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This section includes 5751 Mcqs, each offering curated multiple-choice questions to sharpen your Biology knowledge and support exam preparation. Choose a topic below to get started.
451. |
The following ratio is generally constant for a given species (AIPMT - 2004) |
A. | A + G / C + T |
B. | T + C / G + A |
C. | G + C / A + T |
D. | A + C / T + G |
Answer» D. A + C / T + G | |
452. |
What dose A & B represent ? |
A. | Grycase , Helicase |
B. | Double Stranded Protein, Helicase |
C. | Helicase, Single strand binding protein |
D. | Topoisomerase Helicase |
Answer» D. Topoisomerase Helicase | |
453. |
Which one of the following triplet codes, is correctly matched with its specificity for an aminoacid in protein synthesis or as ‘start’ or ‘stop’ codon ? |
A. | UCG - start |
B. | UUU - stop |
C. | UGU - Leucine |
D. | UAC - Tyrosine |
Answer» E. | |
454. |
State the use of moleculer medicine ? |
A. | Improves diagnosis of diseases |
B. | Used as gene theraphy |
C. | Used to understand several diseass like Alzheimer’s Parkinsons diseases etc., |
D. | All the above |
Answer» E. | |
455. |
In Which of the following DNA not directly involved ? |
A. | Repication |
B. | Transcription |
C. | Translation |
D. | Transformation |
Answer» D. Transformation | |
456. |
DNA replication in lagging strand of most of the eukaryotic organis ms is |
A. | conservative and continuous |
B. | semi conservative but discontinuous |
C. | conservative and semi - discontinuous |
D. | semi conservative but continuous |
Answer» C. conservative and semi - discontinuous | |
457. |
In the genetic code dictionary how many codons are used to code for all the 20 essential amino-acids?(AIPMT - 2003) |
A. | 20 |
B. | 64 |
C. | 61 |
D. | 60 |
Answer» C. 61 | |
458. |
Transcription begins when one of the following enzymes binds to promotor site. |
A. | DNA polymerase |
B. | RNA polymerase |
C. | helicase |
D. | Gyrase |
Answer» C. helicase | |
459. |
State the process and mention the labelled protion. |
A. | Process of Translation - X-RNA Polymerase- Y-DNA Template- Z-m RNA Transcript |
B. | process of Transcription - X-RNA Polymerase- Y-DNA Template- Z- RNA Transcript |
C. | process of Translocation - X-DNA polymerase- Y- Template- Z- Transcript |
D. | Process of Transformation - X - DNA polymerase- Y - RNA template- Z - RNA transcript |
Answer» C. process of Translocation - X-DNA polymerase- Y- Template- Z- Transcript | |
460. |
What does “Lac” refer to in what we call the lac operon ? (AIPMT - 2003) |
A. | Lactose |
B. | Lactase |
C. | Lac insect |
D. | The number 1,00,000 |
Answer» B. Lactase | |
461. |
E coli cells with a mutant z gene of the lac operon cannot grow in medium containing only lactose asthe source of energy because (AIPMT - 2005) |
A. | the lac operon is constitutively active in these cells |
B. | they cannot synthesize functional beta galactosidase |
C. | in the presence of glucose E Coli cell do not utilize lactose |
D. | they cannot transport lactose from the medium into the cell |
Answer» C. in the presence of glucose E Coli cell do not utilize lactose | |
462. |
Which one of the following makes use of RNA template to synthesize DNA (AIPMT - 2005) |
A. | DNA polymerase |
B. | RNA polymerase |
C. | Reverse transcriptase |
D. | DNA dependant RNA polymerase |
Answer» D. DNA dependant RNA polymerase | |
463. |
The two strands of a DNA molecule are separted and one of them is analysed for its A + T / G + C ratio, This is found to be 0.2 What is the A + T / G + C ratio of the other strand |
