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Answer for the number 7 question. Thank you. No need for long explanation.
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- Metabolism of Lipids (R - PowerPoint 动画 EndNote XB < lipid Officel 开始 入 Qit a A a E 國 园 食 @ 圈 文档优化智能號 四G 登录 发送到微字体一色彩的一段统一三地折西西中西虚化图形图 機板 主题板 色彩選密表数密表图标 图片 PPTRS Office em 余源余源Live 編入素材 AITR 37 The peton patwyy Misban Rai 129 ATP Pay attention to : Shahzil Ch 129 how many ATPS are produced during Maryam Shanzadi: 129 ATP Amadooo: 129 ATP one molecule of palmitic acid! MuhammaCT Azhar: 129 ATP Go Live produced SAJJAD AHMAD: 7 FADH2, 7 NADH2, 8 40 108 Pay attention to : Say something veed during ... 单击此处添加备注 41 OT片第40张,共175张 02 E 小餐注 注 87%46 2 H 8 Is M (n Ge M In In VI LE Q s M In b M 7 x + uv A canvas.northseattle.edu/courses/2086259/quizzes/5876542/take (a) Saturated fatty acid (b) Unsaturated fatty acid Account Dashboard Palmitic acid Linoleic acid 200L anauer Aemciates Ine. Courses Shown here are examples of saturated and unsaturated fatty acids. Explain what is meant by the term saturated fatty acid, and explain, at the molecular level, why Groups saturated fatty acids remain solid at higher temperatures than unsaturated acids do. Calendar Edit View Insert Format Tools Table В T? v: 12pt v Paragraph v U A Inbox History Help 8:183:43 ll LTE O moellim.riphah.edu.pk Biochemistry-l Dashboard / My courses / Biochemistry-II | General / QUIZ 1 Question 7 Not yet answered Marked out of 1.00 P Flag question ADP will the regulatory enzymes of TCA cycle? Select one: O a. None O b. Both depending on condition O c. Inhibit O d. Activate Next page Course Outline
- Blue numbers (3, 10, 11, 14, 16, 17) refer to a biochemical molecule Orange numbers (1, 2, 5, 6, 7, 8, 9, 12, 15) refer to reaction(s) in the pathway/process.Black numbers (4, 13 ) refer to an entire pathwayEach arrow may indicate multiple biochemical reactions required Match the description of lipid metabolism to the number on the figure (1-17): Inhibits b-hydroxy-b-methylglutaryl reductase Produced during the oxidation of malate to pyruvate…Part 1: Assess the following partial results section below by editing it for brevity by omitting any unnecessary parts (1 point), explain why you decided to remove certain sections (1 point): To evaluate inhibitory effects of the selected molecules, 10mM stock solutions of each molecule were prepared in DMSO. A reaction mixture (200μl) was prepared with the same formula optimized for the enzyme activity assay (0.1 M Tris-HCl ph 8, 0.1 M KCI, 25 mM NaCl, 0.25 mM ATP, and two units of inorganic yeast pyrophosphatase) with 10 µM of the sample molecule. The reaction mixture was incubated for 20 minutes at ambient temperature. Enzymatic reaction was triggered by addition of the substrate B (0.2 mM) and the absorbance of the product was monitored at 290 nm for 10 minutes. Six out of 15 sample molecules showed appreciable inhibition at 10 μM (Figure 5). Three of the molecules, A3, A6, and A7 exhibited more than 50% inhibition of the enzyme activity and were further diluted to find the minimal…PTP1B Substrate kcat Km. kcat/Km UM 10-7 x (s-1 M) DADEPYLIPQQG DADAPYLIPQQG DAAEP YLIPQQG AAAAPYLIPQQG 44.6 + 1.8 39.8 + 0.32 3.9 + 0.9 13.7 + 0.46 1.1 + 0.25 0.29 + 0.01 35.3 + 0.22 6.6 + 0.22 0.53 + 0.02 34.7 + 0.25 52.7 + 0.7 0.066 + 0.001 ) The units for kcat/KM in the above are given according to standard scientific notation. On this (d) ( basis what is the value of this kinetic parameter for the DADEPYLIPQQG substrate?
