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- EcoRI --- 5' G - AATTC 3' 5' AGAATTCCGACGTATTAGAATTCTTAT CCGCCGCCGGAATTCT CATCA 3' 3' TCTTAAGGCTGCATAATCTTAAGAATAGGCGGCGGCCTTAAGAGTAGT 5' Number of pieces of DNA , and type of fragment .Alternative splicing Template strand S F yIGUide1,mq00:c-00: Replisome Transforming principle Origin of replication (or)eleb al msxS ain Coding strand Transcription factors Leading strand Single nucleotide polymorphism Okazaki fragment Telomerase M Nucleoside RNA Polymerase I RNA Polymerase II RNA Polymerase IIIon & of qu 9ven UoY Insertion mutagenesis Spliceosome Transcription Unit SNP Reverse transcriptase 1 Seminal work by Oswald Avery and colleagues demonstrated that DNA is what Frederick Griffiths called this etniog OS dotsM bioW 1-2kb of newly synthesized DNA strands are called this ainiog PS Assembly of the replisome is an orderly process that begins at these precise sites Snoiteeu 4 Transcribes ribosomal RNA genes in eukaryotes 5. A large nucleoprotein complex that coordinates activity at the replication fork Single base pair differences between homologous genomic regions isolated from different members of a population Complex of proteins and snRNAs catalyzing the removal of…BM4_DNA AND PROTEIN S X /1FAIPQLSDP_g5B-629FSHNpGnTMIEppLS4A71zBd4vcUBqNUILubXONw/formResponse 4. What is the nitrogen base pair of Adenine in transcription? O Cytosine O Uracil O guanine O thymine 5. The central dogma of Molecular Biology states that There are four nitrogen bases in DNA, two purines (adenine and guanine) and two pyrimidines (cytosine and thymine). Which process is not included in the central dogma? duplication transcription translation O translocation Leadple
- he bases of one of the strands of DNA in a regionwhere DNA replication begins are shown at the endof this problem. What is the sequence of the primerthat is synthesized complementary to the bases inbold? (Indicate the 5′ and 3′ ends of the sequence.)5′ AGGCCTCGAATTCGTATAGCTTTCAGAAA 3′pcc300ATAAADATATAOOTTAA 1. Use the genetic code table and the information in the diagram below to determine the amino acids that would make up the portion of the polypeptide shown. Include information for a key as well. DNA template 3' G CATA ACAGAGGATT-5' al bnsua AMAm pniwollot erfT E transcription s yd bnsita ebitgeqylog s sidmeaze of beae RNA strandUU UAOUOUU A-emoaodin 5'-CGUA AUUGUC UCCUUA- 3' J J JL erit o elinW (s) translation bluow terdt aspnso sigootiwsone polypeptide viemetis ns ebivo19 (d) ent ot etslanT Key:nd 2 minutes): Any RNA polymerase in any organism: O A Synthesizes RNA chains in the 3 to-5" direction O B. Binds tightly to a reqion of DNA located thousands of base pairs away from the transcobed rogion of the DNA OC Has proofreading activity O D. Separates DNA strands throughout a long region of DNA (up to tinousands of base pairs) and then copies one of them. OE Has a subunit called A (lambda), which acts as a proofreading ribonuclease OF. Can initiate synthesis of a new RNA chain without a primer
- Labeling DNA Replication Directions: Drag the lahels from the left tn corrary derti theinats of ar=rlicating strandarA 24 Newly Created Strand of ONA Replication Carke Original DNA Strand Replication Direction of Origin of Replication Directions: Bolow is a more in-depth look at a replication bubble. A.l of the psrts are still tne came, but#4 BamI --- 5’ CCTAG ↓G 3’ 5’ ACGCCTAGGACGTATTATCCTAGGTAT CCGCCGCCGT CATCA 3’ 3’ TGCGGATCCTGCATAATAGGATCCATAGGCGGCGGCAGTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut:Cynt Classifying mutations A certain section of the coding (sense) strand of some DNA looks like this: $-ATGTATATCTCCAGTTAG-3" It's known that a very small gene is contained in this section. Classify each of the possible mutations of this DNA shown in the table below. mutant DNA 5- ATGTATCATCTCCAGTTAG-3' S-ATGTATATCTCCAGTTAG-3 5- ATGTATATATCCAGTTAG-3' type of mutation (check all that apply) insertion deletion point silent noisy insertion O deletion point silent noisy insertion O deletion point silent Onoisy X G
- "DNA polymerase I serves a secondary function in vivo,now believed to be critical to the maintenance of fidelityof DNA synthesis" Explain this statement ?This is part of the Escherichia coli DNA sequence that contains an inverted repeat. (Note: top strand is the coding strand). 5'-AACGCATGAGAAAGCCCCCCGGAAGATCACCTTCCGGGGGCTTTATATAATTAGC-3' 3'-TTGCGTACTCTTTCGGGGGGCCTTCTAGTGGAAGGCCCCCGAAATATATTAATCG-5' (i) Draw the structure of hairpin loop that will be formed during the end of transcription. (ii) Describe the function of the hairpin loop during transcription.Genetic Code-Reference Second Letter First Letter C Third Letter UAU] UACS **UAA Stop UGU] Cys UGC U UUU Phe UUC UCU Туг UCC U Ser UUA) Leu UUG **UGA Stop UGG Trp UCA UCG **UAG Stop CUU CCU CAU) CGU His CACJ CAA) CUC CCC CGC Leu Pro Arg CUA ССА CGA A Gln CUG CCG CAGJ CGG AAU] AGUSer U AUU ACU AACAsn AAA Lys AUC Ile ACC AGC C Thr AGA] Arg AUA ACА AAGJLYS GAU] GACASP *AUG Met/Start ACG AGG U GUU GCU GGU Asp GUC GCC GGC Val Ala Gly GAA) Glu GAGJ GUA GCA GGA GUG GCG GGG