A. | 0.02 |
B. | 0.08 |
C. | 0.8 |
D. | 0.2 |
Answer» E. | |
464. |
Match the following using salient features of Human genome project A B |
A. | (P - iv) (Q - iii) ( R - ii) ( S - i) |
B. | ( P - iv) (Q - ii) (R - iii) (S - i) |
C. | (P - iv) (Q - i) (R - ii) (S - iii) |
D. | (P - i) (Q - iii) (R - iv) (S - ii) |
Answer» B. ( P - iv) (Q - ii) (R - iii) (S - i) | |
465. |
What would happen if in a gene encoding polypeptide of 50 aminoacids 25th codon (UAU) is mutated to UAA ? (AIPMT - 2003) |
A. | A Polypeptide of 24 aminoacids will be formed |
B. | Two polypeptides of 24 and 25 aminoacids will be formed |
C. | A polypeptide of 49 aminoacids will be formed |
D. | A polypeptide of 25 aminoacids will be formed |
Answer» B. Two polypeptides of 24 and 25 aminoacids will be formed | |
466. |
During transcription if the nucleotide sequence of the DNA strand that is being coded is ATACGthen the nucleotide sequence in the m RNA would be (AIPMT - 2004) |
A. | TATGC |
B. | T C T G G |
C. | U A U G C |
D. | U A T G C |
Answer» D. U A T G C | |
467. |
What do P, Q, R and S regions of t RNA ? |
A. | P - Anticodon loop |
B. | P. D LoopQ - Variable loop Q - T ψ c loopR - T ψ c loop R - Variable loopS - D Loop S - Anticodon loop |
C. | P - T ψ c loop |
D. | P - Anticodon LoopQ - D loop Q - T ψ c loop R - Anticodon loop R - D loop S - Variable loop S - Variable loop |
Answer» B. P. D LoopQ - Variable loop Q - T ψ c loopR - T ψ c loop R - Variable loopS - D Loop S - Anticodon loop | |
468. |
Among the following which is used for separation of DNA fragments ? |
A. | centifugation |
B. | Cell fractionation |
C. | Cell homogenation |
D. | Electrophoresis |
Answer» E. | |
469. |
What does ‘X’ represent |
A. | gene |
B. | segment of DNA |
C. | seqment of DNA coding for specific protein |
D. | Both A & C |
Answer» E. | |
470. |
Aminoacid Sequence in protein synthesis is decided by the sequence of (AIPMT - 2006) |
A. | r RNA |
B. | t- RNA |
C. | m RNA |
D. | c DNA |
Answer» D. c DNA | |
471. |
Write the codon for the anticodon on the t - RNA |
A. | AGU |
B. | UGU |
C. | UGA |
D. | ACU |
Answer» B. UGU | |
472. |
How many bases consist in an average gene ? |
A. | 3, 00, 000 |
B. | 3000 |
C. | 4, 00, 000 |
D. | 4000 |
Answer» C. 4, 00, 000 | |
473. |
Excretory matter of Cockroach is mainly .......... . (UP, PMT 2009) |
A. | Uric acid |
B. | Urea |
C. | Ammonia |
D. | Amino acid |
Answer» B. Urea | |
474. |
Which state is represent by the above model |
A. | Repressed state of lac operon |
B. | Inactive state of Lac operon |
C. | Active state of Lac operon |
D. | Induced state of Lac operon |
Answer» B. Inactive state of Lac operon | |
475. |
Select the correct option from following columns is case of Cockroach ? |
A. | p, q, r, s, = i, iii, ii,iv |
B. | p, q, r, s = ii, iv, i, iii |
C. | p, q, r, s = i, iv, iii, ii |
D. | p, q, r, s = i, ii, iii, iv |
Answer» C. p, q, r, s = i, iv, iii, ii | |
476. |
In Cockroach how many stuctures named after malpighian are present ? |
A. | One |
B. | Two |
C. | Three |
D. | Four |
Answer» C. Three | |
477. |
Which animal secrete uric acid ? (AIPMT 2009) |
A. | Frog |
B. | Man |
C. | Earthworm |
D. | Cockroach |
Answer» E. | |
478. |
Which statement is true for Cockroach ? (NCERT) |
A. | Ten ovirioles in ovary |
B. | Nymph is catterpillar form |
C. | Anal cerci are absent in female animal |
D. | It is ureatelic animal |
Answer» D. It is ureatelic animal | |
479. |
