Human somatic cells are diploid (2n) and contain 46 chromosomes.


This set of 46 chromosomes consists of 44 autosomes and 2 sex chromosomes (XY in males).


Spermatogenesis is the process of sperm formation, which involves meiosis.


A primary spermatocyte is a diploid cell (44 + XY) that undergoes the first meiotic division (Meiosis I).


Meiosis I is a reductional division, where homologous chromosomes separate, resulting in two haploid (n) cells called secondary spermatocytes.


Each secondary spermatocyte receives half the number of chromosomes: 22 autosomes and one sex chromosome.


Therefore, one secondary spermatocyte will have the composition 22 + X, and the other will have 22 + Y.


The question asks for the number of autosomes, which is 22 in a secondary spermatocyte.
Quick Tip: Remember the ploidy changes in meiosis. Meiosis I is the reductional division (2n -> n), so a primary spermatocyte (2n) becomes a secondary spermatocyte (n). A secondary spermatocyte has a haploid number of chromosomes, which for humans is 23 (22 autosomes + 1 sex chromosome).

The provided diagram shows a longitudinal section of a monocot seed, specifically a maize grain.


Part I points to the outermost layer of the grain. In cereals, the seed coat is fused with the fruit wall (pericarp). This combined layer is the pericarp.


Part II indicates the large, bulky tissue that stores food reserves. This nutritive tissue is the endosperm.


Part III points to the large, shield-shaped cotyledon characteristic of monocot embryos. This structure is called the scutellum.


Matching these identifications with the given options confirms that option (A) is the correct identification.
Quick Tip: In monocot seeds (like maize or wheat), remember the key parts: Pericarp (outermost protective layer, fused with seed coat), Endosperm (large food storage tissue), and the Embryo which includes the Scutellum (cotyledon), Coleoptile (sheath covering plumule), and Coleorhiza (sheath covering radicle).

The diagram shows a molecule of heterogeneous nuclear RNA (hnRNA), the primary transcript in eukaryotes.


Exons are the coding sequences, represented by the straight-line segments of the RNA strand.


Introns are the non-coding intervening sequences that are spliced out, represented by the looped-out segments.


By counting the looped-out structures (introns), we find there are 7 of them.


By counting the straight-line segments (exons) from the 5' to the 3' end, we find there are 8 of them.


The question asks for the number of introns and exons, respectively.


Therefore, the count is 7 introns and 8 exons.


This corresponds to the option (D).
Quick Tip: In a linear gene or hnRNA molecule, the number of exons is typically one more than the number of introns. Exons are like "posts" and introns are the "fences" between them. If you have 'n' introns, you will have 'n+1' exons. In this diagram, 7 introns separate 8 exons.

E. coli divides in 20 minutes. The question asks for the ratio after the III generation (which is 3 x 20 = 60 minutes).


Generation 0: All DNA is heavy (\(N^{15}/N^{15}\)). Let's start with 1 DNA molecule.


Generation I (after 20 mins): The 1 heavy DNA molecule replicates to form 2 hybrid (\(N^{15}/N^{14}\)) DNA molecules.


Generation II (after 40 mins): The 2 hybrid molecules replicate to form 2 hybrid (\(N^{15}/N^{14}\)) and 2 light (\(N^{14}/N^{14}\)) molecules. Total 4 molecules.


Generation III (after 60 mins): The 2 hybrid molecules form 2 hybrid and 2 light molecules. The 2 light molecules form 4 light molecules.


Total molecules = 8.


Number of heavy (\(N^{15}/N^{15}\)) molecules = 0.


Number of hybrid (\(N^{15}/N^{14}\)) molecules = 2.


Number of light (\(N^{14}/N^{14}\)) molecules = 2 + 4 = 6.


The ratio of \(N^{15}/N^{15}\) : \(N^{15}/N^{14}\) : \(N^{14}/N^{14}\) is 0 : 2 : 6.


Simplifying this ratio by dividing by 2, we get 0 : 1 : 3.
Quick Tip: In the Meselson-Stahl experiment, after the first generation in \(N^{14}\) medium, the number of pure heavy (\(N^{15}/N^{15}\)) DNA molecules will always be zero. The number of hybrid (\(N^{15}/N^{14}\)) molecules will always be two. The number of light (\(N^{14}/N^{14}\)) molecules after 'n' generations (for n>0) will be \(2^n - 2\).

Let's analyze the given information:


Mother's blood group is AB, so her genotype is definitively \(I^A I^B\).


Father's blood group is B, so his genotype can be \(I^B I^B\) or \(I^B i\).


Child's blood group is A, so the child's genotype can be \(I^A I^A\) or \(I^A i\).


The child must inherit one allele from each parent. The child has an \(I^A\) allele.


Since the father has blood group B (no \(I^A\) allele), the child must have inherited the \(I^A\) allele from the mother.


The mother has genotype \(I^A I^B\), so she passed the \(I^A\) allele to the child.


The child inherited the other allele from the father. This allele cannot be \(I^B\), because then the child's blood group would be AB.


Therefore, the child must have inherited the recessive \(i\) allele from the father.


This makes the child's genotype \(I^A i\).


For the father to provide an \(i\) allele, his genotype must be \(I^B i\).


Thus, the genotypes are: Father (\(I^B i\)), Mother (\(I^A I^B\)), Child (\(I^A i\)).


