit 5 atm. The result will be:

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no movement of WATER
equilibrium between the two
movement of water from cell A to B
movement of waterfrom cell B to A

Answer :C
3622.

Question : Two blood sampleH A and B picked up from the crime scene were handed over to the forensic depertment for gentic finger printing . Describehow tle technique of genetic finger printing is carried out. How will it be confirmed whether the samplebelonged to the same individual or two differnent individunls?

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Solution :DNA Fingerprinting : This technique was DISCOVERED by Ales Jeffreys in 1985.
The technique mvolved following steps
Isolation and EXTRACTION of DNA from the ceJJ by centrifugation .
By the help of enzyme RESTRICTION endonuclease, DNA molecules are digested. The FRAGMENT also contains VNTRs.
The small DNA fragments are separated through gel electrophoresis set-up tha contains agarose polymer gel.
The separated DNA fragments are transfered from electrophoresis plate to synthetic membranes like nitrocellulose or nylon membrane sheet called Southern blotting.
The DNA probes are added which target a specific n~c~eot~de sequence which is complementary to them and this process is called hybridisation.
The nylon membrane is EXPOSED to an X-ray film and dark orange coloured bands developed at sites where probes have bound to the DNA fragments. This is known as autoradiography. On comparing the DNA print of blood samples A and B, it can be confirmed that the blood sample picked up from the crime scene belongs to the same individual or to different individuals.
3623.

Question : Two blood samples A and B picked up from the crime scene were handed over to the forensic department for genetic finger printing. Describe how the technique of genetic finger printing is carried out.

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Solution :Technique For DNA Fingerprinting. The DNA is extracted from the nuclei of white blood cells or of spermatozoa or of the hair follicle cells that cling to the roots of hairs that have fallen, or been pulled out.(ii) The DNA molecules are first BROKEN with the help of enzyme restrictionendonuclease (called chemical knife) that cuts them into fragments.The fragments of DNA also contain the VNTRSThe fragments are separated according to SIZE by gel electrophoresis.
(iv) Fragments of a particular size having VNTRs are multiplied throughPCR technique. They are treated with alkaline chemicals to split them into single stranded DNAS
(v) The separated fragments of single stranded DNA are transfered onto a nylon membrane
(vi) Radioactive DNA probes having repeated base sequences complementary to POSSIBLE VNTRs are poured over the nylon membrane. Some of them will bind to the single stranded VNTRs. The method of hybridization of DNA with probes is called Southern Blotting, after the name of the inventor, E.M. Southern (1975). The nylon membrane is washed to remove EXTRA probes.
(vii) An X-ray film is exposed to the nylon membrane to mark the places where the radioactive DNA probes have bound to the DNA fragments These places are marked as dark bands when X-ray film is developed. This is known as autoradiography.
(viii) The dark bands on X-ray film represent the DNA fingerprints (= DNA PROFILES).
3624.

Question : Two black Guinea pigs are mated together on several occasions. Their offsprings are invariably black. However, when their black offsprings are, mated wit white Guinea pigs, half the matings results in litters containing black babies only and the other half equal numbers of black and white babies. If colour is determined by a single pair of alleles, the genotypes of the two parents are

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HOMOZYGOUS DOMINANT and heteroxygous
Homozygous dominant and homzygous RECESSIVE
both homozygous dominant
both HETEROZYGOUS

ANSWER :A
3625.

Question : Two bacteria most useful in genetic engineering are

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Rhizobium and AZOTOBACTER
Escherichia and AGROBACTERIUM
Rhizobium and Diplococcus
NITROSOMONAS and KLEBSIELLA

Solution :Escherichia and Agrobacterium
3626.

Question :Two allelic genes are located on the :

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Same chromosome
Two HOMOLOGOUS chromosomes
Two NON - homologous
Any two DIFFERENT chromosomes chromosomes

Answer :B
3627.

Question : Twisted bones of nasal chambers are

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Choanae
Conchae
RIMA glottidis
Mediastinum

Answer :B
3628.

