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Inheritance, Variation and Evolution -- Diagnostic Tests

Inheritance, Variation and Evolution — Diagnostic Tests

Unit Tests

UT-1: DNA and Genetics

Question: (a) Describe the structure of DNA. (b) Explain the relationship between DNA, genes, and chromosomes. (c) In humans, the gene for earlobe attachment has two alleles: free earlobes (F, dominant) and attached earlobes (f, recessive). A heterozygous man and a woman with attached earlobes have a child. Calculate the probability that the child has free earlobes using a genetic cross diagram. (d) Explain what is meant by genotype and phenotype.

Solution:

(a) DNA (deoxyribonucleic acid) is a double helix structure made of two complementary strands. Each strand is a polymer of nucleotides, where each nucleotide consists of: a deoxyribose sugar, a phosphate group, and one of four bases: adenine (A), thymine (T), cytosine (C), or guanine (G). The two strands are held together by hydrogen bonds between complementary base pairs: A—T (two bonds) and C—G (three bonds). The sugar-phosphate backbone forms the outside of the helix.

(b) A chromosome is a long coiled molecule of DNA found in the nucleus. A gene is a short section of DNA on a chromosome that codes for a specific protein (and therefore a specific characteristic). Each gene exists in different forms called alleles, which produce variations of the same characteristic. Humans have 23 pairs of chromosomes (46 total), containing approximately 20,000 genes.

(c) Man: Ff (heterozygous). Woman: ff (attached earlobes, must be homozygous recessive).

Punnett square:

Ff
fFfff
fFfff

Genotypes: 50% Ff (free earlobes), 50% ff (attached earlobes). Probability of free earlobes =50%= 50\%.

(d) Genotype: The combination of alleles an organism has for a particular gene (e.g., FF, Ff, or ff). Phenotype: The observable physical characteristic resulting from the genotype and environment (e.g., free earlobes or attached earlobes). The phenotype depends on which alleles are dominant/recessive and whether the genotype is homozygous or heterozygous.

UT-2: Natural Selection and Evolution

Question: (a) Describe the process of natural selection. (b) Peppered moths exist in two forms: light (dominant) and dark (recessive). In a polluted forest, 80% of moths are dark. After pollution controls, only 20% are dark. Explain this change using natural selection. (c) Explain the difference between natural selection and selective breeding. (d) Describe how antibiotic resistance in bacteria arises and why it is a concern.

Solution:

(a) Natural selection: (1) Genetic variation exists within a population due to mutations and sexual reproduction. (2) Individuals with characteristics better suited to the environment have a survival advantage. (3) These individuals are more likely to survive, reproduce, and pass on their advantageous alleles. (4) Over many generations, the frequency of advantageous alleles increases in the population. (5) This leads to evolution — a gradual change in the inherited characteristics of a population over time.

(b) In the polluted forest: trees were darkened by soot. Light moths were more visible to predators (birds), while dark moths were camouflaged. Dark moths survived and reproduced more, passing on the dark allele. The dark form became more common (80%).

After pollution controls: trees became lighter again. Now dark moths were more visible and light moths were camouflaged. Light moths had a survival advantage, and the light form became more common (80% light, 20% dark). The selection pressure reversed.

(c) Natural selection occurs in nature without human intervention — the environment selects for advantageous traits. Selective breeding (artificial selection) is when humans deliberately choose organisms with desirable characteristics to breed, increasing the frequency of those characteristics in the population (e.g., breeding crops for higher yield, dogs for specific traits).

(d) Antibiotic resistance arises because: (1) Random mutations in bacterial DNA can produce alleles for resistance. (2) When antibiotics are used, susceptible bacteria are killed, but resistant bacteria survive. (3) Resistant bacteria reproduce rapidly (binary fission), passing on resistance alleles. (4) The resistant population grows, making the antibiotic ineffective. This is concerning because: resistant infections are harder to treat, leading to longer illnesses, higher mortality, and increased healthcare costs. Overuse of antibiotics accelerates this process.

UT-3: Genetic Engineering

Question: (a) Describe the steps involved in genetic engineering to produce human insulin using bacteria. (b) Explain three advantages of producing human insulin by genetic engineering compared to extracting it from animals. (c) Describe two potential risks of genetic engineering. (d) Explain the difference between genetic engineering and gene therapy.

