MCQOPTIONS
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MCQOPTIONS
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.
| 4101. |
Which is the process of biological control,which is for controlling to spread the diseare caused by mosquitoes.? |
| A. | Masquito net |
| B. | Spreading & chemicals |
| C. | Fish like Gambusia |
| D. | Net in doors & windows |
| Answer» D. Net in doors & windows | |
| 4102. |
Chhaganbhai is suffering from fever, then which kind of organism responsible for it ? |
| A. | Plasmodium vivex |
| B. | Plasmodium malarie. |
| C. | Plasmodium falciparum |
| D. | All of hte given. |
| Answer» D. All of hte given. | |
| 4103. |
Which is the most appropriate method for conservation of wild life ?(JharkhandCEE-2008) |
| A. | Vaccination |
| B. | Hybridization |
| C. | conservation in natural habitat |
| D. | Killing of predator |
| Answer» D. Killing of predator | |
| 4104. |
Give the correct sequence from pre erythrocytic cycle ? |
| A. | Saliva - Sporozoites-Blood-liver-Cryptoschizont-Cryptomerozoites. |
| B. | Sporozoites-Saliva-Blood-Liver-Cryptoschizont-Cryptomerozoites. |
| C. | Saliva-Cryptoschizont-Blood-Liver-Sporozoites-Cryptomerozoites. |
| D. | Saliva-Sporozaites-Liver-Blood-Cryptoschizont-Cryptomerozoites. |
| Answer» B. Sporozoites-Saliva-Blood-Liver-Cryptoschizont-Cryptomerozoites. | |
| 4105. |
Generally streptococcus pneumoniae causes pneumonia but which bacteria is responsible for this diseare ? |
| A. | Pneumococcu |
| B. | Tuberculosis. |
| C. | Haemophilus influenzae |
| D. | Salmonella typhi |
| Answer» D. Salmonella typhi | |
| 4106. |
Match proper pair : |
| A. | (P - i)(Q - iii)(R - ii)(S - iv)(T - v) |
| B. | (P - iii)(Q - iv)(R - v)(S - ii)(T - i) |
| C. | (P - iii)(Q - iv)(R - ii)(S - v)(T - i) |
| D. | (P - i)(Q - ii)(R - iii)(S - iv)(T - v) |
| Answer» D. (P - i)(Q - ii)(R - iii)(S - iv)(T - v) | |
| 4107. |
Which one is odd ? |
| A. | Cannabinoid |
| B. | Marjuana |
| C. | Ganja |
| D. | Morphin |
| Answer» E. | |
| 4108. |
Which method of asexual reproduction can be said as method of regeneration ? |
| A. | Binary fission |
| B. | Sporulation |
| C. | Budding |
| D. | Fragmentation |
| Answer» E. | |
| 4109. |
For which plants layering method of vegetative propagation is used ? |
| A. | Lemon, Grape |
| B. | Sugarcane, Rose |
| C. | Mango, Apple |
| D. | Guava, Litchi |
| Answer» B. Sugarcane, Rose | |
| 4110. |
Which of the following is the most widely classification method of amino acid: |
| A. | Whittaker method |
| B. | Linnaeus method |
| C. | Ernest chain method |
| D. | Lehninger method |
| Answer» E. | |
| 4111. |
Find out correct option of column I (aids) of family planning and method of family planning . |
| A. | (P - iii) (Q - ii) (R - iv)(S - i) |
| B. | (P - iii) (Q - i) (R - iv)(S - ii) |
| C. | (P - iii) (Q - ii) (R - iv)(S - i) |
| D. | (P - ii) (Q - iii)(R - iv)(S - i) |
| Answer» C. (P - iii) (Q - ii) (R - iv)(S - i) | |
| 4112. |
Generally protein and carbohydarte components are found in cow milk… [KCET 2005] |
| A. | Albumin, Lactose |
| B. | Globulin, Casin |
| C. | Casin, Lactose |
| D. | Casin, Fructose |
| Answer» D. Casin, Fructose | |
| 4113. |
Which method is used to detect the gender of the foetus ? |
| A. | ART |
| B. | IVF |
| C. | AFT |
| D. | GIFT |
| Answer» D. GIFT | |
| 4114. |
According to statements find the correct option : |
| A. | FTFT |
| B. | TTFT |
| C. | TTTT |
| D. | TTTF |
| Answer» B. TTFT | |
| 4115. |
Which method is used for vegetetive reproduction the devlopment of banana plant ? |
| A. | Cutting |
| B. | Layering |
| C. | Grafting |
| D. | Bud Grafting |
| Answer» B. Layering | |
| 4116. |
The correct order of the changes in hormones level at first day to 28th day of menstrual cycle. |
| A. | estrogen and progesterone level is lower --> Estrogen rising --> Progesterone rising. |
| B. | estrogen and progesterone level is higher --> Estrogen lower --> Progesterone rising. |