Question : Twins.

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SOLUTION :(1)The twins are of two type,viz.monozygotic or identical twins and dizygotic twins or fraternal twins.(2)Monzygotic twins are derived froma single zygote that splits at an early stage during development.They have same genetic constitution and are always of the same sex.(3)Dizygotic twins are formed rarely.When two ova are released simutaneousely from the OVARIES and they are in turn FERTILIZED bt two deifferent sperms then such sizygotic twins are formed.(4)Each fertilized ovum develops into embryo.It could be of the same sex or of the DIFFERENT sex.(5)Dizygotic twins are genetically dissimilar.
3629.

Question : Twenty third pair of human chromosomes are known as

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AUTOSOMES
Hetersomes
CHROMATIDS
CHROMOSOMES

ANSWER :B
3630.

Question : TV-29 is a________variety of tea.

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SOLUTION :TRIPLOID
3631.

Question : TW Engelmann’s prism experiments on a green alga, Cladophora led to the discovery of-

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Cyclic and non - cyclic photophosphorylation
First action SPECTRUM of photosynthesis
Two PHOTOSYSTEMS operating simultaneously
Oxidative phosphorylation

Answer :B
3632.

Question : Turner's syndrome and Kinefelter's syndrome.

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SOLUTION :
3633.

Question : Turner's syndrone is characterised by

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ABSENCE of ovaries
Absence of AUTOSOMAL set
Absence of allosome
Presence of trisony

Answer :C
3634.

Question : Turner syndrome is due to

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LOSS of X chromosome `-44 + XO`
loss of any chromosome
it is due to TRISOMY in `21^st`pair
None

Answer :C
3635.

Question : Turgor operated valve in stomatal complex is

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Stoma
Guard cell
Subsidiary cell
All the above

Answer :A
3636.

Question : Turgor pressure is absent/not detected during

Answer»

THISTLE funnel
Animal cell
Flaccid PLANT cell
All the above

Answer :D
3637.

Question : Ture-over number of the fastest enzume is

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`18xx10^(4)`
`10^(4)`
`36xx10^(6)`
`10^(5)`.

SOLUTION :N//A
3638.

Question : tunica media will be absent in the Wall of

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arteries
vena cava
capillaries
veins

Answer :C
3639.

Question : Tunicated bulb is seen is …………… .

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Scilla
Solanum
Allium
Zingiber

Answer :A::B::C::D
3640.

Question : Tunica vaginalis is found in

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OVARIES of female
TESTIS of MALE
Vagina of female
NONE

Answer :B
3641.

Question : Tunica corpus theory was proposed by

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schmidt
C. nageli
Hanstein
clower

Answer :A
3642.

Question : Tunica and corpus are constituents of

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SHOOT apex
Root apex
Leaf apex
All of these

Answer :A
3643.

Question : Tumour cells are irradiated by lethal doses of ___________.

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SOLUTION :RADIATION
3644.

Question : Tuna fish and sword fish are :

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Regulators
Conformers
Migrators
More than ONE option

Answer :D
3645.

Question : Tunica albuginea is related to

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PENIS
OVARY
TESTES
all

ANSWER :D
3646.

Question : Tumour producing plasmid transforms

Answer»

ANIMALS
PLANTS
BACTERIA
FUNGI

SOLUTION :Plants
3647.

Question : Tubular tower fermenter

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SOLUTION :
3648.

Question : Tumor virus contain which as a genetic material ?

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DNA
RNA
PROTEIN
CARBOHYDRATE

SOLUTION :RNA
3649.

Question : Tubular reabsorption occurs in I. proximal convoluted tubule II.collecting ducts III. loop of Henle IV. Bowman's capsule Choose the correct answer.

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I and IV
II and III
I and II
All of these

Answer :C
3650.

Question : Tuberculosis is caused by

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MYCOBACTERIUM SP.
ASPERGILLUS sp.
Clostridium sp.
Vibrio sp.

Solution :Mycobacterium sp.