This matches option (C).
Quick Tip: When solving blood group genetics problems, always start with the individual whose genotype is certain. In this case, the mother with blood group AB has a fixed genotype (\(I^A I^B\)). Use this certainty to deduce the genotypes of the others by tracing the inheritance of alleles.

To determine the correct evolutionary order, we need to arrange the plants from the most ancient to the most recent.


(iii) Zosterophyllum was an ancient, extinct vascular plant from the Silurian-Devonian period. It is considered an early ancestor of land plants.


(i) Ferns are Pteridophytes. They evolved after the early vascular plants like Zosterophyllum and became dominant during the Carboniferous period.


(ii) Ginkgo is a gymnosperm. Gymnosperms evolved from pteridophyte-like ancestors and appeared after the ferns.


(iv) Gnetales are a group of advanced gymnosperms that show some features similar to angiosperms. They are considered more recently evolved than Ginkgo.


Therefore, the correct evolutionary sequence is Zosterophyllum \(\rightarrow\) Ferns \(\rightarrow\) Ginkgo \(\rightarrow\) Gnetales.


This corresponds to the order (iii), (i), (ii), (iv).
Quick Tip: Remember the general timeline of plant evolution: Algae \(\rightarrow\) Bryophytes \(\rightarrow\) Pteridophytes (like ferns) \(\rightarrow\) Gymnosperms (like Ginkgo, Gnetales) \(\rightarrow\) Angiosperms. Zosterophyllum is a very early fossil pteridophyte.

Let's match the eukaryotic RNA polymerases with their primary products.


(a) RNA polymerase I is located in the nucleolus and is responsible for transcribing ribosomal RNAs (rRNAs), specifically the 28S, 18S, and 5.8S rRNA genes. Therefore, (a) correctly matches with (i) 18s rRNA.


(b) RNA polymerase II is responsible for transcribing the precursor to messenger RNA (mRNA), which is heterogeneous nuclear RNA (hnRNA). It also transcribes most small nuclear RNAs (snRNAs). The most specific and primary match in the options is hnRNA. Therefore, (b) correctly matches with (iii) hnRNA.


(c) RNA polymerase III transcribes transfer RNA (tRNA), 5S rRNA, and some snRNAs (like U6 snRNA). From the given options, SnRNAs is the best available match. Therefore, (c) correctly matches with (ii) SnRNAs.


The correct set of matches is (a)-(i), (b)-(iii), and (c)-(ii).


This corresponds to option (D).
Quick Tip: A simple mnemonic for eukaryotic RNA polymerases is "R-M-T for 1-2-3". Pol I makes rRNA. Pol II makes mRNA (via hnRNA). Pol III makes tRNA (and 5S rRNA).

Pedigree charts use standard symbols to represent family relationships and the inheritance of traits.


A horizontal line connecting a square (male) and a circle (female) represents a mating.


A double horizontal line, as shown in the question, specifically indicates a consanguineous mating.


Consanguineous mating means mating between individuals who are closely related, such as first cousins.


Therefore, the symbol represents mating between relatives.
Quick Tip: In pedigree analysis, memorize the key symbols: square for male, circle for female, shaded for affected, unshaded for unaffected, and a double line for consanguineous mating. This symbol is important as it increases the chance of offspring being affected by recessive genetic disorders.

Immunity can be classified as active or passive, and as natural or artificial.


Active immunity involves the body producing its own antibodies after exposure to an antigen.


Passive immunity involves receiving pre-formed antibodies from an external source.


In this case, the fetus receives already-made antibodies from the mother. The fetus's immune system is not producing them. This is therefore a form of passive immunity.


Natural immunity is acquired through normal life processes.


Artificial immunity is acquired through medical intervention (e.g., vaccination, injection of antiserum).


The transfer of antibodies from mother to fetus via the placenta is a natural biological process.


Combining these classifications, the immunity is naturally acquired passive immunity.
Quick Tip: Break down immunity types: - Active: Your body does the work (makes antibodies). - Passive: You get ready-made antibodies. - Natural: Happens without medical help (infection, mother to baby). - Artificial: Happens via medical procedure (vaccine, antiserum shot). The transfer of IgG via the placenta and IgA via breast milk are classic examples of naturally acquired passive immunity.

The number of different types of gametes an organism can produce is determined by the number of heterozygous gene pairs (loci).


The formula to calculate the number of possible gamete types is \(2^n\), where 'n' is the number of heterozygous loci.


In this question, the organism is heterozygous for three loci. Therefore, n = 3.


Substituting the value of n into the formula:


Number of gamete types = \(2^3\).

\(2^3 = 2 \times 2 \times 2 = 8\).


Therefore, the organism can produce 8 different types of gametes.


For example, if the genotype is AaBbCc, the 8 gametes would be ABC, ABc, AbC, Abc, aBC, aBc, abC, and abc.
Quick Tip: The formula \(2^n\) is fundamental in genetics. 'n' represents the number of heterozygous gene pairs. This formula is used to find the number of different gametes an individual can produce, and also helps in determining the number of phenotypes in polygenic inheritance.

Gel electrophoresis is a technique used to separate DNA fragments based on their size.


DNA molecules are negatively charged due to the phosphate groups in their backbone.


When an electric field is applied, these negatively charged DNA fragments move towards the positive electrode (anode).


The agarose gel acts as a molecular sieve, with pores that impede the movement of the DNA.


Smaller DNA fragments can navigate through the pores of the gel matrix more easily and quickly than larger fragments.


As a result, smaller fragments travel a longer distance from the loading wells in a given amount of time.


