DSE Biology Diagnostic: Genetics

Unit Test 1: DNA Structure and Replication

Question

DNA has a double helix structure described by Watson and Crick.

(a) Describe the DNA double helix structure with reference to (i) the arrangement of the two polynucleotide strands, (ii) the types of bonds holding the two strands together, and (iii) the significance of complementary base pairing. [5 marks]

(b) DNA replication is described as semi-conservative. Explain what this means and describe the role of DNA helicase and DNA polymerase in the replication process. [5 marks]

(c) A sample of DNA contains $28\%$ adenine. Calculate the percentage of each of the other three bases (thymine, guanine, and cytosine) in this sample. [2 marks]

---

Worked Solution

(a) DNA double helix structure:

(i) Two polynucleotide strands are coiled into a double helix. The two strands run in antiparallel directions -- one runs $5'$ to $3'$ and the other runs $3'$ to $5'$.

(ii) The two strands are held together by hydrogen bonds between complementary base pairs: adenine (A) pairs with thymine (T) via two hydrogen bonds, and guanine (G) pairs with cytosine (C) via three hydrogen bonds. The sugar-phosphate backbones of the two strands are on the outside of the helix, and the bases are on the inside.

(iii) Complementary base pairing is significant because:

• It ensures that the two strands are exact copies of each other in terms of information content (the sequence on one strand determines the sequence on the other).
• It allows accurate replication -- each strand serves as a template for the synthesis of a new complementary strand.
• It maintains a constant width of the double helix (since A-T and G-C pairs are similar in size).

(b) Semi-conservative replication means that each new DNA molecule consists of one original (parent) strand and one newly synthesised strand. After one round of replication, half of the original DNA is conserved in each daughter molecule.

Roles of enzymes:

• DNA helicase: Breaks the hydrogen bonds between the complementary base pairs, "unzipping" the double helix and separating the two strands. Each separated strand then acts as a template for the synthesis of a new complementary strand.
• DNA polymerase: Catalyses the formation of sugar-phosphate bonds (phosphodiester bonds) between adjacent nucleotides, joining free DNA nucleotides to the growing new strand. It does this by adding nucleotides complementary to the template strand, following the base-pairing rules (A with T, C with G).

(c) Since A pairs with T (complementary base pairing), if A = $28\%$, then T = $28\%$.

The total percentage of A + T = $28\% + 28\% = 56\%$.

Therefore, G + C = $100\% - 56\% = 44\%$.

Since G pairs with C: G = $22\%$ and C = $22\%$.

Base	Percentage
Adenine (A)	28%
Thymine (T)	28%
Guanine (G)	22%
Cytosine (C)	22%

---

Unit Test 2: Protein Synthesis -- Transcription and Translation

Question

(a) Describe the process of transcription, including the role of RNA polymerase and the formation of mRNA from a DNA template. [4 marks]

(b) Explain the role of tRNA and ribosomes in translation. In your answer, describe what is meant by a codon and an anticodon. [4 marks]

(c) A mutation changes a DNA triplet from GAG to GTG. Using the genetic code table below (partial), determine the effect of this mutation on the amino acid sequence.

mRNA codon	Amino acid
CUC	Leucine
CUU	Leucine
GAG	Glutamic acid
GAC	Aspartic acid
GUG	Valine

[2 marks]

---

Worked Solution

(a) Transcription:

1. The enzyme RNA polymerase binds to a promoter region on the DNA template strand.
2. RNA polymerase unwinds the DNA double helix locally and moves along the template strand ($3'$ to $5'$ direction).
3. RNA polymerase catalyses the formation of phosphodiester bonds between adjacent RNA nucleotides, building a complementary pre-mRNA strand in the $5'$ to $3'$ direction.
4. RNA nucleotides pair with the DNA template by complementary base pairing: A with U (uracil replaces thymine in RNA), T with A, C with G, G with C.
5. When RNA polymerase reaches a terminator sequence, it detaches and the pre-mRNA is released.
6. In eukaryotes, the pre-mRNA undergoes post-transcriptional modification: introns (non-coding regions) are removed by splicing, a $5'$ cap is added, and a poly-A tail is added, producing mature mRNA.

(b) Translation:

• Ribosomes are the site of protein synthesis. A ribosome consists of a large and a small subunit. The small subunit binds to the $5'$ end of the mRNA and reads it. The ribosome has three binding sites for tRNA: A (aminoacyl), P (peptidyl), and E (exit).
• A codon is a sequence of three bases on the mRNA that codes for a specific amino acid (or a stop signal).
• tRNA (transfer RNA) molecules carry specific amino acids. Each tRNA has an anticodon -- a sequence of three bases complementary to a specific mRNA codon. The tRNA anticodon binds to the mRNA codon by complementary base pairing (in the ribosome's A site), ensuring that the correct amino acid is incorporated into the growing polypeptide chain.
• The ribosome moves along the mRNA, reading one codon at a time. Each time a new tRNA brings an amino acid, a peptide bond forms between the new amino acid and the growing polypeptide chain. The ribosome continues until it reaches a stop codon (UAA, UAG, or UGA), when the polypeptide is released.

