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MCQOPTIONS
This section includes 14620 Mcqs, each offering curated multiple-choice questions to sharpen your NEET knowledge and support exam preparation. Choose a topic below to get started.
| 1001. |
Which of the following correctly represents the type of fruits given below. |
| A. | A-Berry; B-Caryopsis; C-Drupe; D-Sorosis; E-Aggregate |
| B. | B-Berry; C-Caryopsis; D-Drupe; A-Sorosis; E-Aggregate |
| C. | B-Berry; C-Caryopsis; D-Drupe; E-Legume; A-Aggregate |
| D. | A-Berry; C-Caryopsis; D-Drupe; B-Sorosis; E-Composite |
| Answer» D. A-Berry; C-Caryopsis; D-Drupe; B-Sorosis; E-Composite | |
| 1002. |
Which of the following are schizocarpic fruits? |
| A. | Siliqua & Legume |
| B. | Capsule & Berry |
| C. | Lomentum & Capsule |
| D. | Carcerulus & Lomentum |
| Answer» E. | |
| 1003. |
Catkin is a type of: |
| A. | Flower |
| B. | Inflorescence |
| C. | Stem |
| D. | All of these |
| Answer» C. Stem | |
| 1004. |
A true fruit develops from: |
| A. | Ovary |
| B. | Thalamus |
| C. | Petals |
| D. | Receptacle |
| Answer» B. Thalamus | |
| 1005. |
Match column-I with column-II and select the correct answer using the codes given below. Column-I Column-II A. Entire leaf modified I. Clematis into a spine. B. Leaf except stipules modified into a tendril. II. Citrus C. Stipules modified into a tendril. III. Euphorbia D. First leaf of axillary bud modified into a spine. IV. Lathyrus |
| A. | A-III B-IV C-I D-II |
| B. | A-III B-I C-IV D-II |
| C. | A-II B-III C-II D-IV |
| D. | A-IV B-II C-I D-III |
| Answer» B. A-III B-I C-IV D-II | |
| 1006. |
Which of the following statements are correct? (i) If the stem is jointed with solid nodes and hollow intemodes, it is called caudex. (ii) In Tridax, the stem is decumbent. (iii) Conn is a condensed form of rhizome growing more or less in vertical direction. (iv) Sucker is an underground modificaton of stem. (v) Biparous type ofcymose branching is seen in Saraca. |
| A. | (i), (iv) and (v) |
| B. | (ii) and (iii) |
| C. | (ii), (iii) and (v) |
| D. | (iii) and (iv) |
| Answer» C. (ii), (iii) and (v) | |
| 1007. |
Which is correct pair for edible part? |
| A. | Tomato - Thalamus |
| B. | Maize - Cotyledons |
| C. | Guava - Mesocarp |
| D. | Date palm - Pericarp |
| Answer» E. | |
| 1008. |
Name the family having (9) + 1 arrangement of stamens? |
| A. | Solanaceae |
| B. | Asteraceae |
| C. | Liliaceae |
| D. | Fabaceae |
| Answer» E. | |
| 1009. |
The anthers in solanaceae are: |
| A. | Monothecous, introse |
| B. | Dithecous, extrose |
| C. | Dithecous, introse |
| D. | Monothecous, extrose |
| Answer» D. Monothecous, extrose | |
| 1010. |
Which one of the following is a fibre yielding plant? |
| A. | Crotalaria juncea |
| B. | Cicer arietinum |
| C. | Triticum vulgare |
| D. | Impatiens balsamina |
| Answer» B. Cicer arietinum | |
| 1011. |
In some viruses, RNA is present instead of DNA indicating that: |
| A. | Their nucleic acid must combine with host DNA before replication. |
| B. | They cannot replicate. |
| C. | There is no hereditary information. |
| D. | RNA can act to transfer heredity. |
| Answer» E. | |
| 1012. |
One gene one enzyme hypothesis was proposed by: |
| A. | Khorana and Nirenberg |
| B. | Beadle and Tatum |
| C. | Bateson and Punnet |
| D. | Bridges |
| Answer» C. Bateson and Punnet | |
| 1013. |
Which step of translation does not consume a high energy phosphate bond? |
| A. | Translocation |
| B. | Amino acid activation |
| C. | Peptidyl-transferase reaction |
| D. | Aminoacyl tRNA binding to active ribosomal site |
| Answer» E. | |
| 1014. |
Which one of the following triplet codes, is correctly matched with its specificity for an amino acid in protein synthesis or as 'start' or 'stop' codon? |
| A. | UAC-Tyrosine |
| B. | UCG-Start |
| C. | UUU-Stop |
| D. | UGU-Leucine |
| Answer» B. UCG-Start | |
| 1015. |
E.coli cells with a mutated z gene of the lac operon cannot grow in medium containing only lactose as the source of energy because: |
| A. | The lac operon is constitutively active in these cells. |
