Gene mutations

Mutations during DNA replication can involve a single nucleotide base being deleted from the genetic code.

Which term describes this mutation?

A Frameshift mutation

B Insertion

C Silent mutation

D Substitution

6 ‑thalassemia is an inherited blood disorder in which the ‑globin chain of haemoglobin is changed or absent in a person’s red blood cells.

Many different DNA mutations in the ‑globin gene can cause ‑thalassemia, including these two mutations:

Each mutation occurs in a triplet in the template strand of DNA.

The transcribed mRNA is complementary to the template strand.

The diagram on the opposite page is a representation of the genetic code. It shows the amino acid encoded by each mRNA codon.

The letters in the centre of the diagram represent the first base in each codon.

State the amino acids that are encoded by the original triplet and the mutated triplet in mutation 1.

Original amino acid........

State and explain the effect of mutation 2 on the structure of the ‑chain of haemoglobin.

* In 2023, a gene editing treatment for ‑thalassemia was approved for use in the UK.

Gene editing is a form of gene therapy.

Standard gene therapy:

• uses a vector to deliver a functional, normal allele to cells in a person’s body

• does not remove the faulty, disease‑causing allele from the person’s cells.

Gene editing:

• can correct the faulty, disease‑causing allele so that only the functional, normal allele is present at that locus

• can be performed on stem cells that have been removed from a person’s body; the stem cells are then transferred back into that person.

Discuss the advantages of gene editing compared to standard gene therapy and discuss the ethical issues shared by both forms of gene therapy.

Extra answer space if required.

What is a direct consequence of the genetic code being degenerate?

A It is possible that a mutation could change the sequence of amino acids but leave the function of the protein unchanged.

B It is possible that a mutation could leave the primary structure of the protein unchanged.

C It is possible that a mutation could leave the sequence of bases unchanged.

D It is possible that a mutation could substitute one amino acid for another with similar properties.

Haemoglobin is a globular protein.

Fig. 16.1 shows the structure of haemoglobin.

There are four levels of protein structure.

Name the bond present in the primary structure of a protein.

Name the structure labelled X.

State one feature, visible in Fig. 16.1, that shows that haemoglobin is a globular protein.

Sickle cell disease is a genetic disease that results from a substitution mutation in one of the genes that code for haemoglobin.

Fig. 16.2 below shows part of the mRNA sequence that codes for normal haemoglobin and the corresponding sequence of amino acids.

Fig. 16.2

Fig. 16.3 is a representation of the genetic code.

Use Fig. 16.3 to identify the missing amino acids 1 and 2 in Fig. 16.2.

Outline the role of RNA polymerase in the production of the mRNA sequence in Fig. 16.2.

In sickle cell disease, the haemoglobin contains the amino acid valine in one of the positions normally occupied by glutamic acid.

State the base sequence on the anticodon of a tRNA molecule that brings valine to the ribosome.

The sweet pea plant has been used to study inheritance since the nineteenth century. The seeds of the sweet pea can vary in colour and shape.

The gene that controls colour has two alleles:

• Y is dominant and produces yellow seeds.

• y is recessive and produces green seeds.

The gene that controls shape has two alleles:

• R is dominant and produces round seeds.

• r is recessive and produces wrinkled seeds.

In the nineteenth century, Gregor Mendel crossed a pea plant that was heterozygous for both seed colour and shape with a pea plant that had green and wrinkled seeds.

List the gametes that would be produced by a sweet pea plant that was heterozygous for both seed colour and shape.

List the genotypes of the offspring that were produced from Mendel’s cross and state the corresponding phenotypes.

When Mendel crossed two pea plants that were heterozygous for both seed colour and shape, the ratio of phenotypes in the offspring was:

• 9 yellow round

• 3 green round

• 3 yellow wrinkled

• 1 green wrinkled.

Some students tried to recreate this investigation using a modern variety of plant that showed the same phenotypic variation in seed colour and shape.

The students crossed two of the modern plants that were heterozygous for both seed colour and shape. The results of this cross were:

• 58 yellow and round

• 31 green and round

• 21 yellow and wrinkled

• 2 green and wrinkled

The students used the chi-squared test to compare their data to the expected 9: 3: 3: 1 ratio.

Table 17 shows a |² probability table.

After analysing the results, the students stated that the inheritance of the seed colour and shape in their investigation was different from that in Mendel’s investigation.

Using Table 17, discuss whether the results of the investigation and the chi-squared test support the students’ statement.

A ratio that is different from the expected 9: 3: 3: 1, in a cross such as this, can be the result of epistasis.

Suggest and explain one reason, other than epistasis, why the phenotype ratio might not be 9: 3: 3: 1.

The yellow colour in peas is the result of an enzyme that breaks down chlorophyll, which is green.

• The Y allele codes for the production of an enzyme that breaks down chlorophyll.

• The y allele is the result of a mutation in the Y allele.

• The y allele codes for an inactive form of this enzyme.