(c) Original DNA: GAG. The template strand is transcribed to mRNA with complementary bases: the DNA template GAG produces mRNA codon CUC (since the coding strand would be CTC, and mRNA matches the coding strand with T replaced by U).

The DNA triplet GAG is the coding strand (sense strand). The mRNA codon is GAG (GAG contains no T, so no substitution is needed), which codes for Glutamic acid.

After mutation, the coding strand becomes GTG. The mRNA codon becomes GUG, which codes for Valine.

The mutation changes the amino acid from glutamic acid to valine. This is a missense mutation (a single base substitution that changes one amino acid to another). This is the exact mutation that causes sickle cell anaemia.

---

Unit Test 3: Sex Linkage and Dihybrid Cross

Question

In humans, red-green colour blindness is caused by a recessive allele ($c$) on the X chromosome. The normal allele is $C$.

(a) A woman with normal vision whose father was colour blind marries a man with normal vision. Determine the possible genotypes and phenotypes of their children. Show your working with a genetic diagram. [5 marks]

(b) Explain why sex-linked conditions such as colour blindness are more common in males than in females. [2 marks]

(c) In cats, black fur (B) is dominant to brown fur (b), and short hair (S) is dominant to long hair (s). A dihybrid cross is performed between two cats, both with genotype BbSs. Calculate the probability of obtaining a cat with brown fur and short hair among the offspring. [3 marks]

---

Worked Solution

(a) Let: $X^{C}$ = normal vision (dominant), $X^{c}$ = colour blind (recessive), $Y$ = male chromosome.

Woman's genotype: She has normal vision but her father was colour blind ($X^{c}Y$). Her father must have passed his $X^{c}$ chromosome to her. Therefore, her genotype is $X^{C}X^{c}$ (carrier).

Man's genotype: Normal vision male = $X^{C}Y$.

Genetic cross:

	$X^{C}$ (mother)	$X^{c}$ (mother)
$X^{C}$ (father)	$X^{C}X^{C}$ (normal female)	$X^{C}X^{c}$ (carrier female)
$Y$ (father)	$X^{C}Y$ (normal male)	$X^{c}Y$ (colour blind male)

Possible offspring:

• $X^{C}X^{C}$: Normal female -- 25%
• $X^{C}X^{c}$: Carrier female (normal vision) -- 25%
• $X^{C}Y$: Normal male -- 25%
• $X^{c}Y$: Colour blind male -- 25%

Phenotypic ratio: 3 normal vision : 1 colour blind (all colour blind offspring are male).

(b) Males have only one X chromosome ($XY$). A male needs only one copy of the recessive allele ($X^{c}$) to express the condition, since there is no second X chromosome with a potentially dominant allele to mask it.

Females have two X chromosomes ($XX$). A female must have two copies of the recessive allele ($X^{c}X^{c}$) to express the condition. A female with one copy ($X^{C}X^{c}$) is a carrier with normal vision. Therefore, the condition is much less likely to be expressed in females.

(c) Dihbrid cross: $BbSs \times BbSs$

For brown fur and short hair, the genotype must be $bbS\_$.

Using the forked line method (or Punnett square):

For fur colour ($Bb \times Bb$): $\frac{1}{4} BB$, $\frac{1}{2} Bb$, $\frac{1}{4} bb$ -- probability of $bb$ = $\frac{1}{4}$.

For hair length ($Ss \times Ss$): $\frac{1}{4} SS$, $\frac{1}{2} Ss$, $\frac{1}{4} ss$ -- probability of $S\_$ ($SS$ or $Ss$) = $\frac{3}{4}$.

Since the genes are on different chromosomes (assumed independent assortment):

Probability of $bbS\_$ = $\frac{1}{4} \times \frac{3}{4} = \frac{3}{16}$

---

Integration Test 1: Genetic Engineering

Question

Human insulin can be produced by genetically engineered bacteria using recombinant DNA technology.