Solution:

(a) Steps: (1) The human gene for insulin production is identified. (2) A restriction enzyme cuts the insulin gene from human DNA. (3) The same enzyme cuts a bacterial plasmid, creating complementary sticky ends. (4) The insulin gene is inserted into the plasmid and joined by ligase enzyme (recombinant DNA). (5) The recombinant plasmid is inserted into a bacterial cell (transformation). (6) The bacteria are grown in a fermenter, where they reproduce rapidly, producing human insulin. (7) The insulin is extracted and purified.

(b) Three advantages: (1) Exact match: Genetically engineered insulin is identical to human insulin, reducing the risk of allergic reactions compared to pig or cow insulin. (2) Quantity: Bacteria reproduce rapidly in fermenters, producing large quantities quickly and cheaply. (3) Ethical: No animals need to be slaughtered for extraction, reducing ethical concerns and costs.

(c) Two risks: (1) Unintended consequences: The inserted gene might have unexpected effects on the organism’s other genes or on the ecosystem if genetically modified organisms escape. (2) Gene transfer: Modified genes could spread to wild populations through crossbreeding (e.g., herbicide resistance spreading to weeds). (3) Ethical concerns: Some people object to modifying the genetic code of organisms on principle.

(d) Genetic engineering modifies the DNA of an organism (e.g., bacteria, crops) by inserting genes from another species to give it a new characteristic. The modification is heritable.

Gene therapy is a medical treatment that modifies the DNA of a patient’s cells to treat or cure a genetic disorder (e.g., replacing a faulty gene with a functional copy). It targets somatic cells (body cells) and is not heritable (does not affect offspring). Germline gene therapy (modifying egg/sperm cells) is controversial and currently not widely practised.


Integration Tests

IT-1: Inheritance Patterns (with Cell Biology)

Question: In humans, cystic fibrosis is caused by a recessive allele (f). The normal allele is F. A man and woman, both carriers (Ff), have two children. (a) Calculate the probability that both children have cystic fibrosis. (b) Calculate the probability that at least one child is a carrier. (c) Cystic fibrosis is caused by a mutation in the CFTR gene on chromosome 7. Explain how a mutation in DNA leads to a change in protein function. (d) Explain why carriers of cystic fibrosis do not show symptoms.

Solution:

(a) Cross: Ff ×\times Ff. Punnett square: FF (25%), Ff (50%), ff (25%).

Probability that one child has CF =1/4= 1/4. Probability that both children have CF =1/4×1/4=1/16=6.25%= 1/4 \times 1/4 = 1/16 = 6.25\%.

(b) At least one carrier: 1 - P(neither is a carrier). P(not a carrier) == P(FF) =1/4= 1/4. P(neither is a carrier) =1/4×1/4=1/16= 1/4 \times 1/4 = 1/16. P(at least one carrier) =11/16=15/16=93.75%= 1 - 1/16 = 15/16 = 93.75\%.

(c) A mutation is a change in the DNA sequence (e.g., a deletion of three nucleotides). This changes the sequence of amino acids in the CFTR protein (since the DNA code is read in triplets, each coding for one amino acid). The altered amino acid sequence changes the protein’s 3D shape (tertiary structure). The misfolded CFTR protein cannot transport chloride ions across cell membranes effectively. This leads to the production of thick, sticky mucus in the lungs and digestive system — the symptoms of cystic fibrosis.

(d) Carriers are heterozygous (Ff). They have one normal allele (F) that produces enough functional CFTR protein for normal cell function. The recessive allele (f) produces a non-functional protein, but the one dominant normal allele compensates. The carrier is phenotypically normal because one functional copy of the gene is sufficient.

IT-2: Evolution and Speciation (with Ecology)

Question: Two populations of a beetle species become separated by a mountain range. Population A lives in a tropical forest, and population B lives in a temperate grassland. (a) Explain how the two populations might evolve into different species over time (speciation). (b) If the allele for heat tolerance (H, dominant) has a frequency of 0.3 in population A and 0.7 in population B, calculate the frequency of the recessive allele (h) in each population. (c) Explain the role of genetic drift in small, isolated populations. (d) After 10,000 years, the mountain range erodes and the populations meet again. Explain why they may not be able to interbreed.