Therefore, the smaller the fragment size, the farther it moves. Option (A) is correct.
Quick Tip: Remember the principle of gel electrophoresis as "DNA's race". It's a race towards the positive finish line, and the smallest, lightest runners (fragments) run the farthest. DNA is always negatively charged, so it will always move towards the positive pole.

The blood cholesterol-lowering agents are a class of drugs called statins.


Statins are produced by a type of yeast called Monascus purpureus.


These statins work by competitively inhibiting the enzyme HMG-CoA reductase, which is responsible for the synthesis of cholesterol in the body.


Let's look at the other options:


Trichoderma polysporum is a fungus that produces the immunosuppressant cyclosporin A.


Saccharomyces cerevisiae is brewer's yeast, used in baking and fermentation.


Aspergillus niger is a fungus used for the commercial production of citric acid.


Therefore, Monascus purpureus is the correct answer.
Quick Tip: Associate key microbes with their products for exams: - Monascus purpureus \(\rightarrow\) Statins (cholesterol). - Trichoderma polysporum \(\rightarrow\) Cyclosporin A (immunosuppressant). - Streptococcus \(\rightarrow\) Streptokinase (clot buster). - Aspergillus niger \(\rightarrow\) Citric acid.

Assertion (A) states that the 'XX - XY' type of sex determination is an example of male heterogamety.


In this system, females have two X chromosomes (XX) and produce only one type of gamete (containing an X chromosome). They are homogametic.


Males have one X and one Y chromosome (XY) and produce two different types of gametes (one with X and one with Y). This is known as male heterogamety. So, Assertion (A) is true.


Reason (R) states that in birds, male heterogamety is observed.


Birds have a 'ZZ - ZW' type of sex determination system.


In this system, males have two Z chromosomes (ZZ) and are homogametic, producing only one type of gamete (with Z).


Females have one Z and one W chromosome (ZW) and are heterogametic, producing two types of gametes (one with Z and one with W).


Therefore, birds exhibit female heterogamety, not male heterogamety. Reason (R) is false.


Since Assertion (A) is true and Reason (R) is false, the correct option is (C).
Quick Tip: Remember the two main types of chromosomal sex determination: - XY system (Humans, Drosophila): Male is heterogametic (XY). - ZW system (Birds, Reptiles): Female is heterogametic (ZW). The "heterogametic" sex is the one that produces two different types of gametes and determines the sex of the offspring.

Assertion (A) describes the phenomenon of industrial melanism in the peppered moth (Biston betularia).


Before industrialization, light-colored, lichen-covered trees were common, and white-winged moths were well-camouflaged, so their population was high.


After industrialization, soot and pollution killed the lichens and darkened the tree barks. This made the white-winged moths easily visible to predators, leading to a decrease in their numbers. So, Assertion (A) is true.


Reason (R) states that the effects of industrialization were more marked in rural areas.


This is incorrect. The effects, such as smoke and soot from factories, were concentrated in and around industrial cities and towns.


Rural areas, being farther away from the industrial centers, remained relatively unpolluted, and the trees there often retained their light-colored bark and lichens. So, Reason (R) is false.


Since Assertion (A) is true and Reason (R) is false, the correct option is (C).
Quick Tip: Industrial melanism is a classic example of natural selection in action. The environment (tree bark color) changed, which changed the selective pressure (predation). The moth population evolved in response, with the better-camouflaged variant becoming more common.

Assertion (A) states that the secondary immune response is quicker and stronger than the primary one.


The primary response is the immune system's first encounter with a specific pathogen. It is relatively slow and of low intensity as the body needs time to identify the antigen and produce specific antibodies.


The secondary (anamnestic) response occurs upon subsequent exposure to the same pathogen. It is much faster, more intense, and longer-lasting. So, Assertion (A) is true.


Reason (R) attributes this to immunological memory.


The acquired immune system has the characteristic of memory. During the primary response, memory B-cells and memory T-cells are produced and they persist in the body.


Upon a second encounter, these memory cells quickly recognize the pathogen and proliferate, leading to a rapid and massive production of antibodies and effector T-cells.


This memory is the direct reason for the quicker and stronger secondary response. So, Reason (R) is true and is the correct explanation for (A).


Therefore, option (A) is correct.
Quick Tip: The concept of immunological memory is the basis for vaccination. Vaccines introduce antigens from a pathogen to induce a primary immune response and create memory cells, without causing the actual disease. If the real pathogen ever enters the body, the secondary response is so quick and strong that it prevents illness.

Assertion (A) describes a treatment for ADA (Adenosine Deaminase) deficiency.


ADA deficiency is a genetic disorder that severely damages the immune system. One of the treatment approaches is a form of gene therapy where lymphocytes are removed from the patient's blood, a functional ADA gene is introduced into them using a retroviral vector, and these genetically-engineered cells are then returned to the patient. So, Assertion (A) is true.


Reason (R) states that lymphocytes are not immortal and have a limited life span.


This is also a true biological fact. Most cells in our body, including lymphocytes, undergo apoptosis (programmed cell death) after a certain period.


The reason the infusion of corrected lymphocytes must be periodic is precisely because these cells are mortal. As the infused cells die off, the patient's symptoms would return. Therefore, repeated infusions are necessary to maintain a population of functional lymphocytes.


Thus, Reason (R) is true and it correctly explains why the treatment described in Assertion (A) needs to be periodic.


