Genetic Information, Variation & Relationships between Organisms Exam Qs Flashcards
A mutation of a tumour suppressor gene can result in the formation of a tumour. Explain how. (2 marks)
- (Tumour suppressor) gene inactivated / not able to control / slow down cell division;
Ignore: references to growth - Rate of cell division too fast / out of control.
1 and 2 Accept: mitosis 1 and 2
Reject: meiosis
Not all mutations result in a change to the amino acid sequence of the encoded polypeptide.
Explain why. (1 mark)
- (Genetic) code degenerate;
Accept: codon for triplet
Accept description of degenerate code, e.g. another triplet codes for the same amino acid - Mutation in intron.
Accept: mutation in non-coding DNA
Some cancer cells have a receptor protein in their cell-surface membrane that binds to a hormone called growth factor. This stimulates the cancer cells to divide.
Scientists have produced a monoclonal antibody that stops this stimulation.
Use your knowledge of monoclonal antibodies to suggest how this antibody stops the growth of a tumour. (3 marks)
- Antibody has specific tertiary structure / binding site / variable region;
Do not accept explanations involving undefined antigen - Complementary (shape / fit) to receptor protein / GF / binds to receptor protein / to GF;
Ignore: same shape as receptor protein / GF - Prevents GF binding (to receptor).
In the diagram above, the first codon is AUG. Give the base sequence of:
-the complementary DNA base sequence
-the missing anticodon
(2 marks)
TAC
UAC
Aspartic acid and proline are both amino acids. Describe how two amino acids differ from one another. You may use a diagram to help your description. (1 mark)
Have different R group
accept in diagram
Aspartic acid GAC, GAU
Proline CCA, CCG, CCC, CCU
Deletion of the sixth base (G) in the sequence shown in the diagram above would change the nature of the protein produced but substitution of the same base would not. Use the information in the table and your own knowledge to explain why. (3 marks)
- Substitution would result in CCA / CCC / CCU;
- (All) code for same amino acid / proline;
- Deletion would cause frame shift / change in all following codons / change next codon from UAC to ACC.
Messenger RNA (mRNA) is used during translation to form polypeptides. Describe how mRNA is produced in the nucleus of a cell. (6 marks)
- Helicase;
- Breaks hydrogen bonds;
- Only one DNA strand acts as template;
- RNA nucleotides attracted to exposed bases;
- (Attraction) according to base pairing rule;
- RNA polymerase joins (RNA) nucleotides together;
- Pre-mRNA spliced to remove introns.
Describe the structure of proteins (5 marks)
- Polymer of amino acids;
- Joined by peptide bonds;
- Formed by condensation;
- Primary structure is order of amino acids;
- Secondary structure is folding of polypeptide chain due to hydrogen bonding;
Accept alpha helix / pleated sheet - Tertiary structure is 3-D folding due to hydrogen bonding and ionic / disulfide bonds;
- Quaternary structure is two or more polypeptide chains.
Name the fixed position occupied by a gene on a DNA molecule. (1 mark)
Locus/loci;
Describe how a gene is a code for the production of a polypeptide. Do not
include information about transcription or translation in your answer. (3 marks)
- (Because) base/nucleotide sequence;
- (In) triplet(s);
- (Determines) order/sequence of amino acid sequence/primary structure (in polypeptide);
Define the term exon. (1 mark)
Base/nucleotide/triplet sequence coding for polypeptide/sequence of amino acids/primary structure;
Describe how a phosphodiester bond is formed between two nucleotides within a DNA molecule. (2 marks)
- Condensation (reaction)/loss of water;
- (Between) phosphate and deoxyribose;
- (Catalysed by) DNA polymerase;
Reject if DNA polymerase joins AT/GC
OR
complementary nucleotides/bases
OR
forms hydrogen bonds
The two DNA strands of a particular gene contain 168 guanine bases between them. The relationship between the numbers of guanine bases (G), adenine bases (A), thymine bases (T) and cytosine bases (C) in these two strands of DNA is shown in the following equation.
G = 4(A + T) – C
Use this information and your understanding of DNA structure to calculate the maximum number of amino acids coded by this gene.
Show your working. (2 marks)
Correct answer for 2 marks = 70;;
Accept for 1 mark,
A = 42 and T = 42
OR
420 (total bases in gene)
OR
210 (bases in template strand)
Name the protein associated with DNA in a chromosome. (1 mark)
Histone;
Scientists investigated the genetic diversity between several species of sweet potato. They studied non-coding multiple repeats of base sequences.
Define ‘non-coding base sequences’ and describe where the non-coding multiple repeats are positioned in the genome. (2 marks)
- DNA that does not code for protein/polypeptides
OR
DNA that does not code for (sequences of) amino acids
OR
DNA that does not code for tRNA/rRNA;
Accept the idea of not transcribed for ‘does not code for’.
Do not credit ‘DNA that does not code for an amino acid’.
Ignore reference to introns. - (Positioned) between genes;
Reject (positioned) ‘in introns’ or ‘between exons’.
Accept ‘(Positioned) at the end of chromosomes’ or ‘(Positioned) in the telomeres’.