| C. | estrogen and progesterone level is lower --> Estrogen rising --> Progesterone lower. |
| D. | estrogen and progesterone level is higher --> Estrogen rising --> Progesterone rising. |
| Answer» B. estrogen and progesterone level is higher --> Estrogen lower --> Progesterone rising. | |
| 4117. |
One plant group kept for 12 hrs in day and 12 hrs in night and flowering is observed in it. Another plant group is kept similarly for day - night period and for very short time light is given during dark period it do not flower. What is such type of plant called ? (CBSE 2004) |
| A. | Long-day plant |
| B. | Day-neutral plants |
| C. | Medium-day plants |
| D. | Short-day plants |
| Answer» E. | |
| 4118. |
Choose the correct option for the prosses of spermatoyenesis from column I, column II |
| A. | P-i, Q-iv, R-iii, S-ii |
| B. | P-ii, Q-iii, R-iv, S-i |
| C. | P-iii, Q-ii, R-iv, S-i |
| D. | P-iv, Q-i, R-ii, S-iii |
| Answer» E. | |
| 4119. |
Which enzyme is needed to digest food reserve in pea seeds? |
| A. | Lipase |
| B. | Nuclease |
| C. | Proteases |
| D. | Amylase |
| Answer» D. Amylase | |
| 4120. |
. In a plant, red fruit (R) is dominant over yellow fruit (r) and tallness (T) is dominant over shortness (t). If a plant with RRTt genotype is crossed with a plant that is rrtt. AIPMT - 2004 |
| A. | 25% will be tall with red fruit |
| B. | 50% will be tall with red fruit |
| C. | 75% will be tall with red fruit |
| D. | All the offspring will be tall with red fruit. |
| E. | cessive |
| Answer» C. 75% will be tall with red fruit | |
| 4121. |
Match the terms in column-I with suitable terms in column-II. |
| A. | P iv Q iii R ii S i |
| B. | P iv Q iii R i S ii |
| C. | P iii Q i R iv S ii |
| D. | P iii Q iv R ii S i |
| Answer» C. P iii Q i R iv S ii | |
| 4122. |
. In pea plants, yellow seeds are dominant to green, If a heterozygous yellow seeded plant isCorssed with a green seeded plants, what ratio of yellow and green seeded plants, would you expect in F1 generation? AIPMT - 2007 |
| A. | 9 : 1 |
| B. | 1 : 3 |
| C. | 3 : 1 |
| D. | 50 : 50 |
| E. | cessive |
| Answer» E. cessive | |
| 4123. |
Which type of dormancy is induced in seeds due to adverse condition and high temperature ? |
| A. | Exogenous dormancy |
| B. | Endogenous dormancy |
| C. | Combinational dormancy |
| D. | Secondary dormancy |
| Answer» E. | |
| 4124. |
In the maize grain , the starchy food is stored in: |
| A. | Cotyledon |
| B. | Coleoptile |
| C. | Aleurone layer |
| D. | Endosperm |
| Answer» E. | |
| 4125. |
From which plant and from which part of that plant Cocain drug is obtained ? |
| A. | Ergot-fruit |
| B. | Erythroxylum coca-leaf |
| C. | Papaver somniferum-leaf |
| D. | Erythroxylum coca-Flower |
| Answer» C. Papaver somniferum-leaf | |
| 4126. |
The name Melvin Calvin is associated with |
| A. | synthesis of ATP during photosynthesis |
| B. | release of water during photosynthesis |
| C. | carbon fixation during photosynthesis |
| D. | capture light energy during photosynthesis |
| Answer» D. capture light energy during photosynthesis | |
| 4127. |
It is advantageous to use a water plant to demonstrate photosynthesis other than a land plant because. |
| A. | it photosynthesize rapidly |
| B. | it respires slowly |
| C. | it does not transpire |
| D. | O2 bubbles from cut and can be collected over H2O. |
| Answer» E. | |
| 4128. |
The non-polar part of chlorophyll is |
| A. | phytol |
| B. | porphyrin |
| C. | pyrrol |
| D. | none above |
| Answer» B. porphyrin | |
| 4129. |
Calvin cycle utilize for fixation of 3 molecules of CO2 |
| A. | 9 ATP and 6 NADPH2 |
| B. | 8 ATP and 8 NADPH2 |
| C. | 9 ATP and 3 NADPH2 |
| D. | 6 ATP and 6 NADPH2 |
| Answer» B. 8 ATP and 8 NADPH2 | |
| 4130. |
The volume of O2 librated in photosynthesis has the following ratio to CO2. |
| A. | O2/CO2 = 1 |
| B. | O2/CO2 = 1/2 |
| C. | O2/CO2 = 2/1 |
| D. | O2/CO2 = 3/1 |
| Answer» B. O2/CO2 = 1/2 | |
| 4131. |
The inhibiting effect of oxygen in C3 plants on photosynthesis is |