Therefore, option (A) is correct.
Quick Tip: This method of gene therapy for ADA deficiency is a good example of ex vivo gene therapy (cells are modified outside the body). It is not a permanent cure because it does not correct the defect in the stem cells in the bone marrow. A more permanent cure involves introducing the gene into early embryonic cells or bone marrow stem cells.

(i) Identification of Hormones:


The graph shows the levels of ovarian hormones during a typical menstrual cycle.


Hormone 'P' level peaks just before day 14 (ovulation). This hormone is responsible for the proliferation of the endometrium and triggers the LH surge. This hormone is Estrogen.


Hormone 'Q' level rises after ovulation (during the luteal phase) and remains high. It is responsible for maintaining the endometrium for implantation. This hormone is Progesterone.


Therefore, P is Estrogen and Q is Progesterone.


(ii) Source of Hormones:


The source of Estrogen (P) is the growing ovarian follicles (Graafian follicle).


The source of Progesterone (Q) is the Corpus Luteum, which is formed from the remnants of the Graafian follicle after ovulation.
Quick Tip: Remember the sequence in the menstrual cycle: FSH stimulates follicular growth, growing follicles secrete Estrogen, a peak in Estrogen causes an LH surge, the LH surge causes ovulation, the remaining follicle becomes the Corpus Luteum, and the Corpus Luteum secretes Progesterone.

(i) Identification of Hormones:


The graph shows the levels of pituitary gonadotropins during the menstrual cycle.


Hormone 'P' shows a sharp and sudden peak around the 14th day. This is known as the LH surge, which is responsible for inducing ovulation. Therefore, P is Luteinizing Hormone (LH).


Hormone 'Q' stimulates the growth and development of ovarian follicles. Its level is moderately high during the follicular phase. Therefore, Q is Follicle-Stimulating Hormone (FSH).


(ii) Role of Hormone 'P' (LH):


In Females: LH induces the rupture of the mature Graafian follicle, leading to the release of the ovum (ovulation). It also stimulates the transformation of the ruptured follicle into the corpus luteum and maintains it.


In Males: LH acts on the Leydig cells or interstitial cells of the testes. It stimulates these cells to synthesize and secrete androgens, primarily testosterone. In males, LH is also known as Interstitial Cell-Stimulating Hormone (ICSH).
Quick Tip: Associate the hormone peaks with key events. The sharp mid-cycle peak is always the LH surge, which is the direct trigger for ovulation. FSH is primarily for "follicle stimulating," so its action is dominant in the first half of the cycle.

(a) The two closely linked genes that control the synthesis of the alpha globin chain of haemoglobin are HBA1 and HBA2. Both are located on chromosome 16 of each parent.


(b) The differentiation is as follows:


\begin{tabularx{\linewidth{|l|X|X|
\hline
Feature & Thalassemia & Sickle Cell Anaemia

\hline
Nature of Defect & It is a quantitative disorder. & It is a qualitative disorder.

\hline
Effect on Globin & It is caused by the synthesis of a reduced quantity of structurally normal globin chains (\(\alpha\) or \(\beta\)). & It is caused by the synthesis of a structurally abnormal globin chain (\(\beta\)-globin).

\hline
Molecular Cause & Caused by gene deletion or mutation leading to reduced gene expression. & Caused by a point mutation (GAG to GUG) in the \(\beta\)-globin gene, substituting glutamic acid with valine.

\hline
\end{tabularx
Quick Tip: Remember: Thalassemia = Quantity problem (Too little globin). Sickle-cell anaemia = Quality problem (Wrong kind of globin). This fundamental difference is key to understanding many genetic blood disorders.

Injection of pre-formed antibodies provides artificial passive immunity. It is recommended in emergencies where an immediate and potent immune response is required, as the body does not have time to mount its own active immune response.


Two such situations are:


1. Snakebite: When a person is bitten by a venomous snake, the venom can cause rapid and life-threatening damage. An anti-venom injection, which contains pre-formed antibodies against the venom toxins, is administered to neutralize the venom immediately and save the person's life.


2. Tetanus Infection: In case of a deep wound contaminated with soil, there is a risk of infection by Clostridium tetani, which produces a deadly toxin. To provide immediate protection, an anti-tetanus serum (ATS) or Tetanus Immunoglobulin (TIG), containing pre-formed antibodies against the toxin, is injected.
Quick Tip: Passive immunity is like getting a borrowed shield. It's fast and provides immediate protection but is temporary because the borrowed antibodies are eventually cleared from the body and no memory cells are formed. It is used for post-exposure prophylaxis.

There is an incubation period in malaria, meaning symptoms do not appear immediately after the mosquito bite for the following reasons:


1. Initial Site of Infection: The infective stage, sporozoites, injected by the mosquito do not infect red blood cells (RBCs) directly. They travel via the bloodstream to the liver.


2. Asymptomatic Liver Phase (Exo-erythrocytic cycle): Inside the liver cells, the sporozoites multiply asexually to produce thousands of merozoites. This phase of the parasite's life cycle does not cause any symptoms of the disease.


3. Infection of RBCs (Erythrocytic cycle): After the liver phase, the liver cells rupture and release the merozoites into the bloodstream. These merozoites then invade the RBCs.


4. Cause of Symptoms: Inside the RBCs, the parasite multiplies asexually, eventually causing the RBC to rupture. The rupture of RBCs releases a toxic substance called hemozoin, along with new merozoites that infect more RBCs.


