SLE254-T2-2023-EoUA

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SLE254 Genetics and Genomics Trimester 2, 2023 Summative Assessment Task 4 End of Unit Assessment SLE254 Genetics and Genomics Online EoUA 1 Trimester 2, 2023 Special Instructions This EOUA is OPEN BOOK and is open for 24 HOURS. Calculators are ALLOWED. You have 15 minutes reading time You should spend 2 HOURS writing your responses. You have 15 mins to ensure that your exam responses are formatted correctly and uploaded to the relevant assignment dropbox in CloudDeakin. This EUoA comprises 10 questions. You are required to answer ALL questions in short answer format, providing your answers in this document as instructed below. Please use this given template for your written responses making sure that each answer corresponds to the correct question. Please type your answer to each question in a space below the questions; that is, use the Enter key to create space and type your responses in the region indicated “ANSWERs – type here”. Question 1= 6 marks, question 2= 4 marks, question 3= 9 marks, question 4= 4 marks, question 5= 4 marks, question 6= 8 marks, question 7= 10 marks, question 8 = 6 marks, question 9= 4 marks, question 10= 5 marks. Total = 60 marks. This EoUA constitutes 40% of your assessment in this unit. If you encounter any technical issues with CloudDeakin, please immediately contact the IT Service Desk online or via phone (1800 463 888; +61 3 5227 8888 if calling from outside Australia) and record your ticket number as evidence of technical issues during the examination period. The completed EoUA document must be submitted via the relevant assignment dropbox in CloudDeakin. 1

SLE254 Genetics and Genomics Trimester 2, 2023 Summative Assessment Task 4 End of Unit Assessment Question 1. Dragon breeding (6 marks) You are the Master of Dragon Island, and you decide to breed more dragons to keep your island safe from intruders. Whilst breeding your dragons you discover some can breathe fire while others cannot, and some have normal wings and can fly while other have ragged wings and cannot fly. You suspect that fire breathing and ragged wings might be single gene traits that sort independently and so you breed your dragons until you are sure you have true breeding strains of fire breathing, normal wing dragons and no fire breathing, ragged wing dragons. When you cross these two strains all the offspring in this F1 are fire breathing and normal wings. You then cross F1 and the type and number of offspring are listed in the table below. Use the information below to determine if the traits are single gene traits following Mendelian inheritance. You must show all details of your workings out, your null hypothesis and degrees of freedom to get full marks. Normal wings, fire breathing 817 Normal wings, no fire 310 Ragged wing, fire breathing 273 Ragged wings, no fire 114 Fire Breathing No Fire Breathing Total Normal Wings 817 310 1127 Ragged Wings 273 114 387 Total 1090 424 1514 Expected value for Normal Wings and Fire Breathing (Total Normal Wings * Total Fire Breathing) / Total Dragons = (1127 * 1090) / 1514 ≈ 812.38 Expected value for Normal Wings and No Fire Breathing (Total Normal Wings * Total No Fire Breathing) / Total Dragons = (1127 * 424) / 1514 ≈ 315.61 2

SLE254 Genetics and Genomics Trimester 2, 2023 Summative Assessment Task 4 End of Unit Assessment Expected value for Ragged Wings and Fire Breathing: (Total Ragged Wings * Total Fire Breathing) / Total Dragons = (387 * 1090) / 1514 ≈ 278.61 Expected value for Ragged Wings and No Fire Breathing: (Total Ragged Wings * Total No Fire Breathing) / Total Dragons = (387 * 424) / 1514 ≈ 108.38 Chi Square Calculation: χ² = [(817 – 812.38)² / 812.38] + [(310 – 315.61)² / 315.61] + [(273 – 278.61)² / 278.61] + [(114 – 108.38)² / 108.38] χ² ≈ 0.000568 + -314.627 + -277.63 + -107.32 χ² ≈ -699.57 Df=(2-1)*(2-1) Df=1 For df = 1 and α = 0.05, the critical value is approximately 3.841. In this case, χ² (-699.57) is less than the critical value (3.841) for df = 1 and α = 0.05. Therefore, we fail to reject the null hypothesis (H0), suggesting that the traits of fire breathing and wing type assort independently as single gene traits following Mendelian inheritance. Question 2. Nucleic Acid structure (4 marks). The diagram below shows a 2’-deoxyribose sugar, which together with a phosphate group (PO 4 3- ) and either a purine or pyrimidine nitrogen-containing base forms a nucleotide. For each of the letters shown (A-E), number the carbons 1’ to 5’ and then describe the role of each carbon (Is it attached to particular component? Is it involved in polymerisation etc) in a strand of DNA. You may draw on the diagram, enter text boxes or answer below. 3

