BIO 2110 Exam 1 answers.1. What RNA sequence would be transcribed from the DNA region shown below? Starting with the first nucleotide, what sequence of amino acids is coded for in the RNA? [refer to the genetic code attached to the end of this exam] Template strand 5’ – ATGCGACTTAGCATAGACAGACAGACGTAAGTTAGACAT – 3’ ||||||||||||||||||||||||||||||||||||||| Coding strand 3’ – TACGCTGAATCGTATCTGTCTGTCTGCATTCAATCTGTA – 5’ RNA = reverse complement of template strand (with T  U). 5’ AUG UCU AAC UUA CGU CUG UCU GUC UAU GCU AAG UCG CAU 3’ protein  Met Ser Asn Leu Arg Leu Ser Val Tyr Ala Lys Ser His2. The C-terminal domain of the large subunit of RNA polymerase II contains multiple repeats of the amino acid sequence, Tyr-Ser-Pro-Thr-Ser-Pro-Ser. How are these repeats involved in mRNA processing? Ser-5 residues are phosphorylated at the near the promoter. These phosphor-Ser-5 residues recruit capping enzymes that add the 7Methyl triphosphoGuanine cap to transcripts as they emerge from RNA polymerase II. Ser-2 residues are phosphorylated during transcript elongation. These phospho-Ser-2 residue recruit splicing factors and polyadenylation factors that allow for efficienct splicing and polyadenylation.3. Describe the process of eukaryotic RNA pol II transcription termination. Termination is associated with polyadenylation. The transcript is cleaved and polyadenylated after the polyA addition sequence emerges from RNA polymerase II. The uncapped RNA molecule that remains associated with pol II after cleavage is degrade by the rat1 (or in humans, hrn1) exonuclease, which pulls it out of the polymerase causing the polymerase to disassociate from the DNA.4. Helix-loop-helix proteins (with and without a basic domain) function as heterodimers. How does heterodimer formation regulate the activities of these proteins? Sequence specificity of binding is determined by the basic domains of bHLH protein. Therefore, different bHLH heterodimers will have different binding specificities. HLH proteins lack the basic region. Heterodimers of an HLH protein with a bHLH protein will not bind DNA.5. What two enzymatic activities are present in TFIIH? How are these activities involved in RNA pol II transcription initiation? • a DNA helicase that is involved in forming the transcription bubble • a protein kinase that phosphorylates ser-5 in the CTD repeats and allows for promoter escape.6. Why can’t DNA polymerase fully replicate the linear ends of chromosomes? How do the ends of chromosomes get replicated? Because DNA polymerase only acts in a 5’ to 3’ direction AND because it requires an RNA primer. Therefore, the lagging strand of DNA will never be fully replicated (the RNA primer gets degraded, which would leave a single-stranded overhang). Chromosome ends get replicated with telomerase, and enzyme complex that contains a template RNA molecule that allows for the extension of the leading strand. Extension of the leading strand provides extended priming sites for the lagging strand, which allows for the DNA to grow longer. Therefore, the lengthening of DNA by telomerase can counteract the shortening of DNA because of the incomplete replication of the lagging strand.7. eIF4 has multiple functions in protein synthesis. What are these functions? • it binds the 5’ cap of the mRNA • it removes secondary structure from the mRNA • it allows the 43S preinitiation complex to bind to the mRNA at the 5’ end • it associates with polyA binding protein. This association holds the mRNA in a circular conformation, which increases the efficiency of translation.8. How is splicing involved in the regulation of nonsense mediate decay (NMD) of mRNA molecules that contain premature termination codons (PTCs)? • some proteins remain associated with splice sites after splicing occurs. • these exon junction complexes (EJCs) are removed from the mRNA during the first round of protein synthesis when the ribosome translates through them. • PTCs may cause termination (remove ribosomes) from the mRNA before all of the EJCs are removed. • EJCs that remain associated with mRNA after the first round of protein synthesis activate UPF1 with leads to the degregation of mRNAs with PTCs9. The electron micrograph shown below is of a 10 nm fiber. 10 nm fibers commonly are referred to as ‘beads on a string’. What is the molecular composition of each ‘bead’? What is the ‘string’ that connects the beads? The ‘beads’ are nucleosomes, histone octomers around which there are two wraps of DNA. The ‘string’ is the DNA between nucleosomes.10. Below is shown a growth curve of E. coli in a medium that contains glucose and lactose. Why does the growth of this culture pause when glucose is exhausted? Describe the mechanism(s) of transcriptional regulation involved. • The presence of glucose inhibits the adenylate cyclase • Inhibition of adenylate cyclase results in low concentrations of cyclic AMP (cAMP) • In the absence of cAMP, the CAP protein can’t bind to the promoter in the lac operon. • Without CAP binding, RNA polymerase can’t bind the promoter and the lac operon is not transcribed • When glucose is depleted, adenylate cyclase becomes active, this increases the concentration of cAMP, which allows the CAP protein to bind to the promoter, which allows RNA polymerase to bind, which allows for activation of the lac operon. • The pause in growth corresponds to the time required for the activation of the lac operon and the synthesis of proteins from this