| B. | They cannot synthesise functional beta- galactosidase. |
| C. | In the presence of glucose, E.coli cells do not utilize lactose. |
| D. | They cannot transport lactose from the medium into the cell. |
| Answer» C. In the presence of glucose, E.coli cells do not utilize lactose. | |
| 1016. |
In lac operon, structural gene 'z' synthesises: |
| A. | \[\beta \]-galactosidase |
| B. | Galactosidase permease |
| C. | Galactosidase transacetylase |
| D. | None of the above |
| Answer» B. Galactosidase permease | |
| 1017. |
Co-repressor binds with: |
| A. | Promoter gene |
| B. | Operator gene |
| C. | Aporepressor |
| D. | Regulator gene |
| Answer» D. Regulator gene | |
| 1018. |
DNA synthesis can be measured by estimating incorporation of radio-labeled: |
| A. | Uracil |
| B. | Ribose sugar |
| C. | Thymidine |
| D. | Adenine |
| Answer» D. Adenine | |
| 1019. |
Nucleotide arrangement in DNA can be seen by: |
| A. | X-ray crystallography |
| B. | Electron microscope |
| C. | Ultracentrifuge |
| D. | Light microscope |
| Answer» B. Electron microscope | |
| 1020. |
tRNA recognises ribosome by: |
| A. | \[T\psi C\,\,loop\] |
| B. | DHU loop |
| C. | Anticodon |
| D. | AA-site |
| Answer» B. DHU loop | |
| 1021. |
DNA is a genetic material: (i) Due to its stability. (ii) Its ability to replicate. (iii) Its ability to mutate. (iv) Its ability to express itself. Choose the correct combination. |
| A. | (i) and (ii) |
| B. | (i), (iii) and (iv) |
| C. | (iii) and (iv) |
| D. | (i), (ii), (iii) and (iv) |
| Answer» E. | |
| 1022. |
Eukaryotic chromosomes: |
| A. | Are circular and contain origin and terminator sequences. |
| B. | Are linear and have origins and telomeres. |
| C. | Contain coding and non-coding sequences. |
| D. | Both [b] and [c] |
| Answer» E. | |
| 1023. |
Choose the sequence in which the following enzymes take part in DNA replication. (i) Helicase (ii) Primase (iii) SSB (iv) DNA polymerase (v) DNA ligase |
| A. | (i) \[\to \] (v) \[\to \] (iv) \[\to \] (iii) \[\to \] (ii) |
| B. | (i) \[\to \] (ii) \[\to \] (iii) \[\to \] (iv) \[\to \] (v) |
| C. | (i) \[\to \] (iii) \[\to \] (ii) \[\to \] (iv) \[\to \] (v) |
| D. | (i) \[\to \] (iv) \[\to \] (iii) \[\to \] (ii) \[\to \] (v) |
| Answer» D. (i) \[\to \] (iv) \[\to \] (iii) \[\to \] (ii) \[\to \] (v) | |
| 1024. |
Transcription: |
| A. | Starts at initiator region and ends at stop region. |
| B. | Starts at operator region and ends at telomeric end. |
| C. | Starts at promoter region and ends terminator region. |
| D. | Starts at CAAT box and ends at TATA box. |
| Answer» D. Starts at CAAT box and ends at TATA box. | |
| 1025. |
Introns are DNA sequences that: |
| A. | Code for functional domains in proteins. |
| B. | Are removed from pre-mRNA by spliceosomes. |
| C. | Allow one gene to make different gene products depending on which introns are removed during splicing |
| D. | Both [b] and [c] |
| Answer» E. | |
| 1026. |
Chromatin structure must be altered in order for gene expression to occur because: |
| A. | Condensed chromatin is replicated but not transcribed. |
| B. | Condensed chromatin makes most DNA sequence inaccessible to the transcription complex. |
| C. | Decondensed chromatin has more nucleosomes per DNA molecule. |
| D. | Heterochromatin is actively transcribed and euchromatin is not transcribed. |
| Answer» C. Decondensed chromatin has more nucleosomes per DNA molecule. | |
| 1027. |
DNA replication is semi-conservative as: |
| A. | Only non-parent strand acts as template. |
| B. | Both strands of new molecule are synthesized de novo. |
| C. | One of the strand in each new molecule is parental and the other is new. |
| D. | Daughter strands are dispersive. |
| Answer» D. Daughter strands are dispersive. | |
| 1028. |
A short sequence of bases on one strand of DNA is AGTCTACCGATAGT. If this sequence serves as a template for the formation of a new strand of DNA, what will be the corresponding base sequence in the new strand? |
| A. | AGTCTACCGATAGT |
| B. | TCAGATGGCTATCA |
| C. | TGATAGCCATCTGA |
| D. | GACATCGATTCGAT |
