Donor DNA MSF Recipient DNA M*S* F* Transformants Percentage + M*SF* 4.0 ⚫ Individual genes are transformed MS F 4.0 frequently Double transformants are rarer: MS* F* 2.6 MS is most common • SF is rarer MF is rarest MSF 0.41 MF is about as rare as the triple MSF 0.22 Conclusion: M and S are close together, M and F are far MST F 0.0058 apart: M _ S F MSF 0.0071

Donor DNA MSF Recipient DNA MS F* Transformants Percentage + MSF 4.0 ⚫ Individual genes are transformed MS F 4.0 frequently Double transformants are rarer: MS* F* 2.6 MS is most common • SF is rarer MF is rarest MSF 0.41 MF is about as rare as the triple MSF 0.22 Conclusion: M and S are close together, M and F are far MST F 0.0058 apart: M _ S F MSF 0.0071

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter17: Metabolism: An Overview

Section: Chapter Questions

Problem 13P: What Art the Features of the Series of -omes? Define the following terms: a. Genome b. Transcriptome...

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| 四 ii D YouTube W Maps 6 viewbook-for-sim. Aid & Awards - Of.. O 2022-23_UAB_pa. W Student Z - McMa. Transfer your Gma. M MarkBook CON. (a) What type of mutation would be least harmful to an organism: silent, missense, or nonsense? Explain your answer. (b) What is the difference between gene mutations and chromosomal mutations? Paragraph В I. + v ... Lato (Recom... 19px... v 民 Add a File Record Audio Record Video MacBook Pro
y Cours S DIOE The following DNA fragment shows where a number of restriction endonucleases cut sites occur within the fragment. The numbers below indicate the distance between those cut sites in base pairs (bp). Kpn I Sall Eco RI Pst Bam HI Xho Bam HI Pst i Eco RI Kon I 100 200 300 400 500 B00 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 How many total restriction fragments will result if you complete digest the DNA fragment above with Sal I? cross out a. 1 cross out b. 2 cross out O c. 3 cross out d. 4 cross out e. 5 cross out O f. 6 cross out g. 7 cross out Oh. 8 The following DNA fragment shows where a number of restriction endonucleases cut sites occur within the fragment. The numbers below indicate the distance between those cut sites in base pairs (bp). t of Sal Eco RI Pat Xhol Bam HI Psti Eco RI Koni Kpn i Bam H
5. The nucleotide sequences of the DNA molecules in the figure below were obtained from four different individuals, one wild type and three mutants. Wild Type 5'-TTATCCATGATCGGATCGATCCATTAGCCGA-3' 3'-AATAGGTACTAGCCTAGCTAGGTAATCGGCT-5’ Mutant I 5'-ATCCATGATCGGATTGATCCATTAGCCGAAT-3’ 3'-TAGGTACTAGCCTAACTAGGTAATCGGCTTA-5’ Mutant II 5'-CCGTTATCCATGATCGGATAGATCCATTAGCC-3’ 3'-GGCAATAGGTACTAGCCTATCTAGGTAATCGG-5’ Mutant III 5'-CACCGTTATCCATGATCGGAACGATCCATTAGC-3’ 3'-CAGGCAATAGGTACTAGCCTTGCTAGGTAATCG-5’ a) Identify the open reading frames in each sequence of DNA and translate them into proteins. Write down the sequence of amino acids that will be obtained after translation: b) Which of the mutations above would be least likely to cause a change in the function of the protein? Why? c) Which of the mutations above would probably cause a major disruption in the function of the protein? Why?
During the process of genetic transduction in bacteria, two living cells exchange segments of their DNA. True O False MacBook Air DII DD 吕0 000 F9 F6 F7 F5 F3 F4 * & 23 $ 8. 4 5 Y E R
1/V (uM x min-1)-1 1. To the right is a Lineweaver-Burk plot for an enzyme that can cleave both DNA (D) and RNA (R). 2200 2000 1800 a. What is the KM when cleaving DNA? I 1/V R 1600 • 1/V D b. What is the KM when cleaving RNA? 1400 1200 C. What is the Vmax when cleaving DNA? Z 1000 800 d. What is the Vmax when cleaving RNA? 600 e. On the graph, draw an appropriate line for a 400 noncompetitive inhibitor for cleaving DNA. Label clearly. f. On the graph, draw an appropriate line for an uncompetitive inhibitor for cleaving RNA. Label clearly. 200 8 10 11 1/S (µM)-1