(a) Describe the steps involved in producing human insulin using genetically engineered bacteria. Your answer should include the roles of restriction enzymes, vectors, ligase, and host cells. [6 marks]

(b) Explain the importance of using a plasmid that contains both a gene for antibiotic resistance and a promoter sequence recognised by bacterial cells. [3 marks]

(c) After the bacteria have been transformed, not all bacteria will have taken up the recombinant plasmid. Describe how bacteria that have successfully taken up the recombinant plasmid can be identified. [3 marks]

---

Worked Solution

(a) Steps in producing human insulin using genetic engineering:

1. Isolation of the insulin gene: The human gene for insulin is located using gene probes and cut out from human DNA using a restriction enzyme (restriction endonuclease). The restriction enzyme cuts the DNA at specific recognition sequences, producing sticky ends (single-stranded overhangs).

2. Preparation of the vector: A plasmid (from a bacterium) is used as a vector. The same restriction enzyme is used to cut open the plasmid, producing complementary sticky ends.

3. Insertion of the gene: The human insulin gene is mixed with the cut plasmid. The sticky ends of the insulin gene are complementary to the sticky ends of the plasmid, so they base-pair (anneal) by complementary base pairing. The enzyme DNA ligase is then used to catalyse the formation of phosphodiester bonds, sealing the insulin gene into the plasmid to form a recombinant plasmid.

4. Transformation: The recombinant plasmid is introduced into host cells (bacteria, e.g. E. coli) by a process called transformation (e.g. using heat shock or electroporation to make the bacterial membrane permeable to the plasmid).

5. Selection and growth: The bacteria are grown on a selective medium (see part c). Bacteria that have successfully taken up the recombinant plasmid are identified and cultured in large-scale fermenters.

6. Harvesting insulin: The bacteria express the insulin gene and produce human insulin, which is then extracted and purified.

(b) - Antibiotic resistance gene: Acts as a genetic marker (selection marker). It allows the identification of bacteria that have taken up the plasmid. When grown on a medium containing the antibiotic, only bacteria with the plasmid (which carries the resistance gene) will survive.

• Promoter sequence: A promoter is a DNA sequence that RNA polymerase binds to in order to initiate transcription. For the human insulin gene to be expressed (transcribed and translated) in bacterial cells, the plasmid must contain a bacterial promoter sequence upstream of the insulin gene. Without a promoter recognised by bacterial RNA polymerase, the insulin gene would not be transcribed and no insulin would be produced.

(c) Identification of transformed bacteria:

1. Bacteria are plated on a growth medium containing the antibiotic for which the plasmid carries a resistance gene (e.g. ampicillin).
2. Only bacteria that have taken up the plasmid (either recombinant or non-recombinant) will survive and grow on this medium, because they express the antibiotic resistance gene. Bacteria without the plasmid die.
3. To distinguish between bacteria with recombinant plasmids and those with non-recombinant plasmids (plasmids that re-closed without the insulin gene), a second genetic marker can be used. For example, if the insulin gene is inserted within a gene for another resistance (e.g. tetracycline resistance), then:
   • Bacteria with the recombinant plasmid: resistant to ampicillin but susceptible to tetracycline (because the tetracycline resistance gene was disrupted).
   • Bacteria with the non-recombinant plasmid: resistant to both ampicillin and tetracycline.
4. Using replica plating or testing on tetracycline plates, colonies resistant to ampicillin but sensitive to tetracycline are identified as containing the recombinant plasmid.

---

Integration Test 2: Mutations and Their Effects

Question

The table below shows three different types of mutations that can occur in a gene:

Mutation type	Description
Silent mutation	Base substitution that does not change the amino acid coded for
Missense mutation	Base substitution that changes one amino acid to another
Nonsense mutation	Base substitution that creates a premature stop codon

(a) Explain why a silent mutation does not affect the protein produced. Refer to the nature of the genetic code. [2 marks]

(b) A nonsense mutation near the beginning of a gene is generally more harmful than a nonsense mutation near the end of a gene. Explain why. [2 marks]

(c) A gene has the following DNA coding strand sequence (showing only part of the gene):

5' -- ATG CCT GAG GAC TGA -- 3'

Using the information below, determine the effect of a frameshift mutation caused by the deletion of the second nucleotide (T). The original and mutated mRNA sequences and their translations should be shown.

mRNA codons and amino acids: AUG = Met, CCU = Pro, CAG = Gln, ACU = Thr, GA_ = ?

[4 marks]

---

Worked Solution

(a) The genetic code is degenerate (redundant): most amino acids are coded for by more than one codon. A silent mutation changes a base in the DNA to one that still codes for the same amino acid (e.g. changing CCT to CCC, both of which code for proline). Since the amino acid sequence of the protein is unchanged, the protein's structure and function remain normal.

(b) A nonsense mutation creates a premature stop codon, causing translation to terminate early and producing a truncated (shortened) polypeptide.