Which is the correct option is case of Cockroch in following column A and B ? |
A. | P, Q, R, S = i, ii, iii, iv |
B. | P, Q, R, S = iii, iv, ii, i |
C. | P, Q, R, S = iii, iv, i, ii |
D. | P, Q, R, S = iv, iii, i, ii |
Answer» D. P, Q, R, S = iv, iii, i, ii | |
480. |
The vexillm, (stan dard) wings and keel in pea flowers constitute: |
A. | Calyx |
B. | Corolla |
C. | Androecium |
D. | Gynaecium |
Answer» C. Androecium | |
481. |
A catkin of unisexual flower is found in: |
A. | Mulberry |
B. | Wheat |
C. | Onion |
D. | Grass |
Answer» B. Wheat | |
482. |
Inflorescence is : |
A. | Number of flower present on an axis |
B. | Arrangement of flowers on an axis |
C. | Method of the opening of flower |
D. | Type of flower borne on peduncle |
Answer» C. Method of the opening of flower | |
483. |
Stem modified into flattened photosynthetic structure is: |
A. | Phyllode |
B. | Bulbil |
C. | Phylloclade |
D. | Tendril |
Answer» D. Tendril | |
484. |
Nodulated roots occur in: (R.P.M.T 1995) |
A. | Leguminoceae |
B. | Solanaceae |
C. | Malvaceae |
D. | Papilionaceae |
Answer» B. Solanaceae | |
485. |
Insectivorous plants catch insects for obtaining: |
A. | Na - K |
B. | Taste |
C. | Phosphorus |
D. | Nitrogen |
Answer» E. | |
486. |
Opposite decussate phyllotaxy is found in: |
A. | Calotropi |
B. | Mango |
C. | Hibiscus |
D. | Nerium |
Answer» B. Mango | |
487. |
Zig-zag development of inflorescence axis is an example of: |
A. | Helicoid cyme |
B. | Scorpioid |
C. | Umbel |
D. | Compound umbel |
Answer» B. Scorpioid | |
488. |
Flowers are always present in : |
A. | Cryptogamou |
B. | Pteridophytes |
C. | Angiosperms |
D. | Bryophytes |
Answer» D. Bryophytes | |
489. |
From the life cycle point of view the most important part of a plants is: |
A. | Flower |
B. | Leaf |
C. | Stem |
D. | Root |
Answer» B. Leaf | |
490. |
Flower is a : |
A. | Modified cone |
B. | Modified spike |
C. | Modified branch system |
D. | Modified reproductive shoot |
Answer» E. | |
491. |
Vegetative reproduction of Agave occurs through: |
A. | Rhizome |
B. | Stolon |
C. | Bulbils |
D. | Sucker |
Answer» D. Sucker | |
492. |
Petiole is modified into tendril in |
A. | Passiflora |
B. | Gloriosa |
C. | Pisum |
D. | clematis |
Answer» E. | |
493. |
The arrangement of leaves on stem is called: |
A. | Venation |
B. | Vernation |
C. | Phyllotaxy |
D. | Axis |
Answer» D. Axis | |
494. |
The “Eyes” of the potato tuber are : (A.P.M.T.2011) |
A. | Root bud |
B. | Flower buds |
C. | Shoot bud |
D. | Axillary buds |
Answer» E. | |
495. |
floral formula represents : |
A. | number and arrangement of floral parts |
B. | Number of flowers in an inflorescence |
C. | Type of flowers in a family |
D. | None of above |
Answer» B. Number of flowers in an inflorescence | |
496. |
If a raceme inflorescence is branched , it is call? |
A. | Umbel |
B. | spike |
C. | Cymose |
D. | Panicle |
Answer» E. | |
497. |
Thorn is a stem structure because it: |
A. | Develops from trunk |
B. | Develops from apical bud |
C. | modification of bank floralbud |
D. | is pointed |
Answer» C. modification of bank floralbud | |
498. |
Bladder of Utricularia and Pitchers of nepenthes are modifications of: (JKCMEE 2004) |
A. | leave |
B. | stems |
C. | root |
D. | flowers |
Answer» B. stems | |
499. |
In monocot male gametophyte is: (C.B.S.E.1990) |
A. | Megaspore |
B. | Nucleus |
C. | Microspore |
D. | Tetrad |
Answer» D. Tetrad | |
500. |
Stipular tendril modification is found in : (Pb. PMT2001) |
A. | Smilex |
B. | Pea |
C. | Guava |
D. | Mimosa pudica |
Answer» B. Pea | |