Explanation​:
The four stages of mitosis are prophase, where chromosomes supercoil and become visible; Metaphase, where chromosomes line up at the equator of the cell; anaphase, where sister chromatids are pulled apart and telophase, where the nuclear envelope reforms and two new daughter nuclei are formed.

Statement I is incorrect as sister chromatids are not pulled apart during metaphase. They are pulled apart during anaphase.
Choice B is incorrect as both statements II and III are correct and should be included.

Explanation​:
Choice C is correct because meiosis halves the number of chromosomes, mitosis maintains the diploid number and fertilisation restores the diploid number.

Choice A is incorrect because meiosis separates pairs of alleles and produces nuclei with one allele for each gene.

Choice B is incorrect because mitosis produces genetically identical cells that have two alleles for each gene.

Choice D is incorrect because, in fertilisation, two gametes fuse, and each gamete has one allele for each gene.

Explanation​:
Choice A is correct because in both selective breeding and natural selection, favourable characteristics are “selected”, by either the environment (natural selection) or humans (selective breeding). Choice B is incorrect because natural selection doesn’t increase variation, it influences the proportion of existing variants. Choices C and D are incorrect because natural selection is not controlled by humans and does not contribute to the domestication of animals.

Explanation​:

Taq polymerase was isolated from bacteria found in the hot springs at Yellowstone National Park, USA. This enzyme is able to withstand high temperatures used for PCR assays.
All other choices (Choice A, C and D) describe bacteria living in colder locations compared to hot springs which strongly suggests that enzymes present in these bacteria would not be adapted to higher temperatures.

Explanation​:
Crossing over (Choice A), random orientation (Choice C) and separation of homologous chromosomes (Choice D) all occur during meiosis. The formation of two identical cells (Choice B) happens in mitosis, not meiosis.

Explanation:
At fertilisation, two gametes, which are haploid, fuse together to produce a diploid cell. Gametes of a fruit fly have a haploid number of 4, so fertilisation produces a cell with 8 chromosomes (choice A). Choice B is incorrect because 4 is the haploid number. Choice C is incorrect because fruit fly somatic cells have 8 chromosomes and fertilised egg cells have 8 chromosomes. Choice D is incorrect because 16 is twice the diploid number of fruit flies.

Explanation:
A greenhouse gas is a gas that contributes to the greenhouse effect. Greenhouse gases absorb the long wave infrared radiation (heat energy) emitted from Earth’s surface and can reradiate it back to Earth’s surface. So, one factor that determines whether a gas acts as a greenhouse gas is its ability to absorb long wave radiation: gases that are better able to absorb long wave radiation are more likely to act as greenhouse gases. A second factor that qualifies carbon dioxide as a greenhouse gas of concern is its relatively high concentration in the atmosphere: gases that are in higher concentration in the atmosphere have a larger effect on Earth’s temperature.

Explanation​:
Peptide bonds form between amino acids as a result of condensation reactions. Statement II (Choice C) is correct because dipeptides are composed of two amino acids held together by a peptide bond and proteins (Statement III) are composed of one or more polypeptides, which are chains of amino acids held together by peptide bonds. Choice A, B and D are incorrect as individual amino acids (Statement I) do not possess peptide bonds.

Explanation​:
Choices B, C and D are incorrect because as the trees get darker, the camouflage of the peppered moth will be less effective, meaning it is more visible to predators and therefore more likely to be eaten.

Explanation​:
Choice C is correct because the biological mother passes on the X chromosome to a son and the biological father contributes only with Y. For the son to be colour-blind, he must have an allele for it on the X chromosome. Therefore the biological mother must have at least one recessive allele (n). Choices A and B are incorrect because individuals with X and Y are male. Choice D is incorrect because, in this case, no sons would have the condition.

Explanation​:
The question states that heterozygous individuals have the banded pattern on their hairs, which means that the agouti trait is dominant. Two heterozygous parents will produce offspring with the dominant and recessive characteristics in the ratio of 3:1 (Choice B). The offspring of a heterozygous individual and a homozygous agouti individual will all receive at least one dominant allele, so they would all be agouti (Choice A). The offspring of at least one parent that is homozygous for agouti will all have the agouti characteristic (Choices C and D).

Explanation​:
Statement III is not true, as homologous chromosomes often have different alleles (forms of genes), so are generally not identical to each other.

Statements I and II are true, as homologous chromosomes are pairs of chromosomes that have their centromeres and genes at the same positions. Therefore, Choice C is correct.

Explanation:
Negative feedback is a regulatory mechanism in which the body responds to a change in order to return the homeostatic variable back to within its set points. For example, if blood pH increases above the set point, negative feedback mechanisms will occur to bring blood pH back down. Statement I is therefore true and Statement II is therefore false. Choices B and D are not correct.

