Section B
[1 Marks]
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(A) Pericarp, Endosperm, Scutellum
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(B) Pericarp, Endosperm, Coleorhiza
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(C) Scutellum, Pericarp, Coleorhiza
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(D) Coleorhiza, Scutellum, Pericarp
Explanation: The correct option is 'Pericarp, Endosperm, Scutellum'. In a monocot seed, the pericarp is the fruit wall surrounding the seed, the endosperm serves as the food storage for the embryo, and the scutellum is a specialized cotyledon that absorbs nutrients from the endosperm during seed germination.
[1 Marks]
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(A) 44
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(B) 23
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(C) 22
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(D) 46
Explanation: A human secondary spermatocyte is haploid and contains 23 chromosomes in total, which include 22 autosomes and 1 sex chromosome. Therefore, the correct number of autosomes present in a secondary spermatocyte is 22.
[1 Marks]
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(A) Father - Iᴮ Iᴮ , Mother -Iᴬ Iᴮ , Child - Iᴬ Iᴬ
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(B) Father - Iᴮ i , Mother - Iᴬ Iᴮ , Child - Iᴬ i
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(C) Father - Iᴬ Iᴮ, Mother - Iᴬ i , Child - Iᴬ Iᴬ
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(D) Father - Iᴬ i, Mother - Iᴮ i, Child - Iᴬ i
Explanation: The correct option is: Father - Iᴮ i , Mother - Iᴬ Iᴮ , Child - Iᴬ i. The father with blood group B can have genotype Iᴮ i, while the mother with blood group AB has the genotype Iᴬ Iᴮ. The child has blood group A, which can result from inheriting Iᴬ from the mother and i from the father.
[1 Marks]
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(A) affected individuals
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(B) unrelated mating
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(C) Non-identical twins
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(D) mating between relatives (consanguineous mating)
Explanation: A pedigree chart represents mating between relatives (consanguineous mating), as it is used to analyze the inheritance of traits over generations within a family. This allows for tracing the transmission of genetic disorders and other traits, which often occurs when relatives mate.
[1 Marks]
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(A) (i), (iii), (ii), (iv)
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(B) (iii), (i), (ii), (iv)
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(C) (iv), (ii), (i), (iii)
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(D) (i), (ii), (iii), (iv)
Explanation: The correct evolutionary order is (iii), (i), (ii), (iv). Zosterophyllum is considered more primitive and one of the earliest vascular plants. Ferns are pteridophytes that evolved next, followed by gymnosperms like Ginkgo, and finally Gnetales, which are more advanced gymnosperms. This progression aligns with the understanding of plant evolution.
[1 Marks]
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(A) phosphate group and OH of 3’C of a nucleoside.
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(B) phosphate group and H of 5’C of a nucleoside.
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(C) phosphate group and OH of 5’C of a nucleoside.
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(D) phosphate group and H of 3’C of a nucleoside.
Explanation: The correct answer is 'phosphate group and OH of 5’C of a nucleoside.' This is because, as mentioned in the context, the phosphate group forms a linkage with the hydroxyl group (OH) of the 5' carbon in a nucleoside, resulting in the formation of a nucleotide.
[1 Marks]
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(A) 8,8
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(B) 7,7
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(C) 7,8
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(D) 8,7
Explanation: The correct option is '8,7'. In eukaryotic transcription, hnRNA (heterogeneous nuclear RNA) contains both introns and exons. During the splicing process, introns are removed and exons are joined to form a functional mRNA. The context implies that there are distinct numbers of introns and exons in hnRNA, and it indicates the presence of at least several introns and exons due to the complexity mentioned. Therefore, the answer '8,7' likely denotes this ratio.
[1 Marks]
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(A) (a) (ii), (b) (i), (c) (iii)
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(B) (a) (ii), (b) (iii), (c) (i)
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(C) (a) (i), (b) (iii), (c) (ii)
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(D) (a) (i), (b) (ii), (c) (iii)
Explanation: The correct matching is
(a) RNA polymerase I- (i) 18s rRna
(b) RNA polymerase II- (iii) hnRNA
(c) RNA polymerase III- (ii) SnRNAs
[1 Marks]
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(A) Aspergillus niger
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(B) Saccharomyces cerevisiae
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(C) Clostridium butylicum
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(D) Lactobacillus sp.
