MHT CET 2024 22 April Shift 2 question paper is available for download here. MHT CET 2024 question paper comprises 150 MCQs carrying a total weightage of 200 marks. MHT CET 2024 22 April Shift 2 Question Paper for PCB is divided into three subjects- Physics, Chemistry and Biology. The Physics and Chemistry section of MHT CET 2024 22 April Shift 2 question paper consists of 50 questions (10 questions from Class 11 and 40 questions from Class 12th syllabus). Meanwhile, the Biology paper of MHT CET 2024 22 April Shift 2 question paper consists of 100 questions (20 questions from Class 11th and 80 questions from Class 12th syllabus).
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MHT CET 2024 22 April Shift 2 Questions with Solutions
Question 1:
In a family, a man with hemophilia, a genetic disorder where blood doesn't clot properly, marries a woman who is a carrier of the gene but does not express the disorder. What is the probability that their son will have hemophilia?
View Solution
Step 1: How hemophilia is inherited. Hemophilia is an X-linked recessive disorder. A man with hemophilia has the genotype XhY, while a carrier woman has the genotype XHXh.
Step 2: Determining offspring genotypes. A Punnett square shows the possible combinations of their offspring's genotypes:
| Xh | Y | |
| XH | XHXh | XHY |
| Xh | XhXh | XhY |
From the square:
- XHY: Normal son.
- XhY: Son with hemophilia.
- XHXh: Carrier daughter.
- XhXh: Daughter with hemophilia.
Step 3: Calculating the probability. Out of the male offspring, XhY represents half of the possibilities. Thus, the probability is 1/2 = 50%.
Question 2:
Which of the following is a characteristic symptom of Down syndrome?
View Solution
Step 1: Overview of Down syndrome. Down syndrome is a genetic disorder caused by the trisomy of chromosome 21. It results in various physical, intellectual, and developmental abnormalities.
Step 2: Key symptoms of Down syndrome. Common symptoms include:
- Congenital heart defects, such as heart murmurs.
- Flattened facial profile and slanted eyes.
- Intellectual disabilities and developmental delays.
Step 3: Analyzing the options.
- Option (A): Incorrect. Abnormal hair growth is not a symptom.
- Option (B): Correct. Heart murmurs are associated with congenital heart defects in Down syndrome.
- Option (C): Incorrect. Skin rash is not characteristic of Down syndrome.
- Option (D): Incorrect. Eye color changes are unrelated to Down syndrome.
Question 3:
During puberty, how many primary follicles are typically present in the ovaries of a female?
View Solution
Step 1: Understanding primary follicles. Primary follicles are immature ovarian follicles containing an oocyte. A female is born with approximately 1 to 2 million follicles, but this number decreases over time.
Step 2: Follicle count during puberty. By puberty, the number of follicles reduces to around 300,000 to 400,000 due to follicular atresia, a natural degenerative process.
Step 3: Analysis of options.
- Option (A): Incorrect. Two million follicles are present at birth, not during puberty.
- Option (B): Incorrect. One million follicles is an estimate for early childhood.
- Option (C): Correct. Only thousands (300,000–400,000) remain at puberty.
- Option (D): Incorrect. Primary follicles are present during puberty.
Step 4: Conclusion. At puberty, the number of primary follicles in the ovaries reduces to thousands.
Question 4:
The causative agent of malaria is:
View Solution
Step 1: Cause of malaria. Malaria is caused by protozoan parasites of the genus Plasmodium. Among them, Plasmodium falciparum is the most dangerous and severe.
Step 2: Role of the mosquito. The female Anopheles mosquito serves as the vector for transmitting Plasmodium but is not the direct causative agent.
Step 3: Explanation of options.
- Option (A): Correct. Plasmodium falciparum is the primary cause of malaria.
- Option (B): Incorrect. The mosquito is a vector, not the cause.
- Option (C): Incorrect. Trypanosoma brucei causes sleeping sickness.
- Option (D): Incorrect. Entamoeba histolytica causes amoebiasis.
Question 5:
Which virus is responsible for causing AIDS?
View Solution
Step 1: Understanding AIDS. AIDS (Acquired Immunodeficiency Syndrome) is caused by the Human Immunodeficiency Virus (HIV). This virus weakens the immune system by destroying CD4+ T cells, leading to immunosuppression.
Step 2: Explanation of other options.
- Option (A): Incorrect. Hepatitis B virus causes liver infections, not AIDS.
- Option (B): Incorrect. Human papillomavirus (HPV) is associated with cervical cancer, not AIDS.
- Option (D): Incorrect. Influenza virus causes flu, not AIDS.
Question 6:
Which disease is primarily spread by female Anopheles mosquitoes?