| Answer» C. TGATAGCCATCTGA | |
| 1029. |
Identify the correct labelling of A, B, C, D & E. |
| A. | A-Hydrogen bonds, -B-Pyrimidine, C-Hexose (deoxyribose) sugar, D-5' end, E-Purine base |
| B. | A-Hydrogen bonds, B-Purine base, C-Hexose (deoxyribose) sugar,-D-5' end, E-Pyrimidine |
| C. | A-Hydrogen bonds, B-Pyrimidine, C-Pentose (deoxyribose) sugar, D-5' end, E-Purine base |
| D. | A-Hydrogen bonds, B-Purine base, C-Pentose (deoxyribose) sugar, D- 5' end, E- Pyrimidine |
| Answer» E. | |
| 1030. |
Which of the following mechanisms of gene regulation operates after mRNA transcription but before translation of mRNA into protein? |
| A. | mRNA splicing |
| B. | DNA packing |
| C. | Repressors and activators |
| D. | Protein degradation |
| Answer» B. DNA packing | |
| 1031. |
Histones are rich in: |
| A. | Alanine and glycine |
| B. | Lysine and arginine |
| C. | Histidine and serine |
| D. | Cysteine and tyrosine |
| Answer» C. Histidine and serine | |
| 1032. |
Hershey and Chase used radioactive \[^{35}S\] and \[^{32}P\]in experiments to provide evidence that DNA was the genetic material. These experiments pointed to DNA because: |
| A. | Progeny viruses retained \[^{32}P\] but not\[^{35}S\]. |
| B. | Retention of \[^{32}P\] in progeny viruses indicated that DNA was passed on. |
| C. | Loss of \[^{35}S\] in progeny viruses indicated that proteins were not passed. |
| D. | All of the above |
| Answer» E. | |
| 1033. |
Isolation and purification of specific DNA segment from a living organism was achieved by: |
| A. | Nirenberg |
| B. | Lederberg |
| C. | Zacharis |
| D. | Watson & Crick |
| Answer» B. Lederberg | |
| 1034. |
The regulation of tryptophan synthesis in E. coli is an example of affecting gene expression through: |
| A. | Translational control |
| B. | Transcriptional control |
| C. | Homeotic gene control |
| D. | Breaking down mRNA molecules |
| Answer» C. Homeotic gene control | |
| 1035. |
During transcription, the DNA site at which RNA polymerase binds is called: |
| A. | Enhancer |
| B. | Promoter |
| C. | Regulator |
| D. | Receptor |
| Answer» C. Regulator | |
| 1036. |
What effect would you expect if gene expression of the lac operon was completely repressed? |
| A. | The cell would be more efficient without wasting' the energy required for the low level of Lac Z, Lac Y, and Lac A gene expression. |
| B. | Allolactose would accumulate within the cell and become toxic. |
| C. | Lactose would not be converted into the inducer and the operon could not be induced. |
| D. | All of the above |
| Answer» D. All of the above | |
| 1037. |
Which of the following factor is required for the protein synthesis? (i) Initiation codon (ii) GTP and ATP (iii) Peptidyl transferase (iv) tRNA (v) mRNA (vi) Amino acid activating enzyme (vii) rRNA Choose the correct combination. |
| A. | (i), (ii) and (iii) |
| B. | (iii), (iv) and (v) |
| C. | (v), (vi) and (vii) |
| D. | All of these |
| Answer» E. | |
| 1038. |
Enzyme required for removing RNA primer during DNA replication is: |
| A. | Primase |
| B. | Ligase |
| C. | DNA polymerase I |
| D. | DNA polymerase III |
| Answer» D. DNA polymerase III | |
| 1039. |
Experiments by Avery, McLeod, and McCarty supported DNA as the genetic material by showing that: |
| A. | Both protein and DNA samples provided the transforming factor. |
| B. | DNA was not complex enough to be the genetic material. |
| C. | Only samples with DNA provided activity. |
| D. | Even though DNA was molecularly simple, it provided adequate variation to act as the genetic material. |
| Answer» D. Even though DNA was molecularly simple, it provided adequate variation to act as the genetic material. | |
| 1040. |
Which of the following statements about the process of DNA replication is incorrect? |
| A. | Many different enzymes are needed for the process to function properly. |
| B. | Mistakes can be corrected at multiple steps in the process. |
| C. | Uncorrected mistakes introduce mutations into the DNA base sequence. |
| D. | Mistakes in the copying process are very common occurrences. |
| Answer» E. | |
