Text: B0 141: DNA Replication Activity Directions: Use the base pairing rule, filling in the missing base pairs in the parent DNA molecule. Color the base pairs according to the following colors: A = green; C = purple; G = red; T = yellow. Color the P-D and P-D backbones of the parent molecule light blue. The parent molecule will now split down the middle. Follow the same pattern of the parent molecule along each arrow and replicate the base pairs into the daughter DNA molecule. Do not change the order of the parent molecule. Fill the base pairs into the new half of each daughter DNA molecule. Color the new bases using the color key above. Then color the P-D-P-D backbone orange. Use the Identification Key. Parent DNA Molecule Summary: DNA Replication Identification Key Parent DNA molecule sequence: AGCTTACGATCGTACG Daughter DNA Molecules: New nucleotides added: 3rd and 14th How many nucleotides were added to each daughter DNA molecule? In the summary/border above, summarize the DNA replication process you observed. What words or phrases may help you organize it down into smaller steps?
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The parent DNA molecule has the following sequence: A-T-G-C-A-T-G-C Show more…
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DNA Replication Exercise B: DNA Replication (con’t) DNA prior to replication Prior to replication, DNA is a double stranded helix with complementary base pairing. In order to achieve complementary base pairing, the two strands of DNA are antiparallel, or line up in opposite directions (Fig. 4). On one strand the phosphate group is on one end of the DNA backbone while the deoxyribose is at the opposite end. In the second strand, phosphate is on the end in which the deoxyribose of the first strand is located, while the deoxyribose end of the second strand is adjacent to the first strand’s phosphate end. The phosphate end of a DNA strand is called the 5’ end of the DNA backbone, while the deoxyribose end is the 3’ end. In Fig. 4, label the 3’ and 5’ ends of both DNA strands prior to DNA replication. Initiation of replication The replication bubble begins with the enzyme, helicase, breaking the hydrogen bonds between nitrogenous bases of the two parental DNA strands (Fig. 5). In order to prevent the hydrogen bonds from reforming, single-strand DNA-binding proteins (SSBPs) attach to the separated DNA strands (Fig. 6). The breaking of the parental DNA’s hydrogen bonds causes the double helix to relax, creating tension further down the molecule. Another enzyme, topoisomerase, cuts the covalent bonds (phosphodiester linkages) of the double helix further downstream of the replication bubble and immediately reattaches them, relieving the tension. Identify the function of helicase. Fig. 5. Initiation of replication, part 1.
Dr. Anas S.
38. Translation is a process that (a) requires DNA ligase (b) converts RNA to DNA (c) is semi-conservative (d) requires tRNA, mRNA and ribosomes (e) produces a polysacharride 39. What is the order of the anticodon sequence if given the following DNA sequence (AGATTTCGCCCCAAA)? (a)UCUAAAGCGGGGUUU (b) TCTAAAGCGGGGTTT (c)AGATTTCGCCCCAAA (d) TCTUUUGCGGGTTT (e) AGAUUUCGCCCCAAA 40. The mating of F+ strains with F- strains can result in (a) the transfer of antibiotic resistant genes (b) the immediate production of a Hfr strain (c) transferring genes needed for conjugation and transfer to the recipient (d) integration of genes into the donor (e) the production of a capsule 41. Exonucleases are needed to (a) cut out RNA (b) needed for excision repair (c) photoreactivation (d) breakdown DNA-RNA strands (e) none of the above TRUE OR FALSE (2 pts) Mark A for True and B for False 42. Replica plating is a method for direct selection 43. RNA is a double stranded molecule in the case of tRNA. 44. Intercalating agents cause mismatch mutation only on the 5'-3' DNA strand. 45. Plasmids can replicate independently from the bacterial chromosome. 46. DNA polymerase 3and- synthesizes from a 5'-3' direction continuously on the leading strand. 47. Transposon are examples of a mobile gene pool. 48. In the process of transcription, the DNA(+) strand is needed to make mRNA. 49. The enzyme DNA helicase is needed to make a replication fork. 50. The terminator sequence stops the process of translation.
Sri K.
DNA components Pyrimidines, purines phosphates Enzymes and deoxyribonucleic acids Ribosomes RNA pyrimidines purines phosphates groups and release sugars. Ribonucleic acid Messenger RNA Proteins DNA JLO quick: only Which of the following modifications specifically affects resulting protein? 4-base deletion near the cleavage site in Human T-cell leukemia virus Slope color Most impactful effect Base substitution; 20. DNA ligase joins: Uracil kinase methylates nucleotide Uracil fragments nonrepetitive Uracil fragments nonrepetitive DNA pieces Phosphodiester RNA fragments phosphodiester creating bonds Recombinant DNA breaks described as fragments of DNA composed of specific sequences originating from a Palindromic sequence of four nucleotides. Biotechnology loop All of the above choices different The function of the Kidney is to filter nutrients from the food and produce hormones; muscle contractions produce urine. Pituitary gland, from a control zone, regulates urinary output. Following is not a structure of a nephron? Which of the following is not a structure of a nephron? The Loop of Henle Proximal convoluted tubule Bowman's capsule renal corpuscle supplies the cortex with nutrients Ciliary part of the eye that the fovea centralis aqueous humor sclera mucus humor The body maintains balance by the fluid-filled semicircular canals Cochlea the fluid-filled ossicles The hormone aldosterone regulates kidney water reabsorption counter Microscopic reabsorption from the distal tubule and collecting duct produces the medulla of the adrenal glands responsible for maintaining the pH of blood responsible for causing the release of angiotensin
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