2018 Euk-Gene-Exp Organisation Control STQ MS

2018 Gene Expression, Organisation and Control STQ MS2018 / H2 / ACJC PRELIM / P2 Q3 1tRNA molecules have an important role in gene expression. They can be found in the cytoplasm, mitochondria and chloroplasts of eukaryotic cells. Fig. 3.1 shows the structure of a tRNA molecule that is able to carry the amino acid threonine in the cytoplasm of a eukaryotic cell. It is 77 ribonucleotides long. Fig. 3.1(a)(i)With reference to Fig. 3.1, describe how the structure of a tRNA molecule differs from the structure of a mRNA molecule.CategorytRNAmRNA1.Presence of double-stranded regions / Overall structureDouble-stranded regions / clover-leaf structSingle-stranded / Linear structure2.Presence of hydrogen bonds / complementary base pairingPresentAbsent3.3’ end3’ CCA sequence3’ poly-A tail4.5’ modified guanosine capAbsentPresent5.AnticodonHas an anticodonMade up of many codons@1m per valid comparison[2](ii)Explain how these differences allow tRNA and mRNA molecules to perform their roles.tRNAmRNAOverall structure (Presence of hydrogen bonds to 1. Specific 3D conformation / clover-leaf structure is complementary to (active site of) 2. Linear structure allows it to be read by the ribosomes / binding to the small ribosomal subunit;

bring about double-stranded regions)aminoacyl-tRNA synthetase / (A, P-sites of) large ribosomal subunit;3’ end3. 3’ CCA sequence allows binding to the amino acid;4. The 3’ poly-A tail determines the stability of the mRNA / duration of translation of the mRNA;Anticodon5. Anticodon can complementary base-pair to codon for the correct amino acid to be added;(1 tRNA can only carry one amino acid)6. Sequence of codons specify the sequence of amino acids in a polypeptide;@1m per structure-function relationship[2]Fig. 3.2 shows a tRNA molecule that is able to carry the amino acid threonine in mitochondria. It is 62 ribonucleotides long. Fig. 3.2(b)The structures of the tRNA molecules in the mitochondria and the cytoplasm of a eukaryotic cell are different although they are both able to carry the amino acid threonine. Suggest a reason for this difference.1.Mitochondria have 70S ribosomes, while 80S ribosomes are found in the cytoplasm;2.Different tRNA molecules have specific 3D conformation that are complementary to the different A and P sites in mitochondrial and cytoplasmic ribosomes; OR3.They are coded for by the different genes found in the mitochondrial DNA and nuclear DNA respectively;4.Different sequences result in the formation of hydrogen bonds at different locations;5.AVP e.g. evolutionary reasons;[2]

There are 61 unique tRNA molecules in the cytoplasm of a eukaryotic cell. (c)(i)Explain why there are 61 unique tRNA molecules in the eukaryotic cell.1.The codon / anticodon is made up of 3 nucleotides and there are 4 different bases, hence 64 possible unique permutations of triplet codes;2.Stop codons do not code for amino acids/ bind to a release factor;3.*There are 3 stop codons, hence this gives 61 unique codons. 61 tRNA molecules will have anticodons that can complementary base pair to these 61 unique codon sequences;Pt 3 is compulsory to score full marks[2](ii)Scientists found that there are more than 120 gene loci coding for tRNA molecules in the nuclear DNA, many of which are duplicate copies of the same tRNA gene.Suggest an advantage of having more than 120 gene loci coding for the 61 tRNA molecules in the eukaryotic cell.1.Rapid synthesis of tRNA molecules;2.If one copy of the tRNA gene is mutated, there are other / duplicate copies to ensure functional tRNA is synthesised;[1](d)Explain how the structure of RNA polymerase allows for the synthesis of tRNA molecules.1.The active site has a specific/complementary 3D conformation, and it binds to the promoter/template strand of the tRNA genes;2.The specific 3D conformation of the active site is also specific/complementary to ribonucleotides and the DNA template;3.Catalytic residues in the active site help to catalyse the formation of the phosphodiester bonds between the 3’ OH group of the one nucleotide and 5’ phosphate group of the other nucleotide;4.Active site contains catalytic residues that break hydrogen bonds between the strands in DNA molecules to expose the template strand;5.Ref to action of enzymes, e.g. lower activation energy, binding sites for general transcription factors;[3] [Total: 12]

2018 / H2 / AJC PRELIM / P2 Q22RNA molecules play important roles within cells. One of the major types of RNA found in all cells is the ribosomal RNA (rRNA).Fig. 2.1 shows rRNA molecules forming the large ribosomal subunit in eukaryotes. Fig. 2.1(a)Explain why the rRNA molecules must adopt the shapes shown in Fig. 2.1. How each shape is formed ( max 2) Single stranded RNA folds back upon itself due to Complementary base-pairing via hydrogen bonds / presence of complementary stretches of bases that can form hydrogen bonds (Forming a double helical structure Purposeto stabilize the molecule Max 2 marks for binding: to form a binding site whose shape is complementary to ribosomal proteins / small ribosomal subunit / tRNA molecules [3]

(b)Another important RNA molecule found in eukaryotic cells is the telomerase RNA. Telomerase RNA is found within the telomerase enzyme, an enzyme essential for elongating telomeres. (i)Outline how RNA molecules such as telomerase RNA and rRNA are synthesised.Transcription of the gene for telomerase RNA/ rRNA by RNA polymerasecatalyzing formation of phosphodiester bonds between RNA nucleotides which were added via complementary base pairing using ( one of the two strand of ) DNA as templateelongation occurs in the 5’ to 3’ direction of the RNA strand ( ½ m each) [3]Fig. 2.2 shows the mode of action of telomerase. Fig. 2.2

(ii)Describe three visible differences between telomere elongation shown in Fig. 2.2 and translation. Telomere elongationTranslation MonomersDNA nucleotides Amino acidBonds formed between monomers Phosphodiester bonds Peptide bonds Enzyme involved in synthesis the bond between monomersTelomerase/ reverse transcriptasePeptidyl transferase Movement of enzymeMoves a distance of 6 nucleotides each timeMoves a distance of 3 nucleotides / 1 codon each time [3][Total:9]2018 / H2 / AJC PRELIM / P2 Q33Eukaryotes regulate the expression of their genes at various levels of protein synthesis. (a)Describe the effect of histone acetylation on gene expression. Any 2when lysine on histones tails are acetylated their positive charges are neutralized . there is decreased interactions of histone tails with neighbouring nucleosomes / decreased interaction of histone tails with the (negatively charged) DNA / DNA less tightly wound around histones Chromatin structure become less compact/ condensed / forms euchromatin Compulsory pt– to answer “ effect on effect on gene expression) Transcriptional machinery like RNA polymerase (reject DNA polymerase) and transcription factors (must give these two egs) will can bind to the promoter of genes in the acetylated region to / form the transcription inititation complex/ Allow transcription to occur / gene to be transcriptional active [3](b)Eukaryotic gene expression can also be regulated at translation initiation after the mRNA is synthesised.Fig. 3.1 shows translation occurring on an eukaryotic mRNA.

Fig. 3.1(i)Explain the significance of the pattern of translation, labelled A in Fig. 3.1.Identify A: Polyribosomes / polysomes/ multiple ribosomes work on translating one messenger RNA Significance of A: single mature mRNA is used to make many copies of the same / identical polypeptide simultaneously / enables a cell to synthesise many copies of the same type of polypeptide very quicklysingle mature mRNA is used to make many copies of the same (type) of polypeptide simultaneously / enables a cell to synthesise many copies of the same type of po