12)G
13)E
14)I
15)J
16)D
17)F
18)A
19)B
20)H
21)E
22)G
23)C
24)B
25)F
26)H
27)A
28)D
29)B
30)F
31)A
32)D
33)C
34)E
12. Nucleus 13. Point mutation 14. Deletion mutation 15. Exons 16. Translation 17. Nitrogenous bases H...
11. Deoxyribose 12. 13. 14. 15. 16. Nucleus Point mutation Deletion mutation Exons Translation Nitrogenous bases H bonded mRNA A. Molecule that carries instructions for making a protein from a gene in the nucleus to a ribosome in the cytoplasm B. Enzyme that unwinds DNA double helix C. Sugar found in DNA nucleotide D. Process of making a protein E. Substitution of one nucleotide base pair for another F. Rungs (steps) of DNA "ladder" G. Transcription occurs in this part...
UNIT V Chapter 12,13,814 WORKSHEET1. DNA geneA. Complementary to Cytosine2. Transcription 3. Sugar-Phosphate sackbone 4. IntronsB. With proteins makes up ribosomes5. Semi-conservative6. tRNA.E. Uprights (sides) of DNA "ladder"7. DNA Ligase.F. Part of a gene that gets spliced out during RNA processing G. Enzyme that joins lagging strand fragments of DNA or "sticky ends"9. Guanine?10. Anti-codon H. Molecule that carries amino acid to translation site1. New DNA contains one old and one new strand12. Nucleus 13. Point mutation 14. Deletion mutationA....
21. Double helix22. Repressor protein23. Adenine24. Ribosome.25. Promoter26. Replication27. RNA Polymerase.28. CodonA. Enzyme that synthesizes RNAB. Organelle where proteins are assembledC. Complementary to either Thymine or UracilD. mRNA sequence that codes for one amino acidE. Shape of double stranded DNAF. Sequence of DNA that controls gene expressionG. binds an operator and stops gene expression in LAC operon by preventing RNA polymerase from binding gene and transcribing. H. Duplication of DNA in 5 phase of Interphase
DNA DNA Replication: ONA Because DNA Is the ge m Tumes and heart e ine in process called DNA curs in the nucleus of s acest FS Parent strand Parent strand Newly replicated DNA Newly replicated DNA- SA0 Daughter DNA molecule Daughter DNA molecule Figure 8.2: Overview of DNA replication and illustration of complementary base pairing. DNA must replicate before cell division so that each new daughter cell receives an exact copy of the parent DNA. 1. Replication begins when...
Review Questions BIOL 260: Chapters 8-10, 13, 19 1. Consider a mutation involving the deletion of either 1, 2, or 3 nucleotides in the DNA of a bacterium. Which of these mutations (ie., deletion of 1, 2, or 3 nucleotides) would likely have the LEAST impact on the organism? Why? Include in your answer a comparison with the other two options to justify your reasoning. Think carefully about the impact each mutation would have on the ultimate protein coded for...
13. Why are ribonucleoside triphosphates the monomers required for RNA synthesis rather than ribonucleoside monophosphates? A. Only ribonucleoside triphosphates contain the sugar ribose. B. Ribonucleoside triphosphates have low potential energy, making the polymerization reaction endergonic. C. Ribonucleoside triphosphates have high potential energy, making the polymerization reaction exergonic. D. Ribonucleoside monophosphates cannot form complementary base pairs with the DNA template. E. Ribonucleoside triphosphates are not used, rather all use deoxyriboside triphosphates. 14. How is a mutation in a bacterial cell that...
answer all the questions 1) All of the following contribute to promoter binding by RNA polymerase I in bacteria except: a)-10 consensus sequence b)-35 consensus sequence c) rho factor d) sigma factor e) none of the above 2) Common structural changes or lesions found in DNA after exposure to ultraviolet light are: a) thymine dimers b) cytosine dimers c) purine dimers d) adenine dimers e) none of the above 3) What is the function of the sigma subunit in the...
The following statements apply to concepts and material discussed in Chapter 15; identify which statement is TRUE. Answers: A common ancestor of two species on evolutionary trees can be found at the point where the two branches meet. Humans evolved from Neanderthals about 50,000 years ago. Mitochondrial DNA is not particularly useful when trying to determine the movement pattern of humans historically across the globe. Scientists can estimate when species diverged from a common ancestor by comparing their Karyotypes. We...
2. A dominant allele H reduces the number of body bristles that Drosophila flies have, giving rise to a “hairless” phenotype. In the homozygous condition, H is lethal. An independently assorting dominant allele S has no effect on bristle number except in the presence of H, in which case a single dose of S suppresses the hairless phenotype, thus restoring the "hairy" phenotype. However, S also is lethal in the homozygous (S/S) condition. What ratio of hairy to hairless flies...