Bio 30 Unit 3 Lesson 2 Assignment

Biology 30 Unit 3 Lesson 2 Assignment Total /51 1. Which of the following bases is not found in RNA? A. adenine B. uracil C. cytosine D. thymine /1 2. How do purines compare to pyrimidines? Essential constituents of nucleotides, purines and pyrimidines form the cornerstone structures for both DNA and RNA. Purines such as adenine (A) and guanine (G) with their double ring structure, stand larger than the single ring counterparts: the pyrimidines. Adenine pairs exclusively with thymine (T) within DNA strands; conversely guanine forms a complementary duo with cytosine (C) thus underscoring their pivotal role in genetic coding. Thymine (T) and Cytosine (C), representing pyrimidines in DNA; Uracil (U) in RNA demonstrate a single-ring structure, an integral part of their molecular composition: this is indeed its defining feature. The intricate symphony that orchestrates the existence of living organisms results from genetic information storage and transmission a process vital to life itself through complementary base pairing within DNA where Adenine pairs with Thymine, while Guanine aligns perfectly with Cytosine. /2 3. The unit comprised of a deoxyribose sugar, a phosphate group, and a nitrogen base is a(n) A. nucleic acid B. nucleotide. C. base pair. D. amino acid. /1 4. Watson and Crick's model of DNA predicted that the molecule existed as a double helix, with bases opposite one another forming hydrogen bonds. In order for the helix to take on a regular conformation, they proposed that A. purines paired with purines. B. pyrimidines paired with pyrimidines. C. purines paired with pyrimidines.

D. pyrimidines paired with uracil /1 5. Analysis of a sample of DNA reveals that 33% of the nucleotides contain adenine. What is the percentage of nucleotides which contain thymine and cytosine, respectively? A. 33% and 17% B. 17% and 67% C. 33% and 34% D. 33% and 67% /1 6. DNA replication is termed semiconservative because A . daughter double helices contain one parental strand and one new strand. B. one daughter double helix is parental, and the other has two newly synthesized strands. C. new DNA and old DNA coexist on the same strand after replication. D. the newly synthesized strands of DNA are complimentary. /1 7. Adding a single base to the DNA sequence that codes for a protein would result in a A. a nonsense mutation, and would always inactivate the protein involved. B. a missense mutation, and may or may not inactivate the protein. C. a frameshift mutation, and would change the amino acid sequence from that point on. D. a reversion mutation. /1 8. Explain: a) nonsense mutation: This mutation type forms a premature stop codon, which in turn prematurely terminates mRNA translation and produces a truncated usually nonfunctional protein; thus, it results in loss of protein function. b) missense mutation: This mutation type forms a premature stop codon, which in turn prematurely terminates mRNA translation and produces a truncated usually nonfunctional protein; thus, it results in loss of protein function. c) silent mutation: This type of mutation maintains the integrity of the corresponding protein's amino acid sequence without altering it. Often, these mutations occur at a codon's third position where several codons can code for an identical amino acid.

10. An enzyme which builds messenger RNA by reading the DNA sequence. A. RNA ligase B . RNA polymerase C. ribosomal gyrase D. RNA amylase /1 11. Which row correctly identifies the enzyme-directed processes, 1,2,and 3? Row Process 1 Process 2 Process 3 A. ligase restriction enzymes ligase B. restriction enzymes restriction enzymes ligase C. ligase ligase restriction enzymes D. restriction enzymes ligase restriction enzymes /1

12. Although the adenine base composition of DNA from the above organisms is very similar, these organisms vary greatly in their characteristics. The cause of this variation is that the DNA molecules of the respective organisms contain different A. pairings of nitrogen bases B. sugars in their nucleotides C. sequences of nitrogen bases D. proteins in their nucleotides /1 Use the following information to answer the next question The following statements are related to the information available to Watson and Crick before they began to work out the structure of a DNA molecule. 1. The number of adenine molecules always equals the number of thymine molecules. 2. The DNA molecule consists of four nitrogen bases only. 3. The DNA molecule consists entirely of oxyribose. 4. The number of guanine molecules always equals the number of cytosine molecules. 13. Which of the statements are true? A. 1 and 2 B. 1 and 3 C. 3 and 4 D . 1, 2 and 4 /1

15. The results of the experiment described in the information supported the semiconservative model of DNA replication. According to this model, how many bands would be formed after centrifuging and why? A . One band, because each DNA molecule would be a hybrid of intermediate density B. Two bands, because some DNA contained 15 N nucleotides and some contained 14 N nucleotides C. Three bands, because some DNA contained only 15 N nucleotides, some only 14 N nucleotides and some DNA contained both D. More than three different bands, because the 15 N nucleotides were randomly dispersed among the 14 N nucleotides in the DNA /1 16. Refer to the information. How did this experiment demonstrate that the replication of a DNA molecule is semi-conservative? Utilizing 15N labeled bacteria, the experiment demonstrated semi conservative DNA replication: a single replication cycle in a 14N medium followed by centrifugation; this yielded an emerging band. This band was composed of hybrid 15N/14N DNA with intermediate density to showcase its successful synthesis and presence. A second band significantly lighter in density denoted the presence of newly synthesized 14N DNA: thus, manifesting effective replication and growth processes at work. The semi conservative model aligns daughter molecules in a pattern of inheritance, with each carrying one original strand and a newly synthesized one. The absence of bands that alternative models conservative and dispersive replication predict further substantiates the evidence for DNA replication's semi-conservative nature. /2 Use the following information to answer the next 2 questions. Gene for human insulin Structure X 532004 17. The process illustrated in the diagram A. is called deletion