View Solution
Step 1: Role of female Anopheles mosquito. The female Anopheles mosquito is the primary vector responsible for transmitting malaria. It spreads the disease-causing parasite Plasmodium to humans during blood meals.
Step 2: Explanation of other options.
- Option (A): Incorrect. Dengue fever is transmitted by Aedes aegypti mosquitoes.
- Option (C): Incorrect. Zika virus is spread by Aedes mosquitoes.
- Option (D): Incorrect. Chikungunya is also transmitted by Aedes mosquitoes.
Question 7:
What is genomics?
View Solution
Step 1: Defining genomics. Genomics is the branch of biology that deals with the study of the structure, function, and analysis of the genome. It involves understanding the entirety of an organism's genetic material.
Step 2: Explanation of other options.
- Option (A): Incorrect. While genomics involves studying genes, its focus is on the entire genome.
- Option (B): Incorrect. The study of heredity and variation is genetics, not genomics.
- Option (D): Incorrect. Studying gene-environment interactions is the focus of epigenetics.
Question 8:
Molecular scissor of genetic engineering?
View Solution
Step 1: Role of restriction endonucleases. Restriction endonucleases, also called molecular scissors, are enzymes that cut DNA at specific sequences. They are essential in genetic engineering for creating recombinant DNA molecules.
Step 2: Explanation of other options.
- Option (A): Incorrect. DNA ligase joins DNA fragments but does not cut them.
- Option (B): Incorrect. Ligase performs DNA joining, not cutting.
- Option (D): Incorrect. RNA polymerase synthesizes RNA but does not cut DNA.
Question 9:
_____ is also called the terror of Bengal.
View Solution
Step 1: Identifying the terror of Bengal. The term "terror of Bengal" refers to Eichhornia crassipes, commonly known as water hyacinth. This invasive plant disrupts aquatic ecosystems and clogs water bodies, causing severe environmental damage.
Step 2: Explanation of options.
- Option (A): Incorrect. Pistia, or water lettuce, is not referred to as the terror of Bengal.
- Option (B): Correct. Eichhornia is synonymous with the water hyacinth.
- Option (C): Correct. Water hyacinth is the common name for Eichhornia.
- Option (D): Correct. Both (B) and (C) refer to the same plant species.
Question 10:
Which amino acids are histones rich in, facilitating their interaction with DNA?
View Solution
Step 1: Understanding histone structure. Histones are basic proteins that play a key role in DNA packaging. Their positive charge facilitates binding with the negatively charged phosphate groups in DNA.
Step 2: Amino acids responsible for DNA interaction. Histones are rich in lysine and arginine, both basic amino acids. Their positive charges enable strong interaction with DNA.
Step 3: Explanation of other options.
- Option (A): Incorrect. Glycine and proline are not basic amino acids.
- Option (C): Incorrect. Alanine and serine are neutral amino acids.
- Option (D): Incorrect. Aspartic acid and glutamic acid are acidic and repel DNA.
Question 11:
Leaf cutting is done successfully in which of the following plants?
View Solution
Step 1: Understanding leaf cutting propagation. Leaf cutting is a vegetative propagation technique where a leaf or part of a leaf is used to grow a new plant. This method is successful in plants that have the ability to regenerate shoots and roots from leaf tissues.
Step 2: Why Sansvieria is suitable. Sansvieria, also known as snake plant, is a succulent with the ability to propagate through leaf cuttings. Each leaf cutting can produce a new plant because it has meristematic cells that can differentiate into roots and shoots.
Step 3: Explanation of other options.
- Option (A): Incorrect. Roses are propagated through stem cuttings, not leaf cuttings.
- Option (B): Incorrect. Blackberries are propagated through root cuttings or stem cuttings, not leaf cuttings.
- Option (D): Incorrect. Bougainvillea is propagated through stem cuttings, not leaf cuttings.
Step 4: Conclusion. Among the given options, only Sansvieria can successfully propagate through leaf cuttings.
Question 12:
How many ATP molecules are needed as an initial investment in the glycolytic cycle (normal glycolysis)?
View Solution
Step 1: Understanding ATP investment in glycolysis. The glycolytic pathway begins with glucose being phosphorylated to form glucose-6-phosphate. This process requires energy in the form of ATP. A second ATP molecule is used to phosphorylate fructose-6-phosphate into fructose-1,6-bisphosphate.
Step 2: ATP molecules used. Two ATP molecules are consumed during the energy investment phase of glycolysis:
- One ATP is used in the conversion of glucose to glucose-6-phosphate.
- One ATP is used in the conversion of fructose-6-phosphate to fructose-1,6-bisphosphate.
Question 13:
Total genetic content of an organism is called:
View Solution
Step 1: Definition of gene pool. The gene pool is the total set of genetic material, including all alleles for all loci, within a population or species. It represents the diversity of genes available for inheritance.