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SLE254 Genetics and Genomics Trimester 2, 2023 Summative Assessment Task 4 End of Unit Assessment A: Carbon 1' (1' Carbon) - This carbon is the attachment point for the nitrogenous base. It forms a covalent bond with the nitrogenous base. B: Carbon 2' (2' Carbon) - This carbon is involved in the deoxyribose sugar's unique structure. It lacks the oxygen atom found at the 2' position in ribose sugar, hence the name "2'-deoxyribose." C: Carbon 3' (3' Carbon) - This carbon is involved in polymerization during the formation of the DNA strand. It forms a covalent bond with a phosphate group (PO43-) from the adjacent nucleotide in the DNA chain through a phosphodiester bond. This linkage forms the sugar-phosphate backbone of the DNA strand. D: Carbon 4' (4' Carbon) - This carbon is also involved in the deoxyribose sugar's unique structure. E: Carbon 5' (5' Carbon) - This carbon is attached to a phosphate group (PO43-) and forms the other end of the sugar-phosphate backbone in the DNA strand. The 5' carbon of one deoxyribose sugar is linked to the 3' carbon of the next deoxyribose sugar in the DNA chain, creating a linear polymer. Question 3: DNA replication (9 marks) DNA replication occurs naturally in the body while PCR is a laboratory technique used to amplify desired DNA. Below are two figures, one DNA replication (A) and PCR amplification (B). From specific points, arrows lead to numbers. Answer the questions relating to the locations specified by the numbers. DNA replication (A) 1. Which end (5 or 3 ) of the molecule is here ′ ′ (0.5 marks) ? The 5’ end of the molecule is there. 2. Which end (5 or 3 ) of the molecule is here ′ ′ (0.5 marks) ? The 5’end molecule is here. 3. Which enzyme is functioning here to deal with supercoils in the DNA (0.5 marks) ? Topoisomerases 4. Which enzyme is functioning here to unwind the DNA (0.5 marks) ? DNA helicase 5. Is this strand the leading or lagging strand (0.5 marks) ? It is a lagging strand. 6. Which nucleic acid is depicted here? What are the short DNA fragments usually called (1 mark)? The nuclic acid depicted here is primer and the short DNA fragments are called Okazaki Fragments. 7. Which enzyme functions here to couple these two newly synthesized fragments of DNA (0.5 marks) ? 4

SLE254 Genetics and Genomics Trimester 2, 2023 Summative Assessment Task 4 End of Unit Assessment DNA ligase 8. What are the three steps involved in DNA replication (1.5 marks) ? initiation, elongation, and termination. PCR amplification (B) 1. What process separates the DNA strands (0.5 marks) ? Denaturation 2. What occurs in the second step of PCR amplification (0.5 marks) ? Annealing of primers to each original strand for new strand synthesis. 3. What are the fragments attaching to the DNA strand (0.5 marks) ? Okazaki fragments 4. And in what direction are they (5’-3’ or 3’-5’) (0.5 marks)? The direction is going through the 5’3. 5. What enzyme is allowing them to be attached to the target DNA (0.5 marks) ? DNA polymerase 6. What PCR reagent is added that comprises the growing DNA strand (0.5 marks) ? deoxynucleotide triphosphates (dNTPs) ` 7. What occurs in the third stage of PCR amplification (0.5 marks) ? Extension-Once joined together, they form a new complementary strand of DNA (termed extension of the DNA). 5

SLE254 Genetics and Genomics Trimester 2, 2023 Summative Assessment Task 4 End of Unit Assessment Question 4: Transcription/translation (4 marks) The accompanying figure represents simultaneous transcription and translation in E. coli . The direction of the RNA polymerase is given by the arrow. a. Is the letter A nearer the 5 or the 3 end of the molecule? ′ ′ The letter A is nearer to the 5’ end of the molecule. b. Is the letter B nearer the 5 or the 3 end of the molecule? ′ ′ The letetr B is nearer to the 3’ end of the molecule. c. Is the letter C nearer the 5 or the 3 end of the tRNA molecule? ′ ′ The letter C is nearer to the 3’ end of the molecule. d. What is the function of the large rRNA that is closest to the letter D? The ribosome is the large rRNA and its function is to synthesize proteins in accordance with the chains of amino acids. 6

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SLE254 Genetics and Genomics Trimester 2, 2023 Summative Assessment Task 4 End of Unit Assessment Question 5: Telomeres (4 marks) Imagine you are a researcher interested in telomere shortening and aging. You run a series of experiments with the DNA shown below. You stained the telomeres using Fluorescent In situ Hybridisation (FISH), and you also profile histone methyl group modifications between older and younger sources of DNA. However, you fail to generate any results. What is a likely explanation for your experimental failure? How would you improve your experimental outcomes? There are many factors that may affect the results which leads to experimental failure such as DNA quality, Experimental Design and the size of the sample.To improve results, we need to plan, optimize and collaborate to help perceive better results in telomeres and aging. 7

SLE254 Genetics and Genomics Trimester 2, 2023 Summative Assessment Task 4 End of Unit Assessment Question 6: Chromosomal abnormalities (8 marks) Above is an image showing the chromosomes of a healthy adult human male. Below is a selection of karyotypes showing chromosomal abnormalities. For each abnormality, name the type of abnormality (1 mark) and their identifying feature (1 mark) . Please note, some are male, and some are female (8 marks total) . A. 8

SLE254 Genetics and Genomics Trimester 2, 2023 Summative Assessment Task 4 End of Unit Assessment ANSWER – type here B. C. This image has a Karyotype with a polyploidy; 92,XXXX. 9

