Explanation:
Choice A - Excretion is the process wherein waste and excess water in the organism are removed. Paramecia expel waste products such as carbon dioxide through the process of diffusion.

Choice B - Homeostasis pertains to the ability of living things to maintain a stable internal environment. Paramecia maintain a stable internal condition by expelling excess water using the contractile vacuole.

Choice C - Reproduction is the biological process by which organisms produce offspring that are biologically similar to themselves. Paramecia can reproduce sexually through meiosis and asexually through mitosis.

Choice D - Metabolism refers to all the chemical reactions in organisms such as anabolic and catabolic reactions. Paramecia produce enzymes that catalyse different catabolic and anabolic reactions in the cell.

Explanation:
Choice D is correct because homologous structures refer to structures inherited from a common ancestor which are similar but often used for different functions.

Choice A is incorrect, as ancestral structures refer to structures present in an ancestor species, not structures found in present day species. Choice B is incorrect as convergent structures are not a term used in biology. Convergent evolution, on the other hand, leads to analogous structures. Choice C is incorrect as analogous structures refer to structures used for the same function, which are significantly different when closely inspected. Analogous structures also have different evolutionary origins.

Explanation:
Statement II is correct as prokaryotic cells do not have a Golgi apparatus and statement III is correct as prokaryotic cells do not have membrane bound organelles. Choices A, C and D are incorrect.

Explanation:
Louis Pasteur's experiments where he boiled nutrient broth in swan-necked flasks disproved spontaneous generation. Choice B is correct.

Miller and Urey's (Choice A) experiments proved that organic compounds could be produced in Earth’s early atmosphere. Rosalind Franklin’s (Choice C) work was instrumental in the discovery of the structure of DNA. Meselson and Stahl (Choice D) proved that DNA replication is semi-conservative.

Explanation:
The chromosomes of eukaryotes are linear, and the DNA is coiled around histone proteins (Choice C). Choice A is incorrect because eukaryotic DNA molecules are not circular. Choice B is incorrect because eukaryotic cells do not have looped DNA in the nucleus. Choice D is incorrect because the linear DNA molecules in eukaryotic cells are coiled around histone proteins.

Explanation:
Water's thermal properties (Choice A) include a high heat of vaporisation which results in the absorption of a large amount of heat by water before changing its physical state. This means that changing the temperature of the water is slow, and this helps to maintain the internal temperature of organisms between set limits.
Water has cohesive properties (Choice B) which means it sticks to other water molecules.
Water has adhesive properties (Choice C) which means it adheres to charged surfaces.
Water is an excellent solvent (Choice D) which means it dissolves a wide array of polar and ionic substances.

Explanation:
Choice B is not true. Not all genes are expressed in each cell. Only the necessary genes are expressed based on the specialised function of the cell. All other statements are true.

Explanation:
Gravity is responsible for holding water on Earth and is a fundamental factor in water retention within the planet's atmosphere and surface. Choice D is, therefore, correct.

Choice A is incorrect, as although rainfall is a part of the water cycle and is responsible for distributing water, it does not directly contribute to the retention of water on Earth.

Choice B is incorrect, as the moon's gravitational force regulates the tides of large bodies of water but does not directly contribute to water retention on Earth. Choice C is incorrect as water is lost from bodies of water on the surface of the Earth and evaporates into the atmosphere, which doesn't directly address water retention.

Explanation:
Evolution is the change in the characteristics of a population over time; these characteristics can be passed from one generation to another (Choice A). Choice B is incorrect as it describes when organisms are reproductively isolated, which is a mechanism that may lead to evolution in a population. Choice C is incorrect because it refers to learned behaviours which are learned within an organism’s lifetime and are not genetically inherited. Choice D is incorrect as it refers to selective breeding.

Explanation:
Hydrophilic (Choice C) substances dissolve in water through the formation of hydrogen bonds with water.
Non-polar (Choice D) substances are not attracted to water molecules and do not form bonds with them. These substances are hydrophobic (Choice B), making them insoluble (Choice A) in water.

Explanation:
X is pointing to the nucleus and Y is pointing to a mitochondrion. Choice A is incorrect as neither organelle has a single membrane. Choice B is incorrect as only mitochondria contain cristae, the nucleus does not. Choice D is incorrect as neither the nucleus or mitochondria contain starch grains.

Explanation:
Vesicles are formed from a phospholipid bilayer. This phospholipid bilayer forms spontaneously because the phosphate heads are hydrophilic (‘water loving’), and the fatty acid tails are hydrophobic (‘water hating’). Therefore Choice B is correct.

The membrane around the vesicle is not a single layer (Choices A and C) of phospholipids due to the hydrophilic phosphate heads naturally pointing outwards and inwards into the aqueous environments, and the hydrophobic fatty acid tails aligning in the centre to minimise their exposure to water. Vesicles are not induced to form in cells (Choices C and D) as the amphipathic nature of the phospholipids means that the vesicles form through self-assembly.

Explanation:
Uracil and thymine are nitrogenous bases found in two types of nucleic acids: DNA and RNA.

Uracil is found in RNA, whereas thymine is found in DNA. (Choice B)

Uracil pairs with adenine in RNA and thymine pairs with adenine in DNA (Choice D).

In both DNA and RNA, these nitrogenous bases are joined to a five-carbon sugar, also called a pentose sugar which has a ring structure, so Choice A is incorrect.

Both uracil and thymine are pyrimidines which are single-ring structures (Choice C). Adenine and guanine are purines which consist of two rings.

Explanation:
Unlike eukaryotes, prokaryotes lack a nucleus and membrane-bound organelles. However, like eukaryotes, prokaryotes contain ribosomes, a non-membrane-bound organelle required for protein synthesis. So, ribosomes are found in both prokaryotes and eukaryotes.