Explanation: Aspergillus niger is the correct answer because it is specifically mentioned in the context as a fungus that produces citric acid, making it the suitable microbe for commercial production.
[1 Marks]
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(A) 0 : 1 : 3
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(B) 0 : 1 : 8
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(C) 1 : 1 : 0
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(D) 1 : 4 : 0
Explanation: After 80 minutes, which corresponds to three generations (since E. coli divides every 20 minutes), the generated DNA will consist of hybrids and light DNA. By the III generation, the expected ratio of the DNA types will be 0 : 1 : 3. This is because, after one generation in a 14N medium, half will be hybrid (N¹⁵/N¹⁴) and the other half light (N¹⁴/N¹⁴), and after the second generation, the light DNA will increase significantly, leading to the final ratio of 0 : 1 : 3.
[1 Marks]
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(A) PCR is used for isolation and separation of gene of interest.
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(B) The polymerase enzyme joins the gene of interest and the vector DNA.
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(C) Gel electrophoresis is used for amplification of a DNA segment.
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(D) Plasmid DNA acts as vector to transfer the piece of DNA attached to it.
Explanation: The correct statement is 'Plasmid DNA acts as vector to transfer the piece of DNA attached to it.' This is accurate because plasmids are commonly used as vectors in recombinant DNA technology to carry and transfer the gene of interest into host cells for cloning and expression, as indicated in the context provided.
[1 Marks]
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(A) decrease in antigens
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(B) decrease in antibodies
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(C) increase in antibodies
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(D) increase in antigens
Explanation: A decrease in T-Lymphocytes, particularly helper T-lymphocytes, leads to a weakened immune response. This impairment decreases the production of antibodies since T-Lymphocytes play a crucial role in stimulating B-cells to produce antibodies that combat antigens. Therefore, the correct answer is 'decrease in antibodies.'
[1 Marks]
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(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
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(B) (A) is true, but (R) is false.
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(C) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
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(D) (A) is false, but (R) is true.
Explanation: Both (A) and (R) are true and (R) is the correct explanation of (A) because the corpus luteum does secrete progesterone, which plays a critical role in maintaining the endometrium for implantation and support of pregnancy, as stated in the context.
[1 Marks]
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(A) (A) is false, but (R) is true.
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(B) (A) is true, but (R) is false.
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(C) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
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(D) Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation: The statement "The number of white-winged moths decreased after industrialisation in England" is true, but the reason "Effects of industrialisation were more marked in rural areas of England" is false. Therefore, the correct answer is (C): (A) is true, but (R) is false.
Therefore, while the number of white-winged moths decreased due to industrialization, the reason provided (that the effects were more pronounced in rural areas) is incorrect. The change was driven by the environmental impact of industrialization in urban centers. Several educational resources note this.
[1 Marks]
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(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
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(B) (A) is false, but (R) is true.
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(C) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
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(D) (A) is true, but (R) is false.
Explanation: Both (A) and (R) are true, and (R) is the correct explanation of (A). Streptococcus pneumoniae and Haemophilus influenzae are indeed known to cause infectious diseases, and the reasoning correctly describes the transmission method through inhalation of aerosols from infected individuals, as indicated in the context.
[1 Marks]
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(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
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(B) (A) is true, but (R) is false.
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(C) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
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(D) (A) is false, but (R) is true.
Explanation: The assertion is true, but the reason is false. Restriction endonucleases do indeed recognize and cut palindromic sequences in DNA. However, a palindromic sequence doesn't inherently have two unique recognition sites for PstI and PvuI. These are specific restriction enzymes with their own unique palindromic recognition sequences.
Section C
[2 Marks]
Answer: Two crucial changes that seeds undergo while reaching maturity are dehydration and the development of a hard seed coat. Dehydration reduces the water content of the seed, allowing it to enter a dormant state, which is essential for longevity. The hard seed coat provides protection to the embryo from external elements and environmental stressors. The oldest viable seed excavated from the Arctic Tundra is the lupine, Lupinus arcticus, which was found to have germinated after an astonishing 10,000 years of dormancy.