- AA View Tell me Convert to SmartArt W- Picture Shapes - Text Box Arrange Quick d Tue Dec 6 3:13 PM ♫ Share Design Qsn 2 (a) A pathway consists of 6 enzymes (p, q, r, s, t, u) that convert substrate J to product X at a rate of 10 moles/hour. If enzyme q is increased from 5 mmoles to 35 mmoles the amount of product X increases to 20.5 moles/hour. What is the flux control coefficient for enzyme q (C¹)?A Classwork for BTMF_Itani_ Biolo x A S'ERRA STANLEY -protein-synth X O 8 period light reaction - Google S x google.com/document/d/1plhyciPrM-AMsQKmaUHZSZU44EJL85jDuCyDxE01E60/edit - protein-synthesis-worksheet_practice-converted O sert Format Tools Add-ons Help TUR Last edit was 10 minutes ago 0% Normal text Noto Sans BIUA 川。 三E▼=、三三 13 + 田回▼ 1 3 4. 5 8. DNA → TAC AGA CGG CAA CTC TGG GTG CTT TGT TCT CTT CTC AGT ATC mRNA → protein > Son out Co 4. 6. 8. 9.National Board of Medical Examiners Biochemistry Mark 36. In the presence of a metabolite (X), 6-phosphofructokinase is assayed at a fixed concentration of ATP and varying concentrations of fructose 6-phosphate. The resulting data are shown in the table. Fructose 6-phosphate (pM) 5 10 20 40 75 100 200 Velocity umoles/min 0.05 0.15 0.25 0.70 1.7 2.2 3.1 3.1 Velocity (+X) umoles/min 0.006 0.025 0. 10 0.35 1.03 16 2.9 3.1 400 Metabolite (X) is most likely which of the following substances? O A) ADP O B) AMP OC) CAMP D) Citric acid O E) Fructose-2,6-bisphosphate
- E Classwork for BTMF_Itani_ Biolo X E SI'ERRA STANLEY - protein-synth X arch 8 period light reaction - Google S x ocs.google.com/document/d/1plhyciPrM-AMsQKmaUHZSZU44EJL85jDuCyDxE01E60/edit LEY-protein-synthesis-worksheet_practice-converted ☺ Insert Format Tools Add-ons Help Last edit was 5 minutes ago 100% Normal text Noto Sans . BIUA 12 川, 三= + - I 1. 2 . 4. 5 PART C. Use your codon chart to determine the amino acid sequence. Remember to read through the strand and ONLY start on AUG and STOP when it tells you to stop. Follow example below: Example: DNA O AGA CGG TAC CTC CGG TGG GTG CTT GTC TGT ATC CTT CTC AGT ATC mRNA O UCU GCC AUG GAG GCC ACC CÁC GAA CAG ACA UAG GAA GAG UCA UAG protein 0 start - glu – ala -thre - hist – asp -glu-threo-stop acid acid 1. DNA → CCT CTT TAC ACA CGG AGG GTA CGC TAT TCT ATG ATT ACA CGG TTG CGA TCC ATA ATC MRNA → protein → 2. DNA → AGA ACA TAA TAC CTC TTA ACA CTC TAA AGA CCA GCA CTC CGA TGA ACT GGA GCA MRNA → protein → 围 Sign out Co & 6. 7CHEM151/251 Biochemistry 1 a. Determine Km 1. The following data were obtained for a competitive inhibition study in which the [I] = 3 µM for each determination of vo in the presence of inhibitor. The Vmax = 200 μM P/min for both data sets. 200 Vo (UM P/min) 8 8 8 8 8 8 8 8 8 180 160 140 120 100 40 Name (Print)/ID #: the absence of inhibitor. Participation Question # 10 No Inhibitor 50 +Inhibitor 100 [Substrate] (UM) 150 b. Determine Km, app for the data obtained in presence of inhibitor. 200 c. Calculate the value for Ki. Note: a = 1 + [I]/Ki and a¹ = 1 + [I]/Ki*. for the data obtained infor the table just do the results and discussion for phosphatw. Table 2. Presence of Nucleic Acid Hydrolysis Products in DNA Hydrolysate Component Tested Diagnosis (+/-) Purine Bases Ribose Deoxyribose Phosphate Sample DNA test solution Distilled water DNA test solution 1% ribose 1% glucose DNA test solution 1% ribose 1% glucose DNA test solution 1% ribose 1% glucose for the photos: use it for the DNA extractiom results. Glucose Below. results during the experiment 8. Test for phosphate on the DNA hydrolysate Sample DNA hydrolysate Ribose Observations (describe what happened) From clear white to light yellow with white precipitate From clear white to cloudy white and no formation of precipitate. From clear white to pale yellow and no formation of precipitate. Diagnosis (Positive/Negative?) 1. Observations on the extraction of DNA Write your observations on the space below. (no sentence limit) If you have photographs, attach them at the end of the document. The blended frozen peas were…