Module Code: BIO00007C

Transcription

1 Examination Candidate Number: Desk Number: BSc and MSc Degree Examinations Department : BIOLOGY Title of Exam: Genetics Time Allowed: 1 Hour 30 Minutes Marking Scheme: Total marks available for this paper: 50 The marks available for each question are indicated on the paper Instructions: Answer all questions in the spaces provided on the examination paper Materials Supplied: CALCULATOR For marker use only: Office use only: Total as % DO NOT WRITE ON THIS BOOKLET BEFORE THE EXAM BEGINS DO NOT TURN OVER THIS PAGE UNTIL INSTRUCTED TO DO SO BY AN INVIGILATOR Page 1 of 11

2 1) A female from a species with a diploid number of chromosomes 2n=6 is generating oocytes. a) How many chromosomes will be present in a cell going through anaphase of meiosis I? ( 1 mark) 6, in anaphase I homologues are separating but still inside a single cell. b) How many different gametes can this female generate from the random arrangement of tetrads at the metaphase plate? ( 1 mark) 2^3= 8 c) In a particular meiosis, one tetrad failed to attach to the spindle fibres, but meiosis was not interrupted. Predict the chromosomal number of the gametes produced. ( 2 marks) 4 and 2 chromosomes d) The DNA content of a somatic cell in G1 is 35 pg. What would be the DNA content of the cell at anaphase of mitosis? ( 1 mark) 35X2= 70pg 2) In a certain breed of dog, fur color can be black (B) or brown (b) and temperament can be vicious (G) or shy (g). B and G are the dominant alleles. a) What would be the expected phenotypic ratio of the progeny of a GgBb x ggbb test cross? ( 2 marks) (1 mark for ratios, 1 mark for phenotypes) 1:1:1:1 (Phenotypes: Black Vicious: Black shy: brown Vicious: brown shy) Fork-lined method for phenotypes from the GgBb x ggbb cross: Fur Temperament Page 2 of 11

3 1/2 Black ½ Vicious ½ brown ½ Shy ¼ Black Vicious, ¼ Black shy, ¼ brown Vicious, ¼ brown shy b) If the cross above produced 1000 progeny from many parental pairs, determine the expected number of individuals with the following genotypes: ( 2 marks) GGbb: ggbb: ggbb: GgBB: Fork-lined method for genotypes from the GgBb x ggbb cross: Gene 1 Gene 2 1/2 Gg ½ Bb ½ gg ½ bb ¼ GgBb, ¼ Ggbb, ¼ ggbb, ¼ ggbb GGbb: 0 ggbb: 250 ggbb: 250 GgBB: 0 3) In Drosophila crosses involving the two loci A and B (dominant alleles A and B; recessive alleles a and b) located 24cM apart, predict the diploids that will be obtained in 800 progeny from the test cross below. Fill in the gamete genotypes and diploid progeny numbers on the table. (4 marks) Page 3 of 11

4 Ab//aB (female) x ab//ab (male): Female Gametes Ab ab AB ab Male gametes ab diploid number = 608/2=304 diploid number = 608/2=304 diploid number = 12%x800=96 diploid number = 12%x800=96 4) The pedigree below shows a neurodegenerative disease segregating. a) What is the pattern of inheritance? ( 1 mark) Autosomal dominant b) Using D for dominant and d for recessive, what is the probability of 3.1 being: ( 2 marks) dd? Dd? dd? 0 Dd? 2/3 Page 4 of 11

5 5) a) A part of the template strand of a DNA molecule that codes for the 5 end of an mrna has the sequence: 3 -TTTTACGGGAATTAGATGCGCAGGACG-5. What is the sequence of the corresponding mrna and what is the amino acid sequence of the polypeptide encoded by this region assuming that the normal start codon is used? A genetic code table is provided. (2 marks) mrna: 5 -AAAAUGCCCUUAAUCUACGCGUCCUGC-3 Polypeptide: Met-Pro-Leu-Ile-Tyr-Ala-Ser-Cys Mostly correct answers though many started translation at the start of the mrna rather than the AUG and included a Lys at the start. Some missed out the methionine. b) In E.coli, a mutation in a particular trna gene has resulted in a change in the anticodon from 5 -GGG-3 to 5 -GGA-3. How will the mutation affect which codon is recognised by this trna during translation? You should use the genetic code table for your answer. (2 marks) The wild-type trna recognised a Pro codon (5-CCC-3 ) but the mutant version will recognise Ser (5 -UCC 3 ). Many forgot about the requirement for antiparralel paiting and didn t realise that the 5 -GGA-3 anticodon on the trna would pair with 5 -UCC-3. Page 5 of 11

6 6) Provide a short definition for each of the following terms. (5 marks) a) Shine-Dalgarno sequence Ribosome binding site on prokaryotic mrna Mostly correct answers though some confused this with the binding site for RNA polymerase b) Poly(A) tail A chain of ~200 adenine nucleotides that is added to the 3 end of eukaryotic mrnas Mostly correct answers though detail was variable c) Constitutive mutant A mutation that results in constant expression of a gene Some lack of clarity in the explanation but many correct answers d) Polycistronic A mrna that codes for multiple proteins Some lack of clarity in the explanation and confusion with what is an open reading frame. e) Peptidyl transferase Page 6 of 11