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SLE254 Genetics and Genomics Trimester 2, 2023 Summative Assessment Task 4 End of Unit Assessment D. 10

SLE254 Genetics and Genomics Trimester 2, 2023 Summative Assessment Task 4 End of Unit Assessment Question 7: Recombinant DNA technology I (10 marks) Describe the process of creating a genetically modified bacterium using a plasmid as a vector. Describe α- complementation in E. Coli lacZΔM15 mutants in blue white screening and the purpose for this this process. Creating a genetically modified bacterium using a plasmid as a vector is a very common technique in molecular biology. Fisrt thing we need to do is choose a p;lasmid vector that is suitable for your purpose which then the genbe of interest can be inserted into the plasmid and then ligate them together. The transformation is introduced by containing the GOI into the bacterium. The selection and screening is given to identify bacteria that took up the plasmid and then the culturing and expression allows the GOI to be expressed. The harvesting of the genetically modified bacteria is being taken place and then the characterization confirms that there is the presence as well as the expression of the GOI. a-complementation is a technique used in E.coli for blue-white screening to identify recombinant plasmids carrying the GOI. The lacZ gene produces an enzyme that cleaves sugar lactose into glucose and galactose the lacZΔM15 Mutant is deleted making it nonfunctional. In the plasmid a section of the lacZ gene carrying the peptide is inserted. Blue-white screening is used to identify recombinant plasmids containing GOI so they plate the E.coli on agar plates. The blue colonies is bacteria carrying non-recombinant plasmids will produce β-galactosidase where as white colonies that carry recombinant plasmids with GOI produce an inactive β-galactosidase due to the lacZ peptide. The purpose of blue-white screening is to identify colonies that contain recombinant plasmids wityh GOI. This allows researchers to distinguish between colonies that caryy and those that do not making the selection of recombinant clones more effectual in molecular experiments. Question 8: Recombinant DNA technology II (6 marks) The Clustered Regularly Interspaced Palindromic Repeats (CRISPR)/Cas9 system is a revolutionary new gene-editing technology that allows errors in the genome to be corrected and genes in cells and organisms to be turned on or off quickly, cheaply and with relative ease. The technology is adapted from the natural defence mechanisms employed by a wide range of prokaryotes against invading viruses. Describe the process for each of the numbers in the diagram below. You may draw on the diagram, enter text boxes or answer below: 11

SLE254 Genetics and Genomics Trimester 2, 2023 Summative Assessment Task 4 End of Unit Assessment Stage 1: Foreign DNA acquisition-is a piece of DNA that is isolated from the desired organism and inserted into a vector for cloning where it combines with vector DNA. Stage 2:CRISPR processing - creating double-stranded breaks in the DNA and then taking advantage of cellular DNA repair pathways. Stage 3:RNA guided targeting of element- enables a single protein or protein complex to target multiple sites. Question 9: Quantitative genetics (4 marks) The formula used is 1/4 n . In a certain plant, height varies from 3 to 30 cm and is inherited as a polygenetic continuous trait (Quantitative trait). When 3-cm and 30-cm plants are crossed, the ratio of F 2 individuals resembling either of the two extreme phenotypes (short or tall) can be estimated based on the number of polygenes ( n ) contributing. If the ratio of F 2 individuals being small (3 cm) is 3 out of 3072, then how many polygenes are involved? Show your calculations for full marks. n=3 3/3072=1/1024 1/3072=1/4n 4n=1/(3/3072) n-log4(1/(3/3072)=4 Therefore there must have been 3 polygenes involved 12

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SLE254 Genetics and Genomics Trimester 2, 2023 Summative Assessment Task 4 End of Unit Assessment Question 10. Genomics (5 marks). Disciplines in genomics encompass several areas of study, including transcriptomics, comparative genomics, and metagenomics, and have led to an “omics” revolution in modern biology. Imagine, you are a scientist working on a drug for lycanthropy which currently shows limited efficiency across a broad group of patients. How could you improve the current treatment options by making them more personalised. Discuss a personalised response from lab bench to bedside. Personalising the treatment for lycantropy involves the clinical history of the individual patients. The lab bench couls start conducting genetic profiling of patients with lycantropy and the bedside will use the data to categorize patients into subgroups based on their lycantropy-related markers. The lab bench will identify biomarkers specific to lycantropy and the bedside will assess the disease severity and progression in each patient. The targeted therapies on the lab bench develop targeted therapies that address mechanism underlying lycantropy in each patient whereas the bedside can manage the tretmenat to each patient based on their genetic biomarker profiles. The pharmacogenomics on the lab bench couls study lycanthropy treatments to understand drug metabolism and the bedside use pharmacogenomic data to optimize drug selection. The data integration can use clinical data to develp models for trewatment outcomes where as the bedside will use the models to make decisions about the treatment regimes for each patient.The continuous monitoring can be done by the bedside people by tracking the patients progress through various techniques and follow the privacy guidelines when handling patients clinical records. END OF ASSESSMENT 13

SLE254-T2-2023-EoUA

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