[2 Marks]
Answer: Pea flowers can produce assured seed sets primarily due to their apomictic nature, which allows seeds to develop without fertilization. This mechanism ensures genetic consistency and stability in offspring, as the seeds are produced from the maternal tissues. For the second part, regarding polyembryony, embryo ‘P’, which develops from a synergid, is haploid (1n), while embryo ‘Q’, derived from the nucellus, is diploid (2n) since it comes from the ovule tissues. Thus, the ploidy levels are 1n for ‘P’ and 2n for ‘Q’.
[2 Marks]
Answer: The trait described is likely an autosomal recessive trait, which can skip generations. In pedigree analysis, this means that both parents do not show the trait but can be carriers, allowing them to pass it on to their offspring. As a result, both male and female children can express the trait, typical for autosomal recessive inheritance patterns. An example of such a trait in humans is Sickle-cell anaemia, which arises from a mutation affecting hemoglobin.
[2 Marks]
Answer: A medical doctor may recommend the injection of pre-formed antibodies in situations such as tetanus and snake bites. In the case of tetanus, when a person is infected with the tetanus toxin, immediate injection of antitoxin is necessary to provide rapid immunity against the effects of the toxin. Similarly, for snakebites, antivenom containing pre-formed antibodies against snake venom is administered to neutralize the venom quickly, preventing severe damage.
[2 Marks]
Answer: The symptoms of malaria do not appear immediately because the sporozoites introduced by the mosquito must first travel to the liver, where they mature into merozoites. This incubation period typically lasts about 7 to 14 days, allowing the parasites to multiply significantly in number before re-entering the bloodstream. It's only when they infect red blood cells and cause immune reactions that the characteristic symptoms, such as fever and chills, emerge.
[2 Marks]
Answer: The restriction enzyme that can recognize the given palindromic DNA sequence 5' —— GAATTC —— 3' is EcoRI. After the restriction enzyme cuts the palindrome, the resultant sequence will have sticky ends: 5' —— G AATTC —— 3' and 3' —— CTTAA G —— 5'. The ends generated after digestion are called ‘Sticky Ends’ because they have single-stranded overhangs that can easily form hydrogen bonds with complementary sequences, allowing for efficient joining of DNA fragments.
[2 Marks]
Answer: The pyramid described could be any of the ecological pyramids: pyramid of numbers, pyramid of biomass, or pyramid of energy. Two identifying features are that the base of the pyramid represents the producers or the first trophic level, while the apex represents the top-level consumers. The pyramid's shape illustrates the decreasing number, biomass, or energy as one moves from the base to the apex.
[2 Marks]
Answer: In an ideal pyramid of energy with 10,00,000 Joules of sunlight, primary producers (first trophic level) convert 1% into net primary productivity (NPP), yielding 10,000 Joules. The energy available for herbivores (second level) is 10% of the first level, totaling 1,000 Joules. For carnivores (third level), it's 10% of 1,000 Joules, equating to 100 Joules. The fourth level, top carnivores, receives 10% of 100 Joules, resulting in 10 Joules.
Section D
[3 Marks]
Answer: To determine the number of male gametophytes produced by the bilobed dithecous anther, we first note that each microspore mother cell (or microsporocyte) undergoes meiosis to form four microspores. Given there are 200 microspore mother cells in each microsporangium and two microsporangia per lobe, the calculation is as follows: 200 microspore mother cells × 4 microspores each = 800 microspores per microsporangium. Since there are four microsporangia in total, this results in 800 microspores × 4 = 3200 male gametophytes. As for the composition of pollen grains, the intine is mainly composed of cellulose and pectin, while the exine is made up of sporopollenin, a tough and resistant substance that protects the pollen grain's contents.
[3 Marks]
Answer: Copper releasing IUDs are effective contraceptives for a couple of reasons. Firstly, they release copper ions which create an environment in the uterus that is hostile to sperm, thereby inhibiting sperm motility and function, which reduces the likelihood of fertilization. Secondly, these IUDs induce an inflammatory response that increases phagocytosis of sperm within the uterus, effectively preventing sperm from reaching an egg. Regarding oral contraceptive pills, they prevent pregnancy primarily by inhibiting ovulation. These pills contain hormones that regulate the menstrual cycle and prevent the ovaries from releasing eggs. Additionally, oral contraceptives thicken cervical mucus, making it more challenging for sperm to enter the uterus, and they alter the uterine lining to prevent implantation if fertilization occurs.