It is the periodic release of hemozoin with each cycle of RBC rupture that causes the characteristic malarial symptoms of recurring chills and high fever. Therefore, the time taken for the completion of the liver phase is the reason for the delay in the appearance of symptoms.
Quick Tip: Remember that malarial symptoms (fever, chills) are linked to the rupture of Red Blood Cells. The initial phase in the liver is a "silent" multiplication stage. No RBC rupture means no symptoms.

(a) The DNA fragment contains the sequence:

5' ...G A A T T C... 3'

3' ...C T T A A G... 5'

This six-base pair sequence is a palindrome that is the specific recognition site for the restriction endonuclease \textit{EcoRI (from Escherichia coli).


(b) \textit{EcoRI cuts the DNA backbone between the Guanine (G) and Adenine (A) bases on each strand. The cut is as follows:

5'---C A G | A A T T C---T T A---3'

3'---G T C T T A A | G---A A T---5'

This results in two DNA fragments with single-stranded overhangs:

Fragment 1: 5'-C A G-3' and Fragment 2: 5'-A A T T C T T A-3'

Overhangs: The 3' end of the first fragment is 3'-G T C T T A A-5' and the 5' end of the second fragment is 5'-A A T T C-3' and 3'-G-5' overhang. The resulting fragments have single stranded ends: 5'-AATT-3'.

The cut produces ends like: 5'---G and AATTC---3'


(c) The ends generated are called 'sticky ends' or cohesive ends because the single-stranded overhangs (5'-AATT-3' in this case) are complementary to each other. This complementarity allows them to readily form hydrogen bonds with other DNA fragments that have been cut with the same restriction enzyme, causing them to 'stick' together. This property is crucial for creating recombinant DNA.
Quick Tip: A palindromic sequence in DNA is one where the 5' to 3' sequence on one strand is the same as the 5' to 3' sequence on the complementary strand. For EcoRI, the sequence is GAATTC. Reading the complement from left to right (in the 3'->5' direction) it is CTTAAG, but reading it in the 5'->3' direction (right to left) it is GAATTC.

The three trophic levels are:

Trophic Level 3 (T3 - Secondary Consumers): Small Fish

Trophic Level 2 (T2 - Primary Consumers): Zooplankton

Trophic Level 1 (T1 - Producers): Phytoplankton


The pyramid of biomass for this aquatic ecosystem is inverted.




Justification:

The pyramid of biomass in a sea or pond is inverted because the biomass of the producers (phytoplankton) is much less than the biomass of the next trophic level (zooplankton) at any given point in time.


This happens because phytoplankton are microscopic organisms with a very short life span and a high rate of metabolism and reproduction. They are consumed by zooplankton as rapidly as they are produced.


In contrast, the consumers (zooplankton and fish) have longer life spans and accumulate more biomass over time. Therefore, the standing crop (total mass of living organisms at a particular time) of the producers is smaller than that of the primary consumers.
Quick Tip: While pyramids of energy are always upright, pyramids of biomass and numbers can be inverted. The classic example of an inverted pyramid of biomass is the aquatic ecosystem. The classic example of an inverted/spindle-shaped pyramid of numbers is a single large tree supporting many insects.

The trophic levels based on the number of individuals are:

Trophic Level 4 (T4 - Tertiary Consumers): Big birds (few in number).

Trophic Level 3 (T3 - Secondary Consumers): Small birds (more in number than big birds).

Trophic Level 2 (T2 - Primary Consumers): Insects (a very large population).

Trophic Level 1 (T1 - Producer): A single big tree (one in number).


The pyramid of numbers for this ecosystem is spindle-shaped.



Description:

The base of the pyramid, representing the producer, is very narrow as it consists of only one large tree.


The next level, representing the primary consumers (herbivorous insects), is very broad, as thousands of insects feed on the single tree.


The subsequent trophic levels, representing the secondary consumers (small birds) and tertiary consumers (big birds), become progressively narrower as the number of individuals decreases at each higher level.


This results in a pyramid that is narrow at the bottom, wide in the middle, and narrow again at the top, giving it a spindle shape.
Quick Tip: A pyramid of numbers shows the total number of individual organisms at each trophic level. It doesn't account for the size of the organisms. That's why one large producer (like a tree) can support a huge number of smaller primary consumers (like insects), leading to an inverted or spindle-shaped pyramid.

(a) Calculation of male gametophytes:


A bilobed, dithecous anther has 4 microsporangia (pollen sacs).


Number of microspore mother cells (MMCs) per microsporangium = 200.


Total number of MMCs in the anther = Number of microsporangia \(\times\) MMCs per microsporangium.


Total MMCs = 4 \(\times\) 200 = 800 MMCs.


Each diploid MMC undergoes meiosis to produce 4 haploid microspores (pollen grains).


Each microspore develops into one male gametophyte.


Total male gametophytes produced = Total MMCs \(\times\) 4.


Total male gametophytes = 800 \(\times\) 4 = 3200.


(b) Composition of pollen grain layers:


Exine: It is the hard outer layer. It is made of a highly resistant organic material called sporopollenin, which protects the pollen from harsh conditions.


Intine: It is the thin, continuous inner wall. It is composed of cellulose and pectin.
Quick Tip: Remember the formula: Total male gametophytes = (No. of microsporangia) \(\times\) (No. of MMCs per microsporangium) \(\times\) 4. A typical anther is bilobed and dithecous, meaning it has 2 lobes and 4 microsporangia.

(a) Expanded Abbreviations:


(i) ZIFT: Zygote Intra Fallopian Transfer.


(ii) ICSI: Intra Cytoplasmic Sperm Injection.


