Examination Candidate Number: Desk Number: BA, BSc, and MSc Degree Examinations 2017-8 Department : BIOLOGY Title of Exam: Genetics Time Allowed: 1 hour and 30 minutes Marking Scheme: Total marks available for this paper: 50 Instructions: Answer all ques ons in the spaces provided on the examina on paper The marks available for each ques on are indicated on the paper Materials Supplied: CALCULATOR For marker use only: Office use only: 1 2 3 4 5 6 7 8 9 10 11 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 12
Answer all questions in the spaces provided 1. This image below was taken during the practical analysing spermatogenesis in the male Locust (22, X0). a) Specify the phase and subphase(s) this picture may have been taken at. Phase: Prophase I Subphases: These are only seen during transition from pachytene, diplotene and diakenesis (will accept any of them) b) What do the arrows point at and what do they indicate has happened? Chiasmata They indicate points where DNA exchanges between non-sister chromatids may have occurred. c) What would the chromosomal content of the gametes be? 11 or 11, X LO: Describe how chromosomes behave during meiosis page 2 of 12
2. A pure breed of normal size albino pigeons arrive at an island and successfully reproduce with the local population, which is also a pure breed characterized by grey feathers and large body size. F1 individuals all show grey feathers and normal size. a) Using the letters G (grey feathers) and g (albino feathers) to indicate the colour of the feathers and L (normal) and l (large) to indicate the body size, write the genotype of the parents of this cross and the F1. P: ggll X GGll F1: GgLl b) If F1 pigeons mate amongst themselves, assuming these two genes are on different chromosomes, how many albino pigeons of large size would be expected out of 80? Show your working. (2 marks) GgLl x GgLl Albino ¼ X large size ¼=1/16 (1) 80/16=5 (1) c) The colonizers have a unique singing ability and display a fantail whereas the local population do not. All F1 pigeons have the unique singing ability but not the fantail. Using the symbols N (dominant) or n (recessive) for singing ability and D (dominant) or d (recessive) for fantail, write the genotypes for these two genes for the parents (P) and the F1 of this cross. P: NNdd nndd F1: NnDd d) Pigeons from the F1 generation that mated with double homozygous recessive pigeons (nndd) produced 100 progeny pigeons that were all classified as Nndd (50%) and nndd (50%). What would you conclude from this? (2 marks) NnDd X nndd The F1 only produces gametes that are Nd and nd, the same as their page 3 of 12
parents, indicating these genes are linked on the same chromosome at a very short distance from each other. LO: Apply basic probability rules to calculate the likelihood of genetic outcomes in Mendelian Genetics 3. In a test cross in maize between a triple heterozygous plant and a triple homozygous recessive plant 850 plants were obtained. The alleles are arranged on the chromosomes as indicated below. v -----22.3cM------ pr ----------43.4cM----------- bm Considering the genetic distances between the genes: a) Calculate the number of single crossover plants between pr and bm that would be expected from the cross above. Show your working. (3 marks) Probability of double crossovers: 0.223X0.434= 0.096782=9.6782% Percentage of single crossover pr-bm= 43.4-9.6782=33.7218% Number of plants out of 850=850 X 33.7218%=287 b) Write the alleles of the recombinant chromosome of these single crossover classes. v pr+ bm and v+ pr bm+ LO: Explain the relationship between meiotic events and phenotypic classes for linked genes page 4 of 12
4. Study the pedigree below and answer the following questions: a) What is the pattern of inheritance? Autosomal recessive b) Write the genotypes that can be ascertained using letters that can clearly distinguish the mutant from the normal allele. (2 marks) I: Aa, Aa, aa, Aa II: A_, aa, aa, Aa, Aa, aa III: aa, A_ (2 marks for full answer, 0.5 deducted per mistake) c) What is the probability of individual III.2 being a carrier for this disease? 2/3 5. Provide a short definition for each of the following terms: a) Open Reading Frame Sequence of DNA consisting of triplet codons that can be translated into amino-acids starting with an initiation codon and ending with a stop codon Surprisingly few accurate definitions given the number of times the term was used in the module. b) Promoter Region of DNA where RNA polymerase binds to initiate transcription A common mistake. was to suggest the promoter is a protein c) Operon page 5 of 12