[3 Marks]
Answer: To determine the distribution of phenotypes in the first filial generation (F1) following a cross between a homozygous female (AA) and a heterozygous male (Aa), we can use a Punnett Square. The homozygous female (AA) can only produce gametes with the dominant allele (A), while the heterozygous male (Aa) can produce gametes with either the dominant (A) or the recessive (a) allele. Setting this up in the Punnett square, the top will have A and A, and the side will have A and a. This gives us four squares: AA, AA, Aa, Aa. Thus, the phenotypic ratio of the F1 generation will show that 100% will express the dominant phenotype. Therefore, the expected genotypic ratio is 2 homozygous dominant (AA) to 2 heterozygous (Aa), but since both express the dominant trait, the phenotypic ratio is 100% dominant features.
[3 Marks]
Answer: Natural selection influences Hardy-Weinberg equilibrium by causing changes in allele frequencies, indicating evolution. According to the Hardy-Weinberg principle, allele frequencies remain stable in a population under ideal conditions. However, natural selection introduces three forms of selection: stabilizing, directional, and disruptive. In stabilizing selection, individuals with average traits are favored, leading to a more consistent allele frequency. Directional selection shifts the frequency towards one extreme, while disruptive selection favors individuals at both extremes, potentially leading to speciation. Graphically, these selections are represented by normal distribution curves, where the height reflects allele frequency over generations. The disturbance in equilibrium provoked by natural selection implies that evolutionary changes are continuous, correlating with selective pressures in the environment.
[3 Marks]
Answer: Conducting drug tests for sportspersons is essential to maintain fairness and integrity in sports. Performance-enhancing drugs like anabolic steroids, narcotic analgesics, diuretics, and hormones can give athletes an unfair advantage, compromising the spirit of competition. Authorities typically screen for drugs that enhance muscle mass, strength, and endurance. Commonly tested substances include steroids, amphetamines, and cannabinoids. These drugs can be derived from plants; for instance, anabolic steroids are often synthesized from the wild yam (Dioscorea villosa) and hormones like testosterone can be obtained from the sarsaparilla plant (Smilax spp.).
[3 Marks]
Answer: The insulin synthesized in the human body is a natural hormone composed of two polypeptide chains (A and B) linked by disulfide bonds. Eli Lilly's recombinant DNA insulin, however, is produced in E. coli bacteria after inserting human insulin gene sequences into plasmids. Although both forms are chemically identical, the method of production and possible impurities differ, as bacterial systems can introduce contaminants not found in human insulin. In terms of effectiveness, recombinant insulin is more beneficial for patients since it is genetically identical to human insulin, reducing the risk of immunogenic reactions. Regarding animal-sourced insulin, its use is declining due to issues of compatibility and purity; animal insulin can provoke allergic reactions or immune responses in humans. Additionally, with the advancement of biotechnology, recombinant insulin is more accessible and ethically preferable over animal-derived options.
[3 Marks]
Answer: For part (a), the graph of a population reaching its carrying capacity is represented as a logistic growth curve. Initially, the population grows exponentially, but as it approaches the carrying capacity (K), growth slows down and eventually stabilizes at K. In part (b), the logistic growth curve is more realistic for most populations due to environmental constraints and limited resources, which result in competition and survival of the fittest. In part (c), the growth curve for unlimited resources is an exponential curve, represented by the equation dN/dt = rN, where N is the population size and r is the intrinsic growth rate.