(iii) IUT: Intra Uterine Transfer.


(iv) GIFT: Gamete Intra Fallopian Transfer.


(b) Reasons for the statutory ban on Amniocentesis:


Amniocentesis is a prenatal diagnostic technique used to detect genetic abnormalities in the foetus. However, it is banned primarily for two reasons:


1. Misuse for Sex Determination: The technique can reveal the sex of the foetus. This information has been widely misused by parents who prefer a male child, leading them to abort the foetus if it is female.


2. Female Foeticide: The misuse for sex determination directly leads to the illegal and unethical practice of female foeticide, which has dangerously skewed the sex ratio in many parts of the country.
Quick Tip: While there are medical risks associated with amniocentesis (like miscarriage), the primary reason for the legal ban is its social misuse for sex-selective abortions, not the medical risk itself.

Let the allele for the dominant phenotype be 'A' and for the recessive phenotype be 'a'.


The female is homozygous. Let's assume she is homozygous recessive, with the genotype 'aa'.


The male is heterozygous, with the genotype 'Aa'.


The cross is: aa (Female) \(\times\) Aa (Male).


Gametes produced by the female: all will be 'a'.


Gametes produced by the male: 50% will be 'A' and 50% will be 'a'.


The Punnett square for this cross is:


\begin{tabular{c|c|c|
\multicolumn{1{c{ & \multicolumn{1{c{A & \multicolumn{1{c{a

\cline{2-3
a & Aa & aa

\cline{2-3
\end{tabular


From the Punnett square, the resulting genotypes in the F1 generation are:


50% are heterozygous dominant (Aa).


50% are homozygous recessive (aa).


The distribution of the phenotypic feature is:


The progeny with genotype Aa will exhibit the dominant phenotype.


The progeny with genotype aa will exhibit the recessive phenotype.


Therefore, the phenotypic ratio is 1 (Dominant) : 1 (Recessive).
Quick Tip: This type of cross (a cross between a heterozygous organism and a homozygous recessive organism) is known as a test cross. It is used to determine the genotype of an organism showing a dominant phenotype.

The Hardy-Weinberg equilibrium states that allele frequencies in a population remain constant if evolutionary influences are absent.


Natural selection is a key evolutionary force that disrupts this equilibrium by favouring certain alleles over others.


This leads to a change in allele and genotype frequencies over generations.


Natural selection can affect the distribution of phenotypes in a population in three main ways:


1. Stabilizing Selection: In this, individuals with the mean (average) phenotype are favoured. It selects against extreme phenotypes. This reduces variation and the population curve becomes narrower and taller.


2. Directional Selection: In this, individuals with one extreme phenotype are favoured over other phenotypes. The peak of the population curve shifts in one direction.


3. Disruptive Selection: In this, individuals at both extremes of the phenotypic range are favoured over intermediate phenotypes. This leads to the formation of two peaks in the population curve.
Quick Tip: Remember that Hardy-Weinberg equilibrium is a theoretical baseline. The five factors that disrupt it are: Mutation, Gene Flow, Genetic Drift, Non-random Mating, and Natural Selection. Natural selection is the only one that leads to adaptive evolution.

(a) Need for Drug Tests:


Drug tests are conducted to prevent athletes from using performance-enhancing drugs.


This ensures fair competition by preventing any athlete from gaining an unfair advantage.


It also protects the health and well-being of the athletes, as many of these drugs have serious side effects.


(b) Drugs authorities look for:


Authorities typically test for banned substances which include:


- Anabolic Steroids

- Stimulants

- Narcotics

- Diuretics and other masking agents

- Peptide hormones (like Erythropoietin)

- Cannabinoids


(c) Two plant sources of such drugs:


1. Cannabis sativa: The source of cannabinoids like marijuana and hashish.


2. Papaver somniferum (Poppy plant): The source of opioids like morphine, which is chemically modified to produce heroin (smack).
Quick Tip: The misuse of drugs by athletes to enhance performance is called doping. The World Anti-Doping Agency (WADA) and National Anti-Doping Agency (NADA) are the regulatory bodies that oversee these tests.

(a) Differentiation between natural insulin and rDNA insulin:



(b) Reason for disuse of animal insulin:


Insulin extracted from animal sources (like pancreas of slaughtered cattle and pigs) is not widely used anymore because its amino acid sequence is slightly different from human insulin.


This difference can trigger an immune response in some patients, leading to allergies or other types of reactions.


Recombinant human insulin is structurally identical to the insulin produced by the human pancreas, thus avoiding these immunological problems.
Quick Tip: The key difference in production is "proinsulin". The human body makes a single proinsulin molecule and then cuts out the C-peptide. rDNA technology skips this step by making the final A and B chains directly and then linking them.

(a) Graph for a population at carrying capacity (K): This is a Logistic Growth Curve (S-shaped). The graph levels off at the top, indicating that the population size has become stable at the carrying capacity.




(b) More Realistic Growth Curve:


The logistic growth curve (S-shaped) is considered more realistic for most animal populations.


Reason: In reality, resources such as food, water, and space are finite. As a population grows, these resources become limited, competition increases, and the growth rate slows down and eventually stops, leading to the S-shaped curve. The exponential curve assumes unlimited resources, which is not a sustainable condition in any natural habitat.