| A. | solarization |
| B. | photooxidation |
| C. | Warbug’s effect |
| D. | none above |
| Answer» D. none above | |
| 4132. |
Translocation of sugar in angiosperms occur in form of |
| A. | glucose |
| B. | starch |
| C. | lactose |
| D. | sucrose |
| Answer» E. | |
| 4133. |
For synthesis of one glucose molecule, the number of ATP required are |
| A. | 9 ATP for C3 cycle and 20 ATP for C4 cycle |
| B. | 18 ATP for C3 cycle and 30 ATP for C4 cycle |
| C. | 22 ATP for C3 cycle and 35 ATP for C4 cycle |
| D. | 24 ATP for C3 cycle and 36 ATP for C4 cycle |
| Answer» C. 22 ATP for C3 cycle and 35 ATP for C4 cycle | |
| 4134. |
Out of the total light energy that is available for plants is |
| A. | 50 % |
| B. | 75 % |
| C. | 25 % |
| D. | 1––2% |
| Answer» B. 75 % | |
| 4135. |
Photorespiration is favoured by |
| A. | low temperature |
| B. | low light intensity |
| C. | high O2 and low CO2 |
| D. | low O2 and high CO2 |
| Answer» D. low O2 and high CO2 | |
| 4136. |
The most efficient convertor of sunlight is |
| A. | Potato |
| B. | Tomato |
| C. | Sugar cane |
| D. | Papaya |
| Answer» D. Papaya | |
| 4137. |
If thylakoids are removed and kept in culture medium having CO2 and H2O and exposed to light ; they cannot form hexose sugars as end product because. |
| A. | light trapping device absent |
| B. | pigments P-700 and P-680 not linked |
| C. | CO2 assimilating enzymes absent |
| D. | CO2 assimilation cannot occur in light |
| Answer» D. CO2 assimilation cannot occur in light | |
| 4138. |
For synthesis of one gram of hexose, the land plant consumes. |
| A. | Only 1/3 of the CO2 of air |
| B. | Only 1/4 of the CO2 of air |
| C. | Only 2/3 of the CO2 of air |
| D. | none of the above |
| Answer» C. Only 2/3 of the CO2 of air | |
| 4139. |
Cyclic photophosphorylation is confined to |
| A. | Photosystem I |
| B. | Photosystem II |
| C. | both a & b |
| D. | none above |
| Answer» B. Photosystem II | |
| 4140. |
Chloroplast has maximum quantity of....in stroma |
| A. | dehydrogenase |
| B. | RuBP carboxylase |
| C. | pyruvic carboxylase |
| D. | hexokinase |
| Answer» C. pyruvic carboxylase | |
| 4141. |
Who gave chemical compositions of chlorophyll and carotenoids ? |
| A. | Park and Biggin |
| B. | Meyers and French |
| C. | Willstatter and Stahi |
| D. | Arnon and Benson |
| Answer» D. Arnon and Benson | |
| 4142. |
For chlorophyll formation most important are |
| A. | Fe++ and Ca++ |
| B. | Fe++ and Mg++ |
| C. | Mg++ and Ca++ |
| D. | all the above |
| Answer» C. Mg++ and Ca++ | |
| 4143. |
The approximate dimension of chlorophyll porphyrin ring is. |
| A. | 1 Ao square |
| B. | 5 Ao square |
| C. | 10 Ao square |
| D. | 15 Ao square |
| Answer» E. | |
| 4144. |
One photon is blue light contains....kcal and of red light....kcal |
| A. | 70 ; 40 |
| B. | 220 ; 70 |
| C. | 10 ; 90 |
| D. | 90 ; 10 |
| Answer» E. | |
| 4145. |
Quinones are |
| A. | mobile electron carrier |
| B. | enzymes of oxidative phosphorylation |
| C. | enzymes of krebs cycle |
| D. | none of the above |
| Answer» B. enzymes of oxidative phosphorylation | |
| 4146. |
During dark reaction, for the fixation of carbon, the three carbon atoms of each molecule of 3 –phosphogylceric acid (PGA) are derived from |
| A. | RuBP only |
| B. | CO2 only |
| C. | RuBP + CO2 |
| D. | RuBP + CO2 + PEP. |
| Answer» D. RuBP + CO2 + PEP. | |
| 4147. |
Pick up C4 plant. |
| A. | Papaya |
| B. | Potato |
| C. | Maize |
| D. | Pea |
| Answer» D. Pea | |
| 4148. |
Photosynthesis is most active in |
| A. | sunlight |
| B. | yellow |
| C. | red |
| D. | green |
| Answer» B. yellow | |
| 4149. |
Q10 is |
| A. | respiratory coefficient |
| B. | photosynthetic coefficient |
| C. | photosynthetic yield |
| D. | temperature coeffiecient |
| Answer» E. | |
| 4150. |
Intensity of light increase 20 times, rate of photosynthesis will |
| A. | increase |
| B. | not increase |
| C. | decrease |
| D. | increase till feed back inhibition. |
| Answer» E. | |