Step 2: Explanation of other terms.
- Genetic drift: Refers to random changes in allele frequencies in a population.
- Gene frequency: Refers to the proportion of a specific allele in the gene pool.
- Gene mutation: Refers to a permanent change in the DNA sequence of a gene.
Question 14:
Theca interna releases which hormone?
View Solution
Step 1: Function of theca interna. The theca interna is a layer of cells in the ovarian follicle that plays a critical role in hormone production. It synthesizes androgens, which are precursors to estrogen.
Step 2: Hormone production. The granulosa cells, under the influence of FSH, convert the androgens produced by the theca interna into estrogen through aromatization.
Step 3: Explanation of other options.
- Option (A): Incorrect. Progesterone is primarily produced by the corpus luteum, not theca interna.
- Option (C): Incorrect. LH (luteinizing hormone) stimulates theca interna but is not secreted by it.
- Option (D): Incorrect. FSH (follicle-stimulating hormone) regulates granulosa cells, not theca interna.
Question 15:
Which of the following is an example of an outbreeding device in plants?
View Solution
Step 1: Understanding outbreeding devices. Outbreeding devices in plants promote cross-pollination and genetic diversity by preventing self-pollination. They ensure that pollen from one flower fertilizes the ovule of another flower, typically from a different plant.
Step 2: Xenogamy as an outbreeding device. Xenogamy refers to cross-pollination between flowers of different plants. This process promotes genetic recombination and variation, which are crucial for adaptation and survival in changing environments.
Step 3: Explanation of other options.
- Option (A): Incorrect. Cleistogamy involves self-pollination within closed flowers, which reduces genetic diversity.
- Option (B): Incorrect. Autogamy is self-pollination within the same flower, which does not promote outbreeding.
- Option (D): Incorrect. Geitonogamy involves pollination between flowers of the same plant, which does not promote true genetic recombination.
Step 4: Conclusion. Among the given options, only xenogamy represents an outbreeding device, as it involves cross-pollination between different plants.
Question 16:
Which of the following techniques is commonly used to introduce herbicide resistance into plants?
View Solution
Step 1: Role of Agrobacterium in genetic engineering. Agrobacterium tumefaciens is a bacterium that naturally transfers a part of its DNA (T-DNA) into plant cells. This mechanism is exploited in genetic engineering to introduce desired genes, such as those conferring herbicide resistance, into plants.
Step 2: Explanation of other techniques.
- RNA interference (Option A): Used to silence specific genes but not for introducing herbicide resistance.
- CRISPR-Cas9 gene editing (Option B): A precise genome editing tool but less commonly used for herbicide resistance compared to Agrobacterium-mediated methods.
- Polymerase chain reaction (PCR) (Option C): Used for amplifying DNA sequences, not for transferring genes into plants.
Question 17:
Which microorganism is used in yoghurt production?
View Solution
Step 1: Role of microorganisms in yoghurt production. Yoghurt is produced through the fermentation of milk using specific bacteria. Streptococcus thermophilus and Lactobacillus delbrueckii subsp. bulgaricus are commonly used starter cultures.
Step 2: Explanation of other microorganisms.
- Streptococcus penicillium (Option B): Incorrect. This is not a known species used in yoghurt production.
- Penicillium roqueforti (Option C): Incorrect. This fungus is used in the production of blue cheese, not yoghurt.
- Aspergillus niger (Option D): Incorrect. This fungus is used for industrial production of citric acid, not yoghurt.
Question 18:
If *p* is the magnitude of linear momentum of a particle executing a uniform circular motion, then the ratio of centripetal force acting on the particle to its linear momentum is given by:
View Solution
Step 1: Relationship between centripetal force and linear momentum. The centripetal force Fc for a particle in uniform circular motion is given by:
Step 2: Ratio of Fc to p. The ratio of centripetal force to linear momentum is:
Step 3: Conclusion.The ratio of centripetal force to linear momentum is v/r.
Question 19:
An e.m.f of 5 volts is produced by a self-inductance when the current changes at a steady rate from 3A to 2A in 1 millisecond. The value of self-inductance is:
View Solution
Step 1: Formula for e.m.f induced by self-inductance. The induced e.m.f (Ɛ) in a coil is given by:
- Ɛ = 5 V (induced e.m.f),
- ΔI = 3 A - 2 A = -1 A (change in current),
- Δt = 1 ms = 1 × 10-3 s.
Step 2: Calculating self-inductance. Rearranging the formula for L:
Question 20:
Two monkeys of mass 10 kg and 8 kg are moving along a vertical light rope. The former is climbing up with an acceleration of 2 m/s², while the latter is coming down with a uniform velocity. Find the tension in the rope at the fixed support.