Section E
[5 Marks]
Answer: Megasporogenesis is the process by which megaspores are formed from the megaspore mother cell (MMC) in an angiosperm ovule. The MMC, which is located in the micropylar region of the nucellus, is a large cell characterized by dense cytoplasm and a prominent nucleus. During megasporogenesis, the MMC undergoes meiosis to produce four haploid megaspores; however, typically only one megaspore survives while the other three degenerate, marking the beginning of monosporic development. The functional megaspore then undergoes three rounds of mitotic division. The first division yields two nuclei, leading to the formation of a 2-nucleate embryo sac. This is followed by further mitotic divisions that create a 4-nucleate sac and subsequently an 8-nucleate embryo sac, with the nuclei positioned at opposite ends. The mature embryo sac contains seven cells, including one egg cell, two synergids, three antipodal cells, and one central cell consisting of two polar nuclei, all responsible for fertilization and subsequent development. A diagram of the mature embryo sac illustrates these seven cells organized within the structure of the ovule, demonstrating key aspects of female gametophyte formation in angiosperms.
[5 Marks]
Answer: The menstrual cycle is a complex process divided into four phases: the menstrual phase, follicular phase, ovulation, and luteal phase. The menstrual phase lasts about 3 to 5 days, characterized by the shedding of the uterine lining. Estrogen peaks during the follicular phase just before ovulation to stimulate follicle maturation, while progesterone peaks in the luteal phase post-ovulation to maintain the endometrium for potential implantation. The surge of Luteinizing Hormone (LH) is significant because it triggers ovulation, signaling that an ovum is ready for fertilization, marking the transition between the follicular and luteal phases.
[5 Marks]
Answer: Bacterial cells are made 'competent' to take up recombinant DNA through a process known as transformation. This involves treating the cells with a divalent cation, such as calcium chloride, which facilitates the formation of pores in the bacterial cell wall. The increased permeability allows the cells to absorb extracellular DNA, including plasmids containing the desired recombinant DNA sequences. Once competent, these bacteria are exposed to the plasmids, and some cells take up the DNA, leading to the incorporation of the new genetic material. \n\nThe amplification of the gene of interest using the Polymerase Chain Reaction (PCR) technique involves several key steps. First, DNA containing the target gene is denatured by heating, resulting in two single strands. Then, specific primers that flank the target gene are annealed to the DNA strands at a lower temperature. This is followed by the extension phase, where a heat-stable DNA polymerase synthesizes new DNA strands by elongating from the primers. Repeating these cycles (denaturation, annealing, and extension) generates millions of copies of the target gene, significantly amplifying the DNA for further analysis or application.
[5 Marks]
Answer: Transgenic animals are organisms whose DNA has been genetically modified to include a foreign gene, allowing them to express traits that are not naturally theirs. These modifications have various applications that can greatly benefit humanity. One key reason for producing transgenic animals is to enhance our understanding of gene regulation and its impact on normal physiological processes. Additionally, transgenic animals serve as vital models for studying human diseases such as cancer and Alzheimer's, leading to the development of new therapies. They also facilitate the creation of biological products like medicines, which can be essential yet costly to produce otherwise. Lastly, transgenic animals can contribute to agricultural advances by improving livestock traits such as growth rates and disease resistance.
[5 Marks]
Answer: The tropics exhibit the greatest levels of species diversity due to several key factors. Firstly, they have experienced more evolutionary time compared to temperate regions, which have been subjected to glaciation events. This prolonged stability allows for the accumulation of species. Secondly, the relatively constant environmental conditions in the tropics promote niche specialization, allowing various species to thrive in diverse ecological roles. Lastly, the abundance of solar energy in these regions supports high productivity, leading to greater biomass and, consequently, more complex food webs that can support a more diverse array of species. Regarding species-area relationships, the notable German naturalist Alexander von Humboldt documented that species richness tends to increase with the area sampled up to a certain limit. This relationship demonstrates how larger habitats can support a greater number of species, influencing biodiversity patterns globally.
[5 Marks]
Answer: The accelerated loss of biodiversity is primarily due to human activities, leading to an alarming rate of species extinctions. One major cause is habitat loss and fragmentation, primarily due to urbanization and deforestation, which disrupts ecosystems and displaces wildlife. Secondly, over-exploitation, such as overfishing and illegal poaching, severely depletes species populations. Lastly, biological invasions, where non-native species disrupt local ecosystems, threaten native biodiversity. Regarding the 'Ex situ' conservation approach, it involves preserving species outside their natural habitats. Two examples include botanical gardens, which conserve plant species, and seed banks, which store genetic material to protect against extinction. These methods help maintain biodiversity and can aid in restoring ecosystems.