(c) Growth curve with non-limiting resources: This is an Exponential Growth Curve (J-shaped). The equation for this curve is:

dN/dt = rN

Where:

dN/dt = Rate of change in population size

r = Intrinsic rate of natural increase

N = Population size
Quick Tip: Remember the shapes and their conditions: J-shape = Exponential = Unlimited resources (unrealistic for long term). S-shape = Logistic = Limited resources = Carrying Capacity (K) (more realistic).

(a) The two principal types of specialised cells (lymphocytes) that carry out primary and secondary immune responses are B-lymphocytes and T-lymphocytes.


(b) Antibody-mediated immunity is called humoral immune response because the antibodies are found circulating in the body fluids, or "humors", like blood plasma and lymph.


Answer to sub-part (c):


(i) The characteristic of our immune system responsible for graft rejection is its ability to differentiate between 'self' cells (body's own cells) and 'non-self' cells (cells from the foreign graft).


(ii) The type of immune response involved is the cell-mediated immune response (CMI). The specific cells responsible for this are the T-lymphocytes (specifically cytotoxic T-cells).


OR


Answer to sub-part (d):


\begin{tabularx{\linewidth{|l|X|X|
\hline
Feature & Active Immunity & Passive Immunity

\hline
Source & The body produces its own antibodies in response to an antigen. & Ready-made antibodies are transferred from an external source.

\hline
Speed & It is slow and takes time to develop a full response. & It is fast and provides immediate relief.

\hline
Duration & It is long-lasting. & It is temporary and lasts for only a few weeks or months.

\hline
Memory & Immunological memory is present. & No immunological memory is formed.

\hline
Example & Immunity from vaccination or natural infection. & Immunity from anti-venom or mother to foetus.

\hline
\end{tabularx
Quick Tip: Remember: "Humoral" relates to body fluids (blood, lymph), where antibodies roam. "Cell-mediated" relates to direct cell-to-cell combat, led by T-cells, which is crucial for rejecting foreign tissues like organ grafts.

(a) The main enzyme involved in transcription is DNA-dependent RNA polymerase.


(b) Transcription occurs on the DNA strand with 3' \(\rightarrow\) 5' polarity, which serves as the template.

Therefore, the template strand is A (3' \(\rightarrow\) 5').

The strand with 5' \(\rightarrow\) 3' polarity is the coding strand, which is B (5' \(\rightarrow\) 3').


Answer to sub-part (c):


(C) is the Promoter.

Significance: It serves as the binding site for RNA polymerase and defines the start site of transcription.


(D) is the Terminator.

Significance: It is the sequence that signals the termination of the transcription process, causing the RNA polymerase to detach from the DNA.


OR


Answer to sub-part (d):


The promoter (C) and the terminator (D) flank the structural gene in a transcription unit.


(C) The Promoter is located towards the 5'-end (upstream) of the structural gene.


(D) The Terminator is located towards the 3'-end (downstream) of the structural gene.
Quick Tip: Remember that the promoter and terminator define which strand is the template and which is the coding strand. The promoter is always at the 5'-end of the coding strand. The enzyme reads the template strand which has the opposite polarity (3' to 5').

(A) (i) Diagram of a seminiferous tubule:



The labels for the required parts are:

(a) The cell that undergoes spermiogenesis (transformation into a spermatozoon) is the Spermatid.


(b) The cell that nourishes the male gametes is the Sertoli cell.


(c) The cell that undergoes meiosis I is the Primary spermatocyte. The cell that undergoes meiosis II is the Secondary spermatocyte.


(ii) Seminal Plasma and Semen:


Seminal plasma is the fluid part of semen, contributed by the male accessory glands (seminal vesicles, prostate gland, and bulbourethral glands).


Two constituents of seminal plasma are fructose (energy source for sperm) and calcium.


The difference between seminal plasma and semen is that semen is the combination of seminal plasma and the spermatozoa (sperm).


Semen = Seminal Plasma + Sperm.
Quick Tip: Remember the roles of the two key cells inside the seminiferous tubule: Sertoli cells 'support' spermatogenesis (nourishment), while Leydig cells (interstitial cells) are 'outside' the tubules and secrete testosterone under the influence of LH.

(B) (i) Event inducing completion of meiosis:


The secondary oocyte is arrested in the metaphase II stage of meiosis II.


The event that induces the completion of this division is fertilization, specifically the entry of a sperm into the cytoplasm of the secondary oocyte.


This leads to the formation of a haploid mature ovum and a second polar body.


(ii) Journey of the Zygote:


After fertilization in the ampullary region of the fallopian tube, the diploid zygote is formed.


The zygote begins to move through the isthmus of the fallopian tube towards the uterus.


During this journey, it undergoes a series of mitotic divisions called cleavage, forming 2, 4, 8, then 16 daughter cells called blastomeres.


The embryo with 8 to 16 blastomeres is called a morula.


The morula continues to divide and transforms into a blastocyst as it reaches the uterus.


The blastocyst then implants into the inner wall of the uterus, the endometrium. This process is called implantation.
Quick Tip: A key difference between spermatogenesis and oogenesis is the timing of meiotic completion. Spermatogenesis completes fully, producing sperm. Oogenesis pauses at Prophase I (in the foetus) and Metaphase II (after ovulation), only completing the entire process if fertilization occurs.

(A) (i) Making a bacterial cell 'competent':


Since DNA is a hydrophilic molecule, it cannot easily pass through the cell membrane.


To make bacteria competent to take up DNA, they are treated with a specific concentration of a divalent cation, such as calcium (\(Ca^{2+}\)).


This increases the efficiency with which DNA enters the bacterium through pores in its cell wall.


