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- P 12 D 12 D. cold bet.p + Who bet P Frien roblc Hom S STAR + ays/1089903/lessons/1533963/variants/2428136/take/15/ A e * Practice TEXT ANSWER Answer the following three questions based on the diagram below. 1. Which of the pictured organisms is a eukaryote? Explain your answer using evidence from the diagrams. 2. Describe two similarities between the cells of prokaryotes and eukaryotes. 3. Describe one difference between the cells of prokaryotes and eukaryotes. You may not use the evidence you described in the first question. Amoeba Bacterium DNA Nucleus (Na to scales Write your answers using 3 to 4 complete sentences. H. = = = = S x, x E E A A Normal BIUS Enter your answer here 22222222 All Changes Saved ons Answered 52°F ClearRabbits were in injected with [32 P] photsphate intraperitoneally ans asacrificed after 2, 5, 8, 12, 18, 24, 48, or 168 hours of radioisotope administration. DNA, nuclear RNA (nRNA), and cytoplasmic RNA (cRNA) were prepared from the kidneys and their specific activities (radioactivity/ mg of nucleic acid) were determined 1. what can be concluded from the nRNA and cRNA curves and why 2. why didn thte specific acitivies of nRNA's and cRNA's decrease significantly after 2 days of labeling.1:55 .ll 0.52 KB/S 56 1.pdf CHEM 3141: BioChemistry Quiz no. 1- Answer key I.) Draw the following compounds. 1.) Maltose: 2 Glucose, a-(1,4) 2.) Gentobiose: 2 Glucose, B-(1,6) 3.) C-2 epimer of Talose 4.) C-3 epimer of Fructose 5.) C-2 epimer of ribose II. Predict the products of the following reactions, if there is none, write NO RXN. Also indicate, if the reaction is fast or slow он FO HO -H H -OH но. + Resorcinol/dilute HCI 1.) 2.) Sucrose + Copper (II) Acetate/Acetic Acid HO H HO H HO H H OH 3.) HO + Orcinol/HCI
- Rabbits were in injected with [32 P] photsphate intraperitoneally ans asacrificed after 2, 5, 8, 12, 18, 24, 48, or 168 hours of radioisotope administration. DNA, nuclear RNA (nRNA), and cytoplasmic RNA (cRNA) were prepared from the kidneys and their specific activities (radioactivity/ mg of nucleic acid) were determined 1. what process was studied in this expereiement 2. What is concluded about the division of kideny cells during the time of labeling and why.Answer: giycogen Time left 0:4T:- If you have got the following DNA template molecules, which one of them will require more energy to break down the hydrogen bonds between the antiparallel strands? O a. GCGCGCGCGCGCGCGCGCGCG O b. GGGGGCCCCCAATTCCCCCCC O c. AAAAAATTTTTCCCCCGGGGG O d. ТAСТАСАСTGTGGTTAАТТААА Oe. АТАТАТАТСGСGTTAAATTCTА CLEAR MY CHOICE Match the given words with the most suitable words from the given list. Unicellular fungi Hyphae Choose... Охуgen Atom ENG سطح المكتب2.58. Bacteria have equivalent diameters of 2 × 10-6 m and densities of approximately 1 kg/L. Under optimal conditions, bacteria can di- vide every 30 min. Determine the mass of bacteria that would accumulate in 72 hr under continuing optimal growth conditions. Can this occur? Explain.
- for 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…Electrophoresis A protein required 6.8 min to travel 82 cm to the detector in a 96 cm -long capillary tube with 25.4 kV between the ends. Find the apparent electrophoretic mobility.8:47 ll 5G% 4 Chapter 9 DNA worksheet 2022.d... 2. Write the mRNA transcript from the DNA template in question 1. Remember your enzyme for RNA polymerase must add RNA nucleotides starting with the 5' direction and moving to the 3' direction. So for the MRNA copy the strand of DNA that is 3'-5') 3. Refer to the codon chart at the bottom of the page. Translate the mRNA into a polypeptide using the following rules. (2 points) Direction of translation of mRNA is from 5' to 3'. 1. MRNA codons are read in groups of three nucleotides. 3. Translation must begin at the START codon. 4. the STOP codon terminates translation. 2. THIRD LETTER SECOND LETTER U TulUUU Phe UUC Phe UUA Leu UUG Leu A UGU Cys U JUGC Cys C UCU Ser UAU Tyr UCC Ser UAC Tyr UCA Ser UAA STOP UGA STOPA UCG Ser UAG STOPUGG Tp G CCUU Leu CÚC Leu CỦA Leu CUG Leu AJAUU Ile AUC Ile AUA Ile AUG Met STARTACG ThrAAG Lys GGUU Val GUC Val GUA Val GUG Val CCU Pro CAU His CCC Pro CAC His CCA Pro CAA Gln CCG Pro CAG Gln ACU ThrAAU Asn AGU…
- Justify all answers (show all work). 1. Minor changes in primary structure can have pronounced effects on biological function. Tetrameric hemoglobin, a₂³₂, dissociates into pairs of dimers aß to different extents depending on the protein primary structure. The standard Gibb energies of dissociation of normal adult hemoglobin A and of several mutants have been determined and are listed in the table below. Substitution in mutant Hemoglobin A, normal Richmond Kansas Georgia 102B; Asn → Thr 102B; Asn → Thr 95x; Pro → Leu AG® of dissociation (kcal/mole Hb) 8.2 6.0 65 5.1 3.6 (a) Write an equilibrium expression for the dissociation of hemoglobin tetramer into dimers. (b) For normal hemoglobin and its three mutants (variations) calculate the equilibrium constant for their dissociation from a tetramer into dimers. (c) How do the equilibrium constants for the dissociation of the hemoglobin mutants compare to that of normal hemoglobin?Anglaauk/courses/ 16967/external tools/retrieve?display full width&url=https 1962F2Fcanys-anglia-a ukquiz-Iti-dub pp-g-zinb Re Using the class data below (where the results shown are the averages obtained from each set of individual data; also on practical schedule), find the mean mitotic index for Allium cepa. Please give your answer correct to 2 decimal places (just add the number, with no units). Table 2. Total number of cells in a field of view of Allium cepa, with numbers of cells in prophase (P), metaphase (M), anaphase (A) and telophase( T) Total M A 141 15 3 2 145 21 4 3 146 9 146 13 7 7 5 146 25 11 5 2 147 13 7 5 150 15 10 10 4 154 24 12 3 155 27 11 7 159 25 19 6. 160 22 11 161 16 14 7 166 24 171 18 2 6. 172 25 5 5 6. 175 15 6. 4 2. 177 27 14 7. 5 179 14 10 180 28 2 191 13 4 1. 1 194 12 3. 3 Tvne vour answer 8°C Mostly cloudy A O 4) search 99+ 近Practice Question 2 The beginning of the hexose kinase gene's sequence can be found below, the +1 nucleotide is underlined and bolded. It also contains an origin of replication (ORI) which is found at position 30. 1 20 ORI 40 60 3'...TTCGAGCTCTCGTCGTCGAGATACGCGATGATATTACTGGTAATATGGGGATGCACTATC...5' 5'..AAGCTCGAGAGCAGCAGCTCTATGCGCTACTATAATGACCATTATACCCOCTACGTGATAG...3' promoter A) Assume that replication has been initiated at that ORI. Provide the sequence of the primer that is complementary to the DNA in each of the following positions. Site A - binding to the top strand of the DNA at position 20 – 30 5' 3' Site B - binding to the top strand of the DNA at position 31 – 41 5' 3'