Solution:

(a) Speciation by allopatric speciation: (1) Geographic isolation separates the populations — they cannot interbreed. (2) Different environmental conditions (tropical vs temperate) create different selection pressures. (3) In the tropical forest, alleles for heat tolerance, dark colouration (camouflage), and specific food preferences are advantageous. (4) In the temperate grassland, alleles for cold tolerance, light colouration, and different food sources are selected. (5) Over many generations, each population accumulates different genetic changes through natural selection and genetic drift. (6) Eventually, even if the populations are brought back together, they have diverged so much that they can no longer interbreed to produce fertile offspring — they are now separate species.

(b) Using the Hardy-Weinberg principle: p+q=1p + q = 1Where pp is the dominant allele frequency and qq is the recessive allele frequency.

Population A: p = 0.3$$q = 1 - 0.3 = 0.7. Population B: p = 0.7$$q = 1 - 0.7 = 0.3.

(c) Genetic drift is the random change in allele frequencies due to chance events, particularly significant in small populations. In a small isolated population: (1) A random event (natural disaster, disease) could eliminate individuals carrying certain alleles, permanently changing the gene pool. (2) The founder effect — if the population was started by a small number of individuals, it would carry only a subset of the original genetic diversity. (3) Genetic drift can lead to fixation (one allele reaching 100%) or loss (an allele disappearing), reducing genetic diversity.

(d) Reproductive isolation may have occurred through several mechanisms: (1) Behavioural isolation — different mating behaviours or courtship rituals evolved. (2) Temporal isolation — different breeding seasons. (3) Mechanical isolation — physical incompatibility of reproductive structures. (4) Genetic incompatibility — chromosomal differences prevent successful fertilisation or embryo development. (5) Hybrid inviability — even if mating occurs, the offspring are sterile or non-viable (like mules from horse ×\times donkey).

IT-3: Genetics and Ethics (with Organisation)

Question: A couple are both carriers of sickle cell anaemia (recessive condition). They wish to use IVF with pre-implantation genetic diagnosis (PGD) to ensure their child does not have the disease. (a) Explain the process of PGD. (b) Calculate the probability that a conceived child would: (i) have sickle cell anaemia, (ii) be a carrier, (iii) be completely unaffected. (c) Explain the advantage of being a carrier of sickle cell trait in regions where malaria is endemic. (d) Discuss one ethical argument for and one against using PGD for this purpose.

Solution:

(a) PGD process: (1) Eggs are collected from the mother and fertilised with the father’s sperm in vitro (IVF). (2) The embryos are allowed to develop to the 8-cell stage (about 3 days). (3) One or two cells are removed from each embryo (biopsy) for genetic testing. (4) DNA from the removed cells is tested for the sickle cell allele. (5) Only embryos that do not have two copies of the faulty allele are selected for implantation. (6) The selected healthy embryo(s) are implanted in the mother’s uterus.

(b) Cross: Ss ×\times Ss (both carriers). Punnett square: SS (25%), Ss (50%), ss (25%). (i) Sickle cell anaemia (ss): 25%25\%. (ii) Carrier (Ss): 50%50\%. (iii) Completely unaffected (SS): 25%25\%.

(c) In regions where malaria is endemic, carriers of the sickle cell trait (Ss) have a survival advantage. The sickle cell allele provides some protection against malaria: the altered red blood cells are less hospitable to the malaria parasite (Plasmodium). Carriers experience mild or no symptoms of sickle cell disease but have significantly reduced risk of severe malaria. This is an example of heterozygote advantage, which explains why the sickle cell allele remains common in malaria-endemic regions despite being harmful in homozygous form.

(d) For: PGD prevents the suffering of a child born with sickle cell anaemia, a serious condition causing pain, organ damage, and reduced life expectancy. Parents can have a healthy child without the emotional burden of terminating a pregnancy.

Against: PGD involves selecting and discarding embryos based on their genotype, which some consider unethical (it involves the destruction of human embryos). It could be seen as a step towards “designer babies” — selecting embryos for non-medical traits. There are also concerns about accessibility (expensive, not available to all) and the message it sends about the value of people with genetic conditions.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Pitfalls

  • Confusing terminology or concepts that appear similar but have distinct meanings.
  • Overlooking key assumptions or boundary conditions that limit applicability.