In DNA cloning, restriction enzymes are used to cut a DNA at specific sites b protein at specific sites DNA at random sites d. protein at random sites Question 32
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- of estion 9 t of uestion THCA ▶ Sou 100- HC This reaction is Entropy is THC San Leafly ATA RNA polymerase SSSSSSSSSS ATTOGOGACATAA ATGACGGATCAGCCOCAAG UACUOCCUAGUC RNA Transcript TACTOCCTAGTCGGCOTTCOOCTTAACCOCTOTATIT (In this picture, RNA is being made by complementary base pairing with DNA.) This reaction is → Entropy is ◆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:EcoRI --- 5' G - AATTC 3' 5' AGAATTCCGACGTATTAGAATTCTTAT CCGCCGCCGGAATTCT CATCA 3' 3' TCTTAAGGCTGCATAATCTTAAGAATAGGCGGCGGCCTTAAGAGTAGT 5' Number of pieces of DNA , and type of fragment .
- 48 Second letter If any single nucleotide is deleted from the DNA sequence shown below, what type of mutation is this? UUU U UUC UUA UCU Phe UCC UAU UGU Cys ANTISENSE 5' GGACCCTAT3' UAC Tyr UGC UAA Stop UGA Stop UAG Stop UGG Trp Ser UCA Leu UUGL" UCG CUU CU CAU) His CAC) CGU CGC CGA Arg CUC C Leu Pro CAA1 Gin CAG) CUA CCA CUG CG CGG AAU AAC Asn AGC AAA AAG Lys AGG Arg AUU ACU AGU Ser AUC lle ACC Thr AUA ACA AGA AUG Met ACG GCU GCC GUU GAU] GGU GUC Val GAC Asp GGC Ala Gly GUA GCA GAA GGA Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a FRAMESHIFT SILENT NONSENSE MISSENSE Third letter First letterCynt 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 GJILLule should [1] De Tepicated exactly. Figure 2 represents part of a DNA molecule. 5'TT ATGC TT 3' T. C CA G tal: 5] TACG A A G 3' GTC her 5' Figure 2 (c) Show, by means of annotated diagrams, how this piece of DNA is replicated. Distinguish clearly between the original and new strands.
- € 2 A X 1_30*_SP23 - General Biology I (for majors)/11 f us page The anticodon sequence created from the following DNA: TACGGGGCTGAGATT Select one: a. AUGCCCCGACUCUAA O b. Met-Pro-Arg-Leu-STOP O c. UACGGGGCUGAGAUU O d. Tyr-Gly-Ala-Glu-lle 20 000 MacBook Air DIIB 6000 5100 Number of Fragment(s) 4000 100 pX 6000 bp 3000 A A 1500 2000 A B 10. Based on the restriction map of the above plasmid, determine the number of DNA fragment(s) and the size(s) of the fragment(s). Digestion with Enzyme A Enzyme B Enzyme C Enzyme A + B Enzymes A + C Enzyme B+ C Enzyme A + B + C Fragment size(s) in base pairs5’ TAAGCGTAACCCGCTAA CGTATGCGAAC GGGTCCTATTAACGCAC 3’ 3’ ATTCGCATTGGGCGATT GCATACGCTTG CCCAGGATAATTGCGTG 5’ Imagine that the double-stranded DNA molecule shown above was broken at the sites indicated by spaces in the sequence and that before the breaks were repaired the DNA fragment between the breaks was reversed. What would be the base sequence of the repaired molecule? Show the sequence, label the 5’ and 3’ ends and briefly explain the reasoning supporting your answer
- 2. Here is the detailed view of the MCS of the PUC19 plasmid: Sma I KpnI SbfI PstI SacI XbaI ECORI BamHI Sall SphI HindIII agt GAATTCGAGCTCGGTACCCGGGGA TCCTCTAGAGTCGACCTGCAGGCATGCAAGCTTGGcgtaatcatggtcat 400 410 420 430 440 450 460 ...S N S SP VR PDEL TS R CAH LS P T IM T M lacZa translational start Figure 25: MCS of PUC19 A. If the MCS were cut with Kpn I and BamH I, draw the small fragment of DNA that would be cut out. Show both strands. For reference, here are the recognition sequences: 5' G|GAT СС 3' 3' С СТАG|G 5' recognition sequence for BamH I 5' G GTAC|C 3' 3' cl CATG G 5' recognition sequence for Kpn I Figure 26: recognition sequences for BamH I and Kpn I#1 HindII --- 5’ GTC ↓ GAC 3’ 5’ ACGACGTAGTCGACTTATTAT GTCGACCCGCCGCGTGTCGACCATCA 3’ 3’ TGCTGCATCAGCTGAATAATACAGCTGGGCGGCGCACAGCTGGTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut:Given: BamHI, cleaves after the first G: 5’ G GATCC 3’ 3’ CCTAG G 5’ AND BclI cleaves after the first T: 5’ T GATCA 3’ 3’ ACTAG T 5’ THEN -- Given the DNA shown below: 5’ATTGAGGATCCGTAATGTGTCCTGATCACGCTCCACG3’ 3’TAACTCCTAGGCATTACACAGGACTAGTGCGAGGTGC5’ i) If this DNA was cut with BamHI, how many DNA fragments would you expect? ii) If the DNA shown above was cut with the enzyme BclI, how many DNA fragment would you expect?