If the nonsense mutation is near the beginning of the gene, the polypeptide is very short and likely completely non-functional because most of the protein's amino acid sequence is missing. Critical functional domains are absent.

If the nonsense mutation is near the end of the gene, a larger portion of the polypeptide is produced before translation stops. More of the protein's functional domains may be intact, so the truncated protein may retain partial function. The effect is therefore generally less severe.

(c) Original DNA coding strand: 5' -- ATG CCT GAG GAC TGA -- 3'

Original mRNA: 5' -- AUG CCU GAG GAC UGA -- 3'

Original translation: AUG (Met) - CCU (Pro) - GAG (Glu) - GAC (Asp) - UGA (Stop)

Polypeptide: Met -- Pro -- Glu -- Asp (then stop)

After deletion of the second nucleotide (T):

DNA coding strand becomes: 5' -- AG CCT GAG GAC TGA -- 3'

But more accurately, deleting the second nucleotide shifts all subsequent bases:

Original: A-T-G-C-C-T-G-A-G-G-A-C-T-G-A Delete T (position 2): A-G-C-C-T-G-A-G-G-A-C-T-G-A

New mRNA: 5' -- AGC CUG AGG ACU GA -- 3'

Reading in codons: AGC (Ser) - CUG (Leu) - AGG (Arg) - ACU (Thr) - GA... (incomplete)

Mutated translation: AGC (Ser) - CUG (Leu) - AGG (Arg) - ACU (Thr)

The frameshift mutation changes the reading frame from the point of deletion onwards. Almost all amino acids after the mutation are changed, and the protein sequence is completely different from the point of the mutation. This typically produces a non-functional protein.

---

Integration Test 3: Inheritance Patterns and Pedigree Analysis

Question

A pedigree chart shows a family where a rare genetic disorder occurs. The disorder is caused by a dominant allele ($D$) on an autosome. The pedigree is described below:

• Generation I: Father (affected, $Dd$) and Mother (unaffected, $dd$)
• Generation II: Two children -- Daughter 1 (affected) and Son 1 (unaffected)

(a) Explain why the father's genotype must be $Dd$ and not $DD$, given that the disorder is rare. [2 marks]

(b) Calculate the probability that Daughter 1's first child (with an unaffected partner, $dd$) will be affected by the disorder. [2 marks]

(c) A different disorder is X-linked recessive. In a family, a normal woman (whose father had the disorder) has a son with a normal man. Calculate the probability that the son will be affected. [3 marks]

---

Worked Solution

(a) The disorder is rare, meaning the recessive allele ($d$) is much more common in the population than the dominant allele ($D$). If the father were $DD$, both of his parents would need to carry or have the $D$ allele. Since the disorder is rare, it is far more likely that the father is heterozygous ($Dd$), having inherited the $D$ allele from one parent (who may have been affected) and the $d$ allele from the other (unaffected parent). This is confirmed by the fact that Son 1 is unaffected ($dd$), which is only possible if the father contributed a $d$ allele, proving the father is $Dd$.

(b) Daughter 1 is affected. Her father is $Dd$ and her mother is $dd$.

Cross: $Dd \times dd$

Daughter 1's possible genotypes: $\frac{1}{2} Dd$ (affected) or $\frac{1}{2} dd$ (unaffected).

Since Daughter 1 is affected, she must be $Dd$.

Daughter 1 ($Dd$) has a child with an unaffected partner ($dd$):

Cross: $Dd \times dd$

Offspring: $\frac{1}{2} Dd$ (affected), $\frac{1}{2} dd$ (unaffected).

Probability that the first child is affected = $\frac{1}{2}$ (50%).

(c) X-linked recessive disorder: $X^{d}$ = disorder allele, $X^{D}$ = normal allele.

Woman's genotype: Normal, but her father had the disorder ($X^{d}Y$). Her father must have passed his $X^{d}$ chromosome to her. Therefore, the woman's genotype is $X^{D}X^{d}$ (carrier, normal phenotype).

Husband's genotype: Normal male = $X^{D}Y$.

Cross: $X^{D}X^{d} \times X^{D}Y$

	$X^{D}$ (mother)	$X^{d}$ (mother)
$X^{D}$ (father)	$X^{D}X^{D}$ (normal female)	$X^{D}X^{d}$ (carrier female)
$Y$ (father)	$X^{D}Y$ (normal male)	$X^{d}Y$ (affected male)

Probability that the son is affected = $\frac{1}{2}$ (50%) of all sons, or $\frac{1}{4}$ of all children.

Note: If the question asks for the probability given that the child is a son, the answer is $\frac{1}{2}$. If it asks for the probability among all children, the answer is $\frac{1}{4}$.