che World Mona L A Leonardo da Vinci Museum - Flo x dSequenceViewer/14003927urlahttps%3A%2F%2F43160dd0-1a11-43d1-b28d-96e9e2abb6d2.sequences.api.brightspace.com2F1400392%2Factivity2217 1.07 Quiz: Experimentation UIZ: Experimentation Kate gathered three boxes of the same size made of different materials glass, clear plastic, and aluminum painted black She placed them on a window sill in the sun for an hour and then measured the warmth of the air in each box In this experiment, what is the warmth of the air? O an independent variable O a control O a constant O a dependent variable 2 3 45 Next
ABO blood group system is defined by the presence of agglutinogens (A and B molecules) at the surface of red blood cells. Enzyme A which leas to the production of the molecule A is coded by all ele A, while enzyme B which leads to the production of the molecule B is coded by allele B, and enzyme O which cannot lead to the production of any molecule is coded by allele O. A part of the coding DNA strand for enzyme A: GAC GTG CGC GCC A part of the coding DNA strand for enzyme B: GAG GTG GcC GCC 5. Compare the non-transcribed strand coding for enzyme A to that coding for enzyme B. 6. Identify the type of mutation involved in this case. Justify. 7. Write the amino acid sequence for both enzymes. 8. "Mutations can lead to diseases or to genetic diversity"Justify by refering to parts A and B. Second letter A G UCU UCC UCA UUG Leu ucG UUU T Phe UUC UAU1- Tyr UACJ Ser UAA Stop UGA Stop UGU), UGCJ UUA UAG Stop UGG Trp CAUTHIS CCU CC CCA CCG CUU CÚC CÁCJ Pro CAA CGU] CGC Arg FLeu CGA CGG CỦA Gln…
= Menu #2 Mol Bio Mutation QU... X + Create All tools Edit Convert E-Sign Q Search Complete the following tasks. Sign in Find text or tools Q H You are fresh recruits to a molecular biology laboratory, and your new boss has tasked you to study mutations in NRAS, a gene that she suspects is involved in cancer pathogenesis. Task A: Polymerase Chain Reaction Master Mix Your first task is to isolate and amplify the NRAS gene from cDNA extracted from different samples through PCR. You will need to run a total of 7 PCRs: 3 normal samples, 3 cancerous samples, and 1 negative control. To make things easier in the lab, when running multiple reactions, the components are prepared not individually, but as a master mix-all the components for multiple reactions are prepared in bulk, except for the template DNA, which is added separately once the master mix has been distributed into individual tubes. The table below lists the different components for PCR, the available stock concentrations of these…
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
+ Jimenez me O Lab Interac O (71) Alfie Ca G lab concord G quantum m b Success Cor a holodeck * Regression * Page missir G regression O Lab Inte x Ask Your ST + Lab Interac A Not secure | http://lab.concord.org/embeddable.html#interactives/sam/DNA-to-proteins/4-mutations.json Share About DNA Sequence Enter DNA sequence below Original sequence ATGCCAGGCGGCGAGAGCTTGCTAATTGGCTTATAG ATGCCAGGCGGCGAGAGCTTGCTAATTGGCTTATAG Edited sequence Apply Show DNA Transcribe Translate Show protein Continue one step Start/continue model Stop Reset The Concord Consortium 9:45 pm 12/05/2022 30°C ENG 4) O Partly cloudy US (7.
For the chromatogram below, what is the sequence of the template DNA from base 115 to 125? CTGTGTGAAA TTGT TA T C CGC T CACA A T TCCACA CA A CATACGAG CCGGAAG CA T AA 110 120 130 140 150 160 СТТТААСАAТА ТАTTCAATTТС ATAACAATTTC GAAATTGTTAT
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