Cluster of genes under the control of a single promoter and transcribed as a single mrna A common mistake was to confuse operon with operator. LO: Describe the basic mechanisms of transcription and translation in prokaryotes and eukaryotes 6. A segment of a polypeptide chain is Arg-Gly-Ser-Phe-Val-Asp-Arg. It is encoded by the following segment of DNA: -GGCTAGCTGCTTCCTTGGGGA- -CCGATCGACGAAGGAACCCCT- a) Which is the template strand? A genetic code table is provided. b) Label the end of each strand with the correct 5 and 3 polarity. 3 -GGCTAGCTGCTTCCTTGGGGA- 5 5 -CCGATCGACGAAGGAACCCCT-3 LO: Describe the basic mechanisms of transcription and translation in prokaryotes and eukaryotes LO: Solve problems to demonstrate an understanding of how gene expression is controlled. Mostly correct answers. page 6 of 12
7. You are using mutants to study the regulation of the lac operon in E.coli. Provide explanations for the following mutant phenotypes. a) Mutant 1 produces high levels of the lac structural proteins even in the absence of lactose. DNA sequence analysis of the promoter and operator sequences of the lac operon shows no difference to wild-type. (2 marks) This is likely to be a loss-of-function mutation in the laci repressor gene so that it fails to bind to the operator sequence and repress transcription. b) Mutant 2 produces lower levels of lac structural proteins in the presence of lactose and absence of glucose than wild-type. DNA sequence analysis of the promoter, operator and structural genes of the lac operon shows no difference to wild-type. (2 marks) This could be a loss-of-function mutation in the CAP activator protein - the lac operon is released from repression by lactose but fails to be activated by the absence of glucose Mostly correct answers for both parts a and b. Answers that showed logical thinking were credited - even if they were different to the model answer given. LO: Solve problems to demonstrate an understanding of how gene page 7 of 12
expression is controlled. 8. Consider the pedigree below that shows the occurrence of achondroplasia (dwarfism/short stature) in a family. Achondroplasia can be caused by mutations in the FGFR-3 gene. DNA sequencing of the FGFR-3 gene was undertaken for some of the individuals and two different alleles were identified. Partial sequence of the relevant area of these alleles is shown below with the correct reading frame of the coding strand indicated. Allele A: ATC-CTC-AGC-TAC-GGG-GTG-GGC-TTC-TTC Allele B: ATC-CTC-AGC-TAC-AGG-GTG-GGC-TTC-TTC Individuals II-2 and II-3: Both have two copies of allele A Individual III-1: One copy of allele A and one copy of allele B Individual III-2: Two copies of allele A Individuals IV-1 and IV-4: Both have one copy of allele A and one copy of allele B Individuals IV-2 and IV-3: Both have two copies of allele A. a) Which of the following provides the most likely explanation for the mode of inheritance of achondroplasia shown in the pedigree assuming that all parentage is confirmed? i) That inheritance is autosomal recessive page 8 of 12
ii) That inheritance is autosomal dominant but that generations I and II have been diagnosed incorrectly Mostly correct answers iii) That inheritance is autosomal dominant but has arisen spontaneously in individual III-1 b) Does the sequence difference between alleles A and B correspond to a mis-sense, nonsense, frame-shift or silent mutation? The genetic code table provided for question 6 can be used here. Missense. Mostly correct answers c) The normal physiological function of the FGFR3 protein is to negatively regulate bone formation. Do you expect the protein encoded by allele A to be more or less active than that encoded by allele B and therefore should the disease be described as being caused by a gain-of-function or loss-of-function mutation? (2 marks) Allele A will be less active. The disease is gain-of-function. Mostly correct answers though some had not understood that allele B was the disease state. LO: Define what is a gene and understand how genotype relates to phenotype LO: Describe the basic mechanisms of transcription and translation in prokaryotes and eukaryotes LO: Solve problems to demonstrate an understanding of how gene expression is controlled. 