B. is called a mutation C. is called recombinant DNA D. is and example of reverse transcription /1

Use the following information to answer this question. T A T G A A T T C T A A T C A A T T A A G A T T T A T G A A T T C T A A T C A A T T A A G A T T human DNA segment DNA cut by enzyme DNA cut by enzyme X T A T G A A T T C T A A T C A A T T A A G A T human DNA segment plasmid DNA segment DNA molecules linked together DNA molecules linked together Z plasmid DNA segment X 240081 21. In the diagram, the structures labelled X are molecules of A. ligase B. DNA polymerase C. RNA polymerase D . restriction enzymes /1

22. In the diagram, the structure labelled Z is a A. nucleotide B. repair emzyme C. ligase molecule D. restriction enzyme /1 Use the information to answer the questions. Some restriction enzymes do not cut DNA straight across. Instead they create a jagged cut by cleaving each strand a few nucleotides apart. This creates “sticky ends” on each DNA fragment as illustrated below. restriction enzyme + DNA fragments DNA molecule 240127 A restriction enzyme called ECO RI acts upon DNA by cleaving both strands only between adjacent adenine and guanine nucleotides as illustrated below by the arrows. 240128 DNA molecule . . . TATGAATTCCAA . . . Strand 1 . . . ATACTTAAGGTT . . . Strand 2 23. Which of the following sequences illustrates the sticky ends that are formed by this enzyme in strands 1 and 2 respectively? A. AATT and TTAA B. TTAA and AATT C. CCAA and ATAC D. TATG and GGTT /1 24. A reasonable hypothesis to explain these results is that the mutation in the mitochondrial DNA that caused KSS in the man first occurred in the

Gyrase: a crucial function in the process of DNA replication relieves supercoiling tension. This action permits DNA helicase to unwind and divide the double stranded DNA at its origin into two separate single strands. Primase: an Enzyme. The function of this process involves synthesizing a short RNA primer, which is complementary to the DNA template at the replication fork; consequently, it provides an initial point for DNA polymerase. DNA Polymerase III is an enzyme. The function initiates the addition of nucleotides to the 3' end of the RNA primer. It elongates the DNA strand on the leading strand in a continuous 5' to 3' direction. Please provide the sentence or paragraph that needs to be rewritten in active voice with graduate level punctuation, particularly colons, semi-colons and dashes. DNA polymerase I is an enzyme. This function involves the removal of the RNA primer and its replacement with DNA nucleotides; at the same time, it proofreads and corrects errors in the newly synthesized DNA strand. The process known as Lagging Strand Synthesis a fundamental step in DNA replication; involves the construction of the lagging strand complementary to the leading strand using short fragments called Okazaki fragments: these are subsequently linked together by DNA ligase. DNA polymerase III and DNA ligase are enzymes. DNA Polymerase III functions by synthesizing short fragments known as Okazaki fragments on the lagging strand; it does this in a direction away from the replication fork specifically, from 5' to 3'. Following this process, DNA ligase joins these individual pieces and forms an unbroken, continuous strand of DNA. The termination of our professional relationship was unavoidable due to the circumstances. Topoisomerase: the Enzyme. This function aids: in the ultimate relaxation of supercoiled DNA; it resolves any lingering tension generated specifically during the replication process. /10

Use the following diagram and information to answer this question (a) regulator gene (r) operator gene (o) structural gene (s) mRNA (d) repressor molecule attaches to operator site (b) no enzyme synthesis r o s repressor molecules (c) repressor molecules inducers and repressors combine inducer molecules (lactose) enzyme (lactase) synthesis repressors no longer fit operator mRNA r o s r o s 240091 28. The enzyme  -galactosidase digests milk sugar, lactose into glucose and galactose. This enzyme is produced by a bacteria called E-coli only when the inducer molecule, lactose is present. A repressor protein from the regulator gene binds to the operator gene and prevents the structural gene from producing the mRNA that produces galactosidase if lactose is not detected. Explain what will be the result of a mutation in the gene that codes for the repressor protein. A mutation in the gene that encodes the lac repressor protein can significantly alter regulation of the lac operon. Should a mutation result in constitutive expression, effective binding ability of the repressor protein to its operator gene might be lost; this loss could cause continuous transcription of the lac operon even without lactose present. As a result, β-galactosidase an enzyme encoded by a structural gene would undergo continual production independent from lactose availability and potentially overproduce itself. Conversely, the mutation may induce hyperactivity or dysfunction in the repressor protein causing impaired expression. Even with present lactose, this malfunction could inhibit production of mRNA by the structural gene. The lac operon's normal response to lactose would consequently be hindered due to such impairment; it could result in a diminished or non-existent production of β-galactosidase despite abundant presence of lactose thus impacting effective metabolism when available. /2 Use the information to answer the next question.