View Solution
Step 1: Tension due to the first monkey (climbing up). The net force on the first monkey is given by:
- T1 is the tension due to the first monkey,
- m1 = 10 kg,
- g = 9.8 m/s² (acceleration due to gravity),
- a1 = 2 m/s² (upward acceleration).
Step 2: Tension due to the second monkey (coming down). The second monkey is moving down with a uniform velocity (a2 = 0). The tension is:
- T2 is the tension due to the second monkey,
- m2 = 8kg.
Step 3: Total tension in the rope at the fixed support. The total tension in the rope is the sum of tensions due to both monkeys:
Question 21:
"Water is flowing through a horizontal pipe in streamline flow at the narrowest part of the pipe?"
View Solution
Step 1: Bernoulli's principle and its application. Bernoulli's principle dictates that for an incompressible fluid moving in streamline flow, the total mechanical energy (pressure energy, kinetic energy, and potential energy) remains constant. This is mathematically expressed as:
Step 2: Behavior at the narrowest part of the pipe. In a horizontal pipe, Bernoulli's equation simplifies to:
Question 22:
"The height from Earth's surface at which acceleration due to gravity becomes *g*/4, where *g* is acceleration due to gravity on the surface of Earth and *R* is the radius of Earth?"
View Solution
Step 1: Relation of gravity with height. The acceleration due to gravity at a height *h* above the Earth's surface is given by:
Step 2: Substituting gh = g/4.
Question 23:
How many steps are there in the process of glycolysis, the metabolic pathway that converts glucose into pyruvate?
View Solution
Step 1: Overview of glycolysis. Glycolysis consists of 10 enzymatic reactions that convert one glucose molecule (C6H12O6) into two molecules of pyruvate (C3H4O3). It occurs in the cytoplasm and does not require oxygen.
Step 2: Breakdown of the steps.
- The first 5 steps are the energy investment phase.
- The last 5 steps are the energy payoff phase.
- A total of 2 ATP molecules are consumed, and 4 ATP molecules are generated, giving a net gain of 2 ATP per glucose.
Question 24:
Correct sequence in water absorption by root hairs.
View Solution
Step 1: Water absorption process by root hairs. Water absorption occurs in three stages:
- Imbibition: The initial absorption of water by hydrophilic substances such as cellulose in the cell walls.
- Diffusion: Water molecules then move from regions of higher concentration (soil) to lower concentration (root hair cells).
- Osmosis: Finally, water passes through the semipermeable membranes into the root cortex via osmotic gradients.
Step 2: Sequence of events. The correct order is: Imbibition → Diffusion → Osmosis.
Step 3: Explanation of incorrect options.
- Option (B): Incorrect. Osmosis occurs after imbibition and diffusion.
- Option (C): Incorrect. Diffusion cannot precede imbibition.
- Option (D): Incorrect. Osmosis happens after diffusion, not before.
Question 25:
Annealing process of PCR is performed at ___ temperature.
View Solution
Step 1: Explanation of the PCR process. Polymerase Chain Reaction (PCR) involves three main steps:
- Denaturation: The DNA strands separate at high temperatures (90-98°C).
- Annealing: Primers bind to the DNA template at moderate temperatures (40-60°C).
- Extension: DNA polymerase synthesizes the new strand at 70-75°C.
Step 2: Annealing temperature. During the annealing phase, the temperature is set between 40-60°C, optimized for primers to attach to the template DNA without non-specific binding.
Question 26:
Which of the following flowers is most likely to be pollinated by birds?
View Solution
Step 1: Characteristics of bird-pollinated flowers. Bird-pollinated flowers are typically characterized by:
- Bright colors (red, orange, yellow) that attract birds.
- Long, tubular shapes that facilitate nectar feeding by birds.
- No strong fragrance, as birds rely more on sight than smell.
Step 2: Explanation of other options.
- Option (A): Correct. Long tubular flowers with bright colors are designed for bird pollination.
- Option (B): Incorrect. Fragrant flowers are usually pollinated by insects.
- Option (C): Incorrect. Flowers that open at night are usually moth-pollinated.
- Option (D): Incorrect. Small, dull flowers are typically pollinated by wind or water.
Question 27:
Bond present between two nucleotides on a single strand of DNA is:
View Solution
Step 1: Bond formation in DNA. In a single DNA strand, nucleotides are linked by phosphodiester bonds. This bond forms between the 3'-hydroxyl group of one nucleotide and the 5'-phosphate group of the next nucleotide, creating the backbone of the DNA strand.
Step 2: Explanation of other bonds.
- Hydrogen bond (Option B): Forms between complementary bases in double-stranded DNA.
- Glycosidic bond (Option C): Connects the nitrogenous base to the sugar in a nucleotide.
- Covalent bond (Option D): A broad category that includes phosphodiester bonds.
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