While the nervous system plays a significant role in many feedback mechanisms, it is not the only system involved. The endocrine system, which releases hormones into the bloodstream also plays a crucial role in feedback control. Statement III is therefore false, meaning Choice B is incorrect, and Choice A is the correct response to the question.

Explanation​:
Statement I (Choices A and D) is incorrect as learned behaviours are changes that occur during a lifetime, and so do not affect the genetic code and cannot be passed on. Statement III (Choices C and D) is incorrect as natural selection increases the frequency of characteristics that make individuals better adapted.

Explanation:
Choice A is correct because during metaphase I homologous chromosomes align at the equator and are subject to random orientation. Choice B is incorrect as, during Prophase I, homologous chromosomes condense and crossing over occurs. Choice C is incorrect as, during Metaphase II, chromosomes align on the cell's equator, and there is random orientation of the sister chromatids and not homologous chromosomes. Choice D is incorrect as, during Prophase II, chromosomes align on the cell's equator.

Explanation​:
Haemoglobin is a globular protein which is responsible for transporting oxygen around the body (Choice A). Choice B is incorrect as collagen is a fibrous protein. Choice C is incorrect as haemoglobin is not a fibrous protein and does not provide structural support. Choice D is incorrect as collagen does not transport oxygen.

Explanation​:
Telophase is the last phase of mitosis and does not cause mutations (Choice C). Single base substitutions cause a change in the base sequence of DNA, which is a mutation (Choice A). Carcinogens (Choice B) and high energy radiation (Choice D) are mutagens which means they can cause changes in the base sequence of DNA.

Explanation:
Choice D is correct because food is stored in the cotyledon(s). Choice A is incorrect because the function of the radicle is to develop into the root of the seedling. Choice B is incorrect because the function of the plumule is to develop into the shoot of the seedling. Choice C is incorrect because the testa is the outer covering of the seed.

Explanation​:
Choice A is incorrect as autosomes are chromosomes that are not the sex chromosomes and so do not determine biological sex. Choice B is incorrect as linked genes are genes that are found on the same chromosome. Choice C is incorrect as a gene locus is the position of a gene on a chromosome.

Explanation:
The graph shows an overall positive correlation, as maternal age increases, the incidence of non-disjunction increases (Choice A). Choice B is incorrect because there is a positive correlation not a negative correlation. Choice C is incorrect because there is a correlation. Choice D is incorrect because non-disjunction does occur below 35 years, but the incidence is lower when compared to above 35 years.

Explanation​:
Choice B is correct because pollination is the transfer of pollen from an anther to a stigma. Choice A is incorrect because the fusion of a male and female gamete is fertilization. Choice C is incorrect because the transfer of seeds away from the parent plant is dispersal. Choice D is incorrect because pollination can occur without a pollinator. Some plants are wind-pollinated.

Explanation​:
mRNA, also known as messenger RNA, carries the genetic information needed for protein synthesis (Choice B).

Ribosomal RNA is found in ribosomes. This RNA contributes to the structure of the ribosome and does not code for proteins (Choice A).

ATP (Choice C) is a molecule made out of a nucleotide (adenosine tri-phosphate), while nucleic acid is made out of many nucleotides.

tRNA (Choice D), also known as transfer RNA, brings amino acids to the ribosomes and does not code for proteins.

Explanation​:
Sweat (Choice B) is a coolant and helps to cool the body.
Tears (Choice A), urine (Choice C) and saliva (Choice D) do not lower or raise the body temperature.

Explanation​:
Choice A is correct because if the heart were placed in a hypotonic solution (as opposed to isotonic) water would diffuse into the cardiac muscles.

Choice B is incorrect because there would be no net movement into the tissues of the heart in an isotonic solution, not a hypotonic or hypertonic solution. Choice C is incorrect because even if it is placed in a solution other than an isotonic solution, it cannot dry out as it is still bathed in a watery solution. Choice D is incorrect as the tonicity of the solution the heart is stored in has no effect on whether the recipient’s immune system identifies the heart as foreign which would lead to rejection of the heart by the immune system.

Explanation​:
The solution used to bathe organs to be used in a transplant should have an osmolarity equal to that of the organ to be transplanted and so the solution is isotonic compared to the organ’s tissues.

A solution with an osmolarity higher than that of the organ to be transplanted (Choice A) would cause tissue cells to shrivel, while a solution with an osmolarity lower than that of the organ to be transplanted (Choice B) would cause organ cells to burst. There are no described protocols that indicate that the solution should be mixed with cells from the organ to be transplanted (Choice D).