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DNA Replication QUESTIONS1. Below is a single strand of

what is charged trna

DNA Replication: QUESTIONS 1. Below. Show More

DNA Replication: QUESTIONS 1. Below is a single strand of a DNA molecule. -T T C G A G A C T T A G T C G G A T G T G A A G T G G T G A T T - Replicate this strand. 2. DNA replication is described as semi-conservative. Explain why this term is appropriate. 1. What is the difference between transcription and translation. 2. What is the role of ribosomal RNA in protein synthesis. 3.Why are proteins so important in living organisms. 4. Here is a chain of triplets, transcribed to mRNA. AUG GUA UAC CCC AAG UGG GGU ACC CGG UAG For which amino acids do they code? Use the table below. Table 10.1 Codon Table Second First C U A G Third C Proline [Pro, P] Leucine [Leu, L] Histidine [His, H] Arginine [Arg, R] C U Proline [Pro, P] Leucine [Leu, L] Histidine [His, H] Arginine [Arg, R] A Proline [Pro, P] Leucine [Leu, L] Glutamine [Gln, Q] Arginine [Arg, R] Proline [Pro, P] Leucine [Leu, L] Glutamine [Gln, Q] Arginine [Arg, R] G U Serine [Ser, S] Phenylalanine [Phe, F] Tyrosine [Tyr, Y] Cysteine [Cys, C] C U Serine [Ser, S] Phenylalanine [Phe, F] Tyrosine [Tyr, Y] Cysteine [Cys, C] A Serine [Ser, S] Leucine [Leu, L] Stop Stop Serine [Ser, S] Leucine [Leu, L] Stop Tryptophan [Trp, W] G A Threonine [Thr, T] Isoleucine [Ile, I] Asparagine [Asn, N] Serine [Ser, S] C U Threonine [Thr, T] Isoleucine [Ile, I] Asparagine [Asn, N] Serine [Ser, S] A Threonine [Thr, T] Isoleucine [Ile, I] Lysine [Lys, K] Arginine [Arg, R] Threonine [Thr, T] Methionine [Met, M] Lysine [Lys, K] Arginine [Arg, R] G G Alanine [Ala, A] Valine [Val, V] Aspartic acid [Asp, D] Glycine [Gly, G] C U Alanine [Ala, A] Valine [Val, V] Aspartic acid [Asp, D] Glycine [Gly, G] A Alanine [Ala, A] Valine [Val, V] Glutamic acid [Glu, E] Glycine [Gly, G] Alanine [Ala, A] Valine [Val, V] Glutamic acid [Glu, E] Glycine [Gly, G] G 5.What amino acid would be encoded for if a mutation changed UAC (third position) to UAA. 6. What amino acid would be encoded for if a mutation changed AAG (fifth position) to AAA. 7. What would the mRNA message (in nucleotides) for this polypeptide look like? (Where you have a choice for which codon to use, use any one you like.) methionine-tryptophan-proline-tyrosine-cysteine-stop. 8. Fill in the missing nucleotides in the chart below: DNA < ATA. CGC. GCC. ACT. mRNA. UAU. UUU. tRNA. UAC. 9. Which amino acids are coded for here? (Be careful!). 10. Sickle cell disease is caused by an apparently minor change in the sequence of bases in the DNA which codes for the beta chain in hemoglobin. The sequence for the start (the first eight amino acids in a sequence of 146) of the normal hemoglobin B chain is: valine-histidine-leucine-threonine-proline-glutamate-lysine. whereas that for abnormal hemoglobin B (which leads to sickle cell disease) is: valine-histidine-leucine-threonine-proline-valine-glutamate-lysine. What is the minimum change necessary in the mRNA code for this sequence to bring about this change—from glutamate to valine? Indicate the original and mutated codon. 11. Explain this statement: "The secret of protein synthesis lies in nucleotide complementarity." OVERVIEW TO HELP YOU ANSWER THE QUESTIONS Biotechnology: Fingerprint Activity Just like fingerprints, DNA is unique to each person. Also like fingerprints, DNA can be used to identify a person when biological material containing DNA is left at a crime scene. Several techniques are available to generate DNA fingerprints from biological samples. Here we will illustrate the use of restriction enzymes and gel electrophoresis. Restriction enzymes are naturally occurring proteins produced by bacteria. These enzymes function like DNA scissors; they look for specific sequences within a piece of DNA, and where those sequences occur, they cut the DNA through both strands. For example, one restriction enzyme recognizes the sequence -GGATCC-, and will cut a piece of DNA containing that sequence between the A and the T. If this enzyme were applied to the piece of DNA below, it would cut it into three pieces, as shown in figure 11.1 Restriction Enzyme Activity Notice the enzyme cut the DNA only where the sequence -GGATCC- occurred. This cut the original piece of DNA that was 25 nucleotide pairs long into three fragments that were 7, 8, and 10 nucleotide pairs long. The sizes of the fragments produced will vary depending on the source of the DNA; thus, the fragmentation pattern can be used to identify the source of a particular DNA sample. To visualize the fragment pattern produced from restriction enzyme digestion of a DNA sample, forensic scientists use a process called gel electrophoresis (visit for additional information and an interactive virtual experiment). In this procedure, a dense, gelatin-like material is poured between two closely spaced glass plates. Once the material has solidified, DNA samples are loaded into slots cut in the top of the gel, and a strong electric field is applied to it for a