| 1041. |
Match column-I with column-II and select the correct answer using the codes given below. Column-I Column-II A. Helicase I. Joining of nucleotides B. Gyrase II. Opening of DNA C. Primase III. Unwinding of DNA D. DNA polymerase IV. RNA priming |
| A. | A-II; B-I; C-III; D-IV |
| B. | A-II; B-I; C-IV; D-III |
| C. | A-IV; B-III; C-I; D-II |
| D. | A-II; B-III; C-IV; D-I |
| Answer» E. | |
| 1042. |
Which statement is/are correct? (i) Nucleosome contains basic protein. (ii) Tightly packed DNA of chromatin is called heterochromatin. (iii) Loosely packed DNA of chromatin is called euchromatin. (iv) \[{{H}_{2}}A,{{H}_{2}}B\] are linker proteins. (v) DNA is basic in nature. |
| A. | (i),(ii) and (iii), |
| B. | (i) and (ii) |
| C. | (i), (ii), (iii), (iv) and (v) |
| D. | (iv) and (v) |
| Answer» B. (i) and (ii) | |
| 1043. |
What would happen if a mutation occurred in the DNA such that the second codon of a polypeptide, UGC, was changed to a UAG? |
| A. | Nothing. The ribosome would skip that codon and translation would continue. |
| B. | Translation would continue, but the reading frame of the ribosome would be shifted. |
| C. | Translation would stop at the second codon and no functional protein would be made. |
| D. | Translation would continue, but the second amino acid in the protein would be different. |
| Answer» D. Translation would continue, but the second amino acid in the protein would be different. | |
| 1044. |
Mutations which alter nucleotide sequence within a gene are called: |
| A. | Frame shift mutations |
| B. | Base pair substitutions |
| C. | Both [a] and [b] |
| D. | None of the above |
| Answer» D. None of the above | |
| 1045. |
Post-transcriptional regulation includes: |
| A. | Binding of repressor on silencer regions. |
| B. | Transport of messenger RNA into the cytoplasm. |
| C. | Decreasing messenger RNA stability in the cytoplasm. |
| D. | Both [b] and [c] |
| Answer» E. | |
| 1046. |
Transcriptional regulation: |
| A. | Is highly efficient at completely preventing transcription. |
| B. | Allows the cell to only produce proteins that are needed at the time. |
| C. | Can be induced by a repressor protein. |
| D. | All of the above |
| Answer» C. Can be induced by a repressor protein. | |
| 1047. |
If the gene encoding the trp repressor is mutated such that it can no longer bind to tryptophan, will transcription of the trp operon occur? |
| A. | Yes, because the trp repressor can only bin the trp operon and block transcriptional initiation when it is bound to tryptophan. |
| B. | No, because this mutation does not affect the part of the repressor that can bind the operator. |
| C. | No, because the trp operon is repressed only when tryptophan levels are high. |
| D. | Yes, because the trp operon can allosterically regulate the enzymes needed to synthesise the amino acid tryptophan. |
| Answer» B. No, because this mutation does not affect the part of the repressor that can bind the operator. | |
| 1048. |
Genes that are involved in turning on or off the transcription of a set of structural genes are called: |
| A. | Operator genes |
| B. | Redundant genes |
| C. | Regulator genes |
| D. | Polymorphic genes |
| Answer» B. Redundant genes | |
| 1049. |
Operon is a: |
| A. | Sequence of three nitrogen bases determining a single amino acid. |
| B. | Set of closely placed genes regulating a metabolic pathway in prokaryotes. |
| C. | Segment of DNA specifying a polypeptide. |
| D. | Gene responsible for switching on and switching off other genes. |
| Answer» C. Segment of DNA specifying a polypeptide. | |
| 1050. |
Match the enzymes given in column-I with its function given in column-II and select the correct option. Column-I Column-II A. \[\beta \]- galactosidase I. Joining of DNA fragments B. Permease II. Peptide bond formation C. Ligase III. Hydrolysis of lactose D Ribozyme IV. Increase permeability of galactosidase |
| A. | A-II; B-I; C-IV; D-III |
| B. | A-III; B-IV; C-I; D-II |
| C. | A-II; B-IV; C-I; D-III |
| D. | A-I; B-II; C-IV; D-III |
| Answer» C. A-II; B-IV; C-I; D-III | |