7 Catalytic activity of the ribosome that forms peptide bonds during translation Some good definitions but many confused this with other processes 7) Predict the consequence for expression of the lac operon in the following scenarios. Provide a brief explanation for your answer. a) Adenylate cyclase is inactivated. (2 marks) As camp would not be formed there would be no positive regulation (1) and so expression would be low even when glucose levels are low and lactose is high (1). Generally good answers though it was clear that some hadn t revised the lac operon material b) There is a frameshift mutation near the 5 end of the laci gene. (2 marks) This is likely to inactivate the laci repressor (1) and result in constitutive expression of the lac operon (1). Generally good answers though it was clear that some hadn t revised the lac operon material. Some confusion regarding the 5 end of the gene. This is the start not the end. 8) You have identified a new prokaryotic operon that is activated when bacteria encounter temperatures lower than 10 o C. The operon has three structural genes ( brra, brrb & brrc ) and is controlled by an activator protein (ICE), which is encoded by a gene that is separate from the operon. The ICE protein is present in the cell irrespective of temperature levels. a) Where in the operon do you expect the ICE protein to bind? (1 mark) In the promoter region. Generally good answers though some mentioned operator sequences b) Suggest a possible mechanism to explain why expression of the operon is induced at low temperature. (2 marks) The activator protein may adopt different conformations depending on temperature (1) such that it is able to bind to the promoter only at low temperatures (1). Page 7 of 11

8 Generally good answers, some hadn t understood that ICE is an activator and not a repressor though partial credit was given. Some very complicated explanations. c) Suggest a mutation that could result in expression of the brra, brrb & brrc genes at temperatures higher than 10 o C. (2 marks) Gain-of-function mutation (mis-sense mutation) in the ice gene such that the ICE protein is active even in the absence of cold temperatures (1). The amino acid change may mimic the conformational change that the ICE protein undergoes in cold (1). Generally good answers. Credit was given even if ICE was incorrectly identified as a repressor in part (b) 9) You are provided with purified plasmid DNA (cloning vector) and purified human genomic DNA containing the sequence of interest (see pictures below). Page 8 of 11

9 After PCR amplification of your sequence of interest, you insert it into the SmaI restriction site of the cloning vector and transform E.coli with your ligation products. a) Explain how you would identify those bacteria that have incorporated the vector with the insert (your sequence of interest). (6 marks) Plating of the bacteria on agar plates with ampicillin (1) kills bacteria that have not incorporated the vector (1). Blue/white selection allows you to distinguish E.coli colonies that carry an empty vector from colonies that carry a vector with an insert (1). In the empty vector, the lacz gene is functional, whereas its function is disrupted if an insert is present in the multiple cloning site (1). After treatment of the colonies with X-gal, the beta-galactosidase enzyme encoded by the functional lacz gene converts X-gal into a blue product (1), making colonies with an empty vector appear blue, and colonies carrying a vector with insert white (1). Many good answers. In some cases, only blue/white selection was described, but not treatment with ampicillin. Some students suggested restriction enzyme digests, which some credit was given for. b) Digestion with which restriction enzyme would allow you to determine the orientation of the insert in the vector? (1 mark) Page 9 of 11

10 BamHI Many correct answers. c) Which DNA fragment sizes would you obtain for the two insert orientations following restriction enzyme digestion? You can ignore the distances between the restriction sites in the multiple cloning site. (2 marks) Plasmid with insert (one orientation): 4.5 kb kb (1) Plasmid with insert (other orientation): 5.5 kb kb (1) Many correct answers, but several students only listed the fragments for one orientation, or only listed the shorter fragment obtained for each orientation. 10) You carry out Sanger DNA sequencing to confirm the sequence of a DNA fragment. The expected sequence of your DNA template strand and the sequence of the primer you use are shown below: Template strand: 5 TAGCGCATACTAATGCGTTTAACG 3 Sequencing primer: 5 CGTTAAACG 3 For your sequencing reaction, you use small amounts of fluorescently labelled ddntps (dideoxynucleotides). Each of the four ddntps is labelled with a different fluorophore. a) Write down the expected sequences of the three shortest DNA fragments produced in your sequencing reaction. Indicate 5 and 3 ends, and the position of labelled ddntps. (4 marks) 5 CGTTAAACG C 3 5 CGTTAAACGC A 3 5 CGTTAAACGCA T 3 Marks awarded for: - Inclusion of primer sequences (1) - Correct 5 and 3 ends (1) - Correct sequences (1) - Correct positions of labelled ddntps (1) Some correct answers. In many cases, the primer sequence was not included. Some confusions with 5 and 3 ends, and the number of incorporated ddntps. b) Explain the steps following DNA synthesis that allow you to determine the sequence of the template strand. (3 marks) Page 10 of 11

11 DNA gel electrophoresis (1) separates the fragments by size (1), and lasers/detectors are used to determine which labelled ddntp each fragment has incorporated (1). Many answers were awarded full marks. Some confusion with gene expression; several students described aspects of transcription. Page 11 of 11