VITEEE PCBE Mock Test - 6 - JEE MCQ

FSH is responsible for the initiation of spermatogenesis and sperm differentiation. It binds with Sertoli cells and spermatogonia and induces proliferation of spermatogonia .

So, the correct answer is 'FSH'.

Each Sporogenous cell under meiotic division to form tetrad of megaspores. The three megaspores degenerate and only one megaspore develops into embryo sac or female gametophyte.

Statement i is incorrect because benign tumors do not spread; they remain localized.
Statement ii is correct as malignant tumors invade surrounding tissues and can spread to distant sites (metastasis).
Statement iii is correct as benign tumors usually do not cause significant damage and stay localized.
Statement iv is incorrect because malignant tumors are usually fast-growing and not confined to their original location.
Thus, the correct answer is Option B: ii and iii.

Topic in NCERT: Tumors: benign and malignant

Line in NCERT: "benign tumors normally remain confined to their original location and do not spread to other parts of the body and cause little damage. the malignant tumors, on the other hand are a mass of proliferating cells called neoplastic or tumor cells. these cells grow very rapidly, invading and damaging the surrounding normal tissues. cells sloughed from such tumors reach distant sites through blood, and wherever they get lodged in the body, they start a new tumor there. this property called metastasis is the most feared property of malignant tumors."

Zinc is a cofactor for the proteolytic enzyme carboxypeptidase.

First gene therapy was done in 1990 to a 4 year old girl with a ADA deficiency. ADA enzyme is essential for proper functioning of immune system.

Centrosome is an organelle usually containing two cylindrical structures called centrioles. They are surrounded by amorphous pericentriolar materials. Both the centrioles in a centrosome lie perpendicular to each other in which each has an organisation like the cartwheel. They are made up of nine evenly spaced peripheral fibrils of tubulin protein. Each of the peripheral fibrils is a triplet. The adjacent triplets are also linked.

Enzyme may be broadly classified into two types depending on their chemical composition-simple enzymes and conjugated enzymes are wholly made up of proteins and any additional substance or group is absent, e.g., pepsin, trypsin, etc. Conjugated enzymes (or holoenzymes) are formed of two parts -a protein part called apoenzyme and a non-protein part named co-factor. The complete conjugated enzyme consisting of an apoenzyme and a co-factor is called holoenzyme. Holoenzyme is the functional unit of enzyme.

Co-factor may be inorgainc or roganic in nature. Catalytic activity is lost when co-factor is removed from the enzyme which indicates that it plays a crucial role in catalytic activity of enzymes.

In the given scenario, the laboratory population of fruit flies consists of 40 fruit flies, and during a specified time interval, 4 individuals died. To calculate the death rate, we divide the number of deaths by the total population.

Number of deaths = 4
Total population = 40

Death rate = Number of deaths / Total population
= 4 / 40
= 0.1 individuals per fruitfly per week

Option a) "0.1 individuals per fruitfly per week" is the correct answer because it accurately represents the death rate calculated based on the given information.

Topic in NCERT: Population Dynamics

Line in NCERT: "If 4 individuals in a laboratory population of 40 fruitflies died during a specified time interval, say a week, the death rate in the population during that period is 4/40 = 0.1 individuals per fruitfly per week."

Organisms have evolved from simpler forms to complex form. Hence the order of the events are genes first, then heterotrophic prokaryotes, then autotrophic prokaryotes, then eukaryotes and then animals.

Topic in NCERT: Evolution of life forms - a theory

Line in NCERT: Error occcured while getting response from embedding

• Imbricate aestivation is found in Cassia
• Valvate aestivation is found in Calotropis
• Vexillary aestivation is found in Bean
• Twisted aestivation is found in cotton

Given: F = at + bt² where t is time.
For this equation to be dimensionally correct, each term in the right hand side should have the dimensions of F.
Dimensional formula of F = [MLT⁻²]
^⇒ aT = [MLT⁻²]
^⇒ The dimensions of a are [MLT⁻³].
And,
^⇒ bT² = [MLT⁻²]
^⇒ The dimensions of b are [MLT⁻⁴].
Hence, option (4) is correct.

Number of nuclei in 235 g = 6.02×10²³
Hence number of nuclei in 1000 g =
Energy liberated in fission of one nuclei = 200 MeV = 200 × 10⁶ × 1.6 × 10^-19 J
Energy liberated during fission of 1 kg of ₉₂U²³⁵ = 2.56 × 10²⁴ × 200 × 10⁶ × 1.6 × 10^-19 = 8.19 × 10¹³ J
or Energy liberated is 8.2 x 10¹³ J

For the parallel plate capacitor, the capacitance is given by:

When the battery is disconnected, the charge remains constant. As the separation between the plates is increased, the capacitance decreases, and voltage across the capacitor will increase as we have:

Instantaneous induced emf is given by

Also

NΦ = Li

Magnetic flux through each turn,

As the mass distribution is the same around the x-axis, therefore, the x coordinate is zero. For y coordinate, it is given as,  

= 
= a/10

Initially, the force between the charges is given by,

After contact, the charges gets equally distributed between the two particles, such that, the charge on each is 

For a zero-order reaction, - (d[A]/dt) = k
Thus, [A]_t – [A]₀ = –kt

N-ethylbutan-2-amine is formed by replacing two H-atoms of ammonia molecule. Hence, it is a secondary amine.

The Hoffmann bromide reaction is as given below:

For a spontaneous process, ΔG₀ must be negative.
According to thermodynamics, ΔH° and ΔS° may be positive or negative, but ΔG° must be negative for a spontaneous reaction.

Hydrolysis is any chemical reaction in which a molecule of water breaks one or more chemical bonds. The term is used broadly for substitution, elimination, and solvation reactions in which water is the nucleophile. Thus hydrolysis adds water to break down, whereas condensation builds up by removing water.

Phenol ca be converted salicylaldehyde in the presence of chloroform and KOH. This reaction is known as Reimer-Tiemann reaction. In this reaction, the mixture chloroform and KOH gives dichlorocarbene, which acts as an electrophile in this reaction.

Among the given reagents, only NaOH + Br₂ converts −CONH₂ group to −NH₂ group, thus it is used for converting acetamide to methyl amine. 

This reaction is called Hoffman bromide degradation reaction. Amine produced in this reaction has one carbon atom less than that in reactant amide.

According to the given information in the question, diagram can be drawn as,

B is the mother of H. C is the nephew of E. A is the husband of H. A is third to the left of H. Both the neighbours of C are females. F and G are daughters of H.
Thus, option D is the correct answer.

40 SP = 50 CP
SP/CP = 50/40
% gain =
=