The cells are then incubated with the recombinant DNA on ice, followed by a brief heat shock (placing them at 42°C and then back on ice).


This heat shock enables the recombinant DNA to enter the competent bacterial cell.


(ii) Steps of Polymerase Chain Reaction (PCR):


PCR is used to amplify a gene of interest into billions of copies. It involves three main steps repeated in a cycle:


1. Denaturation: The double-stranded DNA is heated to a high temperature (around 94-96°C). This breaks the hydrogen bonds between the two strands, separating them.


2. Annealing: The reaction mixture is cooled (to around 50-65°C). This allows the two small, chemically synthesized DNA primers to bind (anneal) to their complementary sequences on the separated DNA strands.


3. Extension: The temperature is raised again (usually to 72°C), the optimal temperature for the thermostable DNA polymerase (like Taq polymerase). The polymerase adds nucleotides to the primers, extending them to synthesize new complementary strands of DNA.


These three steps constitute one cycle. The cycle is repeated 20-30 times to produce a huge number of copies of the target DNA sequence.
Quick Tip: A simple mnemonic for PCR steps is D-A-E: Denature, Anneal, Extend. Remember that the key enzyme, Taq polymerase, is from a thermophilic bacterium (Thermus aquaticus), which allows it to withstand the high temperatures of the denaturation step.

(B) (i) Transgenic Animals:


Transgenic animals are animals that have had their DNA (genome) manipulated to possess and express an extra, foreign gene from another species.


(ii) Reasons for producing transgenic animals (any four):


1. Study of Normal Physiology and Development: Transgenic animals are designed to study how genes are regulated and how they affect the normal functions of the body and its development. For example, studying complex factors like insulin-like growth factor.


2. Study of Disease: Many transgenic animals are designed to increase our understanding of how genes contribute to the development of diseases. They serve as models for human diseases like cancer, cystic fibrosis, and Alzheimer's, allowing for the investigation of new treatments.


3. Biological Products: Transgenic animals can be created to produce useful biological products. For example, human protein (\(\alpha\)-1-antitrypsin) can be produced in sheep milk to treat emphysema. The first transgenic cow, Rosie, produced human protein-enriched milk.


4. Vaccine Safety: Transgenic mice are being developed for use in testing the safety of vaccines before they are used on humans. This can provide a more reliable alternative to using monkeys for safety testing.
Quick Tip: When thinking about the benefits of transgenic animals, categorize them: 1. As models (for disease/physiology), 2. As bioreactors (for producing useful proteins), and 3. As testing subjects (for vaccines/chemicals). This helps in recalling the different applications.

(A) (i) Reasons for high tropical biodiversity:


1. Long Evolutionary Time: Tropical latitudes have remained relatively undisturbed for millions of years, unlike temperate regions which were subjected to frequent glaciations in the past. This long, stable period has allowed for more time for species diversification.


2. Constant and Predictable Environment: Tropical environments are less seasonal and more constant than temperate ones. This promotes niche specialization and leads to a greater species diversity.


3. High Solar Energy and Productivity: The tropics receive more solar energy, which contributes to higher productivity. This in turn can support a greater diversity of species at different trophic levels.


(ii) Species-Area Relationship:




Naturalist: The German naturalist and geographer Alexander von Humboldt studied this relationship.


Observation: He observed that within a region, species richness increases with increasing explored area, but only up to a certain limit. The relationship is described by the equation S = c\(A^z\), where S is species richness, A is area, c is the Y-intercept, and z is the slope of the line (regression coefficient).
Quick Tip: Remember the three key factors for high tropical biodiversity: Time (stable for longer), Climate (less seasonal), and Energy (more sunlight = more food). For the species-area relationship, the value of 'z' (the slope) is steeper for larger areas like entire continents, indicating that species richness increases more rapidly with area on a larger scale.

(B) (i) Three major causes of biodiversity loss (The Evil Quartet):


1. Habitat Loss and Fragmentation: This is the most important cause. The destruction of natural habitats (e.g., deforestation for agriculture, urbanization) drives many species to extinction. When large habitats are broken up into small, isolated fragments, it affects species with large territories and migratory habits.


2. Over-exploitation: Humans have always depended on nature for food and shelter, but when 'need' turns to 'greed', it leads to over-exploitation. Over-harvesting of many species (e.g., Steller's sea cow, passenger pigeon) has led to their extinction.


3. Alien Species Invasions: When new species are introduced into a habitat, either intentionally or unintentionally, they may become invasive and cause the decline or extinction of indigenous species. For example, the introduction of the Nile perch into Lake Victoria led to the extinction of more than 200 species of cichlid fish.


(ii) 'Ex situ' Conservation:


'Ex situ' (off-site) conservation is an approach where threatened animals and plants are taken out of their natural habitat and placed in special settings where they can be protected and given special care.


Two examples of ex situ conservation are:


1. Zoological Parks (Zoos) and Botanical Gardens: These are dedicated places where threatened species are kept in protected environments, with efforts made to breed them in captivity and increase their numbers.


2. Cryopreservation: This is a technique of preserving gametes of threatened species in a viable and fertile condition for long periods at very low temperatures (-196°C in liquid nitrogen). This can be used for breeding programs in the future.
Quick Tip: To remember the approaches: In situ = "in the original place" (e.g., National Parks, Sanctuaries). Ex situ = "off the original site" (e.g., Zoos, Seed Banks). Ex situ is often the last resort when a species is critically endangered and cannot survive in the wild.