9. You are given a DNA sequence of several thousand kilobases. List four features that would indicate that the sequence is obtained from a prokaryote such as E.coli rather than a higher eukaryote such as a human? (4 marks) Genes would be closely spaced together No introns Presence of operons Lack of repetitive sequences Other possibilities accepted Many got some but not full marks. There were some answers that were too vague to gain credit (eg. just saying less complex without explaining what this means). There were also some alarming errors - eg. suggesting that prokaryotes have single-stranded rather than double-stranded DNA or that page 9 of 12
prokaryotes have uracil instead of thymine). Other answers suggested differences between prokaryotes and eukaryotes that would not be apparent just from the DNA sequence (eg. differences in DNA packaging). LO: Compare and contrast the content and organization of prokaryotic and eukaryotic genome 10. Describe how it was shown that DNA is the transforming agent in bacterial transformation. (4 marks) Two bacterial strains were used, one lethal to mice, one non-lethal (1). When the lethal strain was heat-inactivated and mixed with the non-lethal strain, mice injected with the mix died, showing that transformation of the non-lethal strain had taken place (1). To discover what the transforming agent was, different components of the inactivated lethal strain were destroyed before mixing with the non-lethal strain, and then injected into mice (1). Only when the DNA of the lethal strain was destroyed, the injected mice survived (1). LO: Compare and contrast the content and organization of prokaryotic and eukaryotic genome Many answers were awarded full marks. Some students did not include the inactivation of different components in their answers. Several confusions with experiments on conjugation and transduction. 11. You are provided with purified plasmid DNA (cloning vector) and purified human genomic DNA containing the sequence of interest (see pictures below). page 10 of 12
a) Describe how you could insert the genomic sequence of interest into the cloning vector, so that the region labelled y of the sequence of interest is adjacent to the region labelled x of the cloning vector (BamHI, EcoRI and HindIII produce cohesive ends). (6 marks) Amplification of the sequence of interest by PCR (1), using a primer with an overhang containing a EcoR1 restriction site for the primer at y (1), and a primer with an overhang containing a HindIII site for the other end of the sequence of interest (1). Cutting the plasmid (1) and PCR product with EcoR1 and HindIII (1), ligating PCR fragment into the plasmid using DNA ligase (1). Only very few answers were awarded full marks. Many students suggested insertion into the EcoRI site, but this would allow the insert to integrate in either orientation. Several students suggested using BamHI, which cannot be used as it cuts inside the sequence of interest. b) After transformation of E.coli with the resulting DNA and isolation of plasmid DNA from several E.coli clones, you carry out diagnostic restriction digests. List the DNA fragment sizes that you would obtain. page 11 of 12
(4 marks) Vector + insert, digested with BamHI: Vector only, digested with BamHI: Vector + insert, digested with EcoRI: Vector only, digested with EcoRI: Vector + insert, digested with BamHI: 4.5kb + 1.5kb Vector only, digested with BamHI: 4kb Vector + insert, digested with EcoRI: 6kb Vector only, digested with EcoRI: 4kb Many good answers. Answers to a) were considered when awarding marks. c) Explain how you could use PCR to confirm that the insert was inserted into the vector. (2 marks) Carry out PCR with one primer in the vector, one primer in the insert (1). You would only get a PCR product if the insert is present in the vector (1). Some answers were awarded full marks. Many students suggested blue/white selection or diagnostic digests, but these approaches don t require the use of PCR. LO4 Describe techniques of recombinant DNA technology and solve problems to demonstrate understanding page 12 of 12