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An enzyme is needed for the Krebs Cycle. It will be made following the directions contained in a gene of one of the __29__ found in the __30__ of the cell. A portion of the double strand of __31__ splits apart between the weak__32__bonds holding the bases together. This portion contains the code for building the enzyme. The code message begins with a triplet call a _33__ , followed by the code, then signals the end with a triplet called the _34__. The DNA is composed of groups of three molecules called __35__ . They are made up of ribose __36__, a __37___ for additional support, and a nitrogenous __38___. These match up with complementary ones on the opposite strand of DNA. A cytosine will pair up with __39___. An RNA uracil will pair up with __40___ on the DNA. When they line up and join, a __41__ RNA strand is formed. This process of copying the code into RNA is called __42__. The RNA will move out to the cytoplasm, to an organelle called a __43__. There, __44___ RNA molecules will each pick up their respective __45___ that are floating in the cytoplasm and bring them into line next to the first RNA binding them to codons on the RNA copy. As they join, a __46__ is formed. This part of the process is called __47__.
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- Below is a picture that depicts transcription. Shown is the DNA double helix, the replication bubble, RNA polymerase, and the growing RNA molecule. The ends of each nucleic acid are indicated by the labels I-VI. Which labels correspond to 3' ends? I II DNA V OI and III OI, IV, and V O II, III, and VI OV and VI O I, III, V, and VII RNA VI RNA polymerase III IVarrow_forwardThe 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 5'.TTCGAGCTCTCGTCGTCGAGATACGCGATGATATTACTGGTAATATGGGGATGCACTATC...3' 3'.AAGCTCGAGAGCAGCAGCTCTATGCGCTACTATAATGACCATTATACCCCTACGTGATAG...5' promoter 2a. 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. d 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' 2b. Replication is occurring normally in these cells; would you expect to find a primer in both positions? Why or why not?arrow_forwardYou are studying a bacterial metabolic pathway that results in the synthesis of a product, vitalin, that is essential for survival. In the figure below, each number represents an intermediate product, and each letter represents the enzyme (protein) that catalyzes that step. 3 >5'> Vitalin a Compound -"> 1 – Z. 4 > Vitalin You find that bacteria with mutations in either c or f can survive, but a bacterial cell that has mutations in both c and f cannot survive. For every mutation, assume that the bacteria have other functional proteins in the pathway. Which of the following mutants can grow on media containing only intermediate 3? Select all that apply Mutant f Mutant b Mutant d Mutant carrow_forward
- You have the following DNA sequence: 5'GGT ACG TTG GGG CTC CAT3' This sequence is transcribed and translated. If the G that is underlined changes to to a C the result will be - A) A nonsenese mutation B) A frameshift mutation C) A silent substitution D) A missense mutation You have the following DNA sequence: 5'GGT ACG TTG GGG CTC CAT3' This sequence is transcribed and translated. If the G that is underlined is deleted, then the result will be A) A nonsense mutation B) A frameshift mutation C) A silent substitution D) A missense mutatio If there are 3000 bases in the coding region of a gene, the gene will have A) 3000 amino acids B) 6000 amino acids C) 1000 amino acids D) 3000 codonsarrow_forwardThe type of DNA replication error illustrated in the diagram below is _______________________.arrow_forwardThe following sequence of DNA is part of the normal, wild-type gene. 5'TAC CGG GAC TTG AGC CGA TAG 3' A deletion occurs during DNA replication, causing the guanine shown in red to be removed from the nucleotide strand. What effect is this most likely to have on the final protein? Multiple Choice The deletion of the G will cause a single amino acid substitution in the codon in which it occurs. The deletion of the G will cause a frameshift, so that the first amino acid after the mutation will change but the rest of the protein is unaffected. The deletion of the G will cause a frameshift, resulting in a premature stop codon and a truncated protein. The deletion of the G will cause a frameshift, resulting in the loss of the normal stop codon and an abnormally long protein with an altered amino acid sequence. The deletion of the G will not have an effect on the final protein.arrow_forward
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