Lysosomes (Choice A) and the Golgi apparatus (Choice C) are only found in eukaryotes. The nucleoid (Choice D) is only found in prokaryotes.

Explanation:
All 64 codons of the genetic code are present in all organisms, apart from minor variations indicating that all life evolved from the same original cells (Choice D).

Eukaryotic cells evolved from prokaryotic cells (Choice A) is incorrect as this is the endosymbiotic theory. Choice B is incorrect as mitochondria do not have a cell wall. Choice C is incorrect as bacteria do not have histones.

Explanation:
Bound ribosomes synthesise proteins for secretion outside of the cell, membrane-associated proteins or proteins used in lysosomes (Choice D).

Choice A is incorrect, as free ribosomes synthesise proteins for use primarily within the cytoplasm.
Choice B is incorrect as bound ribosomes do not transport mRNA from the nucleus to the cytoplasm.
Choice C is incorrect as ribosomes do not destroy proteins.

Explanation:
The number of organisms found via the fossil record for any era is much lower compared to the number of organisms found on Earth now and those that continue to be found, so Choice B is correct. Choice A is incorrect as changing levels of atmospheric carbon dioxide provide evidence for humans' impact on carbon dioxide levels. Choices C and D are incorrect as evidence suggests that the rate of evolution and speciation are increasing, which provides evidence for increasing biodiversity.

Explanation:
DNA molecules consist of four nitrogenous bases: adenine (A), cytosine (C), guanine (G) and thymine (T). In DNA molecules, adenine always pairs with thymine and cytosine always pairs with guanine. This is known as complementary base pairing.

Sequences I and III both follow base-pairing rules correctly. Statement II is incorrect because it shows T paired with G and A paired with C.

Explanation:
Species that share a recent common ancestor, such as humans and chimpanzees, are more likely to have similar base sequences. Over time, as species diverge and evolve, mutations accumulate in their base sequences. The longer two species have been separated from their common ancestor, the more mutations will have accumulated in their genomes. Therefore, species that are closely related have fewer accumulated mutations and more similar base sequences (Choice B).

Choice A is incorrect because genome size is not necessarily proportional to the size of the species. The genome size can vary significantly even among organisms of similar size, and organisms with similar-sized genomes, such as the horse and mouse vary greatly in size. Choice C is incorrect because it is possible to have a larger genome size with a comparable, or even smaller, number of genes, depending on the amount of non-coding DNA and repetitive sequences present in the genome. Humans have a larger genome but a similar number of genes to the chimpanzee, mouse and horse. Choice D is incorrect because although the nucleotide base sequence in the table for humans and chimpanzees is the same it does not mean that their whole genome is the same.

Explanation:
Choice A is correct because Chargaff used paper chromatography to show that the proportions of adenine (A) and thymine (T) are the same, as are the proportions of guanine (G) and cytosine (C). There are not equal proportions of A, T, G and C.

Franklin (Choice B) and Watson (Choice D) are credited for their contributions to the overall structure of DNA structure. Hershey (Choice C), along with Chase, is famous for his research which provided evidence for the idea that DNA, rather than protein, carries the genetic code.

Explanation:
The nucleotide strand CCUAGAGCUUACCG contains the bases cytosine (C), uracil (U), adenine (A), and guanine (G). While C, A, and G are found in both DNA and RNA, the base uracil is only found in RNA. So, the strand is an RNA strand.

RNA contains the pentose sugar ribose. Therefore, ribose is found in the nucleotides in the strand.

Choices A and C are incorrect because deoxyribose is only found in DNA, not RNA. The nucleotide strand cannot be DNA because it contains uracil in place of thymine (T).

Choice D is incorrect because the strand is RNA which contains ribose.

Explanation:
Choice D is correct because organisms placed in the same group share common ancestry. Choice A and Choice B are incorrect because morphology and anatomical characteristics are used to describe the structures of organisms which are used to primarily group them together. Choice C is incorrect because modes of nutrition are used to describe organisms but not to classify organisms.

Explanation:
Nucleotides are linked together by condensation reactions to form molecules of DNA (Choice B). The formation of DNA requires the removal of water for the joining of smaller molecules to form a larger molecule.

The joining of nucleotides together can also be classified as an anabolic reaction but not as a catabolic reaction (Choice C). Catabolism is the breaking down of a large molecule into a smaller molecule. An example of catabolism is hydrolysis which is a reaction that breaks down a large molecule into a smaller molecule by adding water (Choice A).

Translation is a process which follows the transcription of DNA (Choice D). It is where mRNA is used as a template by ribosomes to synthesize polypeptides.

Explanation:
Isotopes with long half-lives (Choice C), such as uranium and potassium isotopes, are often used in absolute dating because they decay slowly and are suitable for dating older fossils.

Stable isotopes (Choice A) are not used for fossil dating because they do not undergo radioactive decay. Absolute dating relies on the measurement of the decay of radioactive isotopes into stable daughter isotopes to determine the age of a sample. Isotopes with short half-lives (Choice B) are typically used for dating more recent events, not older fossils. Isotopes with no half-lives (Choice D) cannot be used for fossil dating using the methods based on radioactive decay. Isotopes without half-lives do not undergo radioactive decay, so it is impossible to establish a chronology or estimate the age of a fossil using these isotopes.

Explanation:
Choice A is correct because Eukaryota are the only group with nuclei and membrane-bound organelles. Choice B and choice C are incorrect because organisms in these two domains (Archaea and Eubacteria) do not have membrane-enclosed organelles. Choice D is incorrect because prokaryotes are not the name of a domain.