certain period of time (60 minutes, for example), with the negative pole at the top and the positive pole at the bottom. Because DNA is negatively charged, it will migrate through the gel toward the positive pole. The smaller the fragment of DNA, the faster it can travel through the gel, so fragments of different sizes will separate from each other, with the smaller, faster-moving fragments finishing near the bottom, and larger, slower-moving fragments staying closer to the top. This will produce a banding pattern, like a DNA bar code, that can show the unique differences between individuals. The animation in figure 11.2 summarizes the procedure we have just described: 1. DNA fragments migrate on the gel as electrical current is applied (smaller fragments migrate faster). 2. After a certain time period, the current is shut off and the gel is stained to view the various patterns of the DNA samples. Figure 11.2 Gel Electrophoresis Note that in figure 11.2, lane 2 represents the banding pattern that would be produced from the restriction enzyme activity illustrated in figure 11.1 above (yielding fragments of 7, 8, and 10 nucleotide pairs). The Crime Scene In this activity, you will be given DNA isolated from two samples left at a crime scene. A victim has been found murdered. The forensics team has gathered DNA evidence from blood on a power strip and from material found under the victim's fingernails. Your job is to compare the DNA from the samples with the DNA of six known subjects and determine if there is a match. To process these samples you will perform the following steps: First, examine the DNA sequences for the presence of the restriction enzyme recognition sequence -GGATCC-, and wherever it occurs, "cut" the DNA through both strands between the A and the T. Then, determine the fragment sizes that will be produced from cutting the DNA at these sites by counting the number of nucleotide pairs in each fragment. Next, if you have access to a printer, print the page that includes figure 11.3 and draw the appropriate line markings into lanes 1 and 2. Finally, compare the fragment patterns from the two samples to the known patterns of the suspects to determine the source of the two biological samples. (If you can’t print the figure, list your fragment lengths in increasing order and find the lane whose markings match those for samples 1 and 2.) DNA Samples The DNA sequences from the two samples found at the crime scene are given below. Recall the restriction enzyme recognition site: -GGATCC-. The enzyme cuts between each occurrence of A and T in both samples. Sample 1: Blood from power strip -AAACGCGGGATCCCCGATTAGGATGGATCCGGATCCAGAGGATCCAGAGGATACGC- -TTTGCGCCCTAGGGGCTAATCCTACCTAGGCCTAGGTCTCCTAGGTCTCCTATGCG- Sample 2: Biological material under victim's fingernails -GAAACGCGGGATCCCCGATTAGGATGGATCCGGATCCAGAGGATCCAGAGCATCCGC- -CTTTGCGCCCTAGGGGCTAATCCTACCTAGGCCTAGGTCTCCTAGGTCTCGTAGGCG- "Cut" the strands and enter the resulting fragment sizes into the boxes below. QUESTIONS Analysis and Results 1. Fragment sizes for sample. 2. Fragment sizes for sample. Figure 11.3 DNA Fingerprinting Gel Lane 1: Blood sample from power strip Lane 2: Material from under victim's fingernails Lane 3: Subject A's (Victim's) DNA Lane 4: Subject B's DNA Lane 5: Subject C's DNA Lane 6: Subject D's DNA Lane 7: Subject E's DNA Lane 8: Subject F's DNA 3. Sample found on power strip came from. 4. Sample found under victim's fingernails came from. Critical Thinking Questions & Summary 1. What conclusions can or cannot be drawn from the DNA analysis you conducted in the DNA Fingerprints Activity. 2. Both conventional fingerprints (from your fingers) and DNA fingerprints are ways to identify individuals present at a crime scene. Why is DNA fingerprinting a more reliable way to positively identify those present at a crime scene than using conventional fingerprints. 3. In some cases, juries have failed to find suspects guilty of a crime despite DNA fingerprint evidence showing their presence at the crime scene. Describe two ways DNA evidence can be compromised, leading to uncertainty. 4. Besides forensic applications, what are some other ways in which DNA fingerprinting may be useful? Describe three other ways DNA fingerprinting might be useful in providing positive identification. 5. Summarize, what you have learned from the DNA Fingerprints Activity.

NOTE: The DNA strand on the right-hand side is transcribed into RNA.

Remember: the amino acids are laid out on your screen in a straight line, to make them easier to work with. In reality, somatostatin has a three-dimensional structure, so a change in an amino acid will also result in a change in physical shape.

Now proceed to mutate your molecule, as instructed below.

Mutation 1

Sometimes mistakes occur in nucleotide pairing in DNA, caused by chemical mutagens (pesticides or inhaled smoke, for example). An A-T pair may become reversed to T-A, or a C-G pair might be substituted for an A-T pair.

Assume that something like this has happened in the last DNA triplet (position 14) in the code for somatostatin.

Alter the DNA triplet at random and then determine what happens to the mRNA and thus to the amino acid sequence.

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