To answer this question, review Figures 17.10 and 17.11 and the accompanying text on pages 334-336 of Campbell Biology, 9th edition. 2. If 20% of the DNA in a guinea pig cell is adenine, what percentage is cytosine? Explain your answer, 1£ 20% is adenine, then 20% is thymine. The remaining 60% is composed of cytosine and guanine in equal pércentages so that eich makes up 30% of the DNA. 3. A number of different types of RNA exist in prokaryotic and eukaryotic cells. List the three main types of RNA involved in transcription and translation, Answer the questions to complete the chart. a. ‘Types of RNA ’. Where are they ©. Where and how do they produced? funétion in cells? mRNA Inthe nucleus, from mRNA functions in the specific genes (often called | cytoplasm, where itis structural genes) on the | translated into protein. The DNA’ mRNA carries the information in codons that determine the ordet of amino acids in a protein, tRNA Other genes in the nuclear | {RNA molecules function in| DNA code for tRNA the cytoplasm in translation molecules Hach tRNA molegule can combine with a specific amino acid. Complementary| base pairing of (RNA molecule with a codon in the A site of the ribosome brings the correct amino acid into position in the growing polypeptide chain. TRNA Still other genes inthe. | rRNA molecules combine : nuclear DNA code for with protein to form the 1RNA molecules: ribésomes, which serve as the base for interactions between mRNA codons and tRNA anticédons in traslation in the cytoplasm, (See Figure 17.17, page 340)b. For what sequence of amino acids does this mRNA, code? (Assume it does not contain introns.) i ‘Sequence of amino acids: methionine-arginine-serine-leucine-tryptophan- Jeucine-leucine . The chart lists five point mutations that may occur in the original strand of DNA. ‘What happens,to the amino acid sequence or protein produced as a result of each mutation? (Note: Position 1 refers to the first base at the 3” end of the transcribed strand. The last bese in the DNA strand, at the 5’ end, is at position 21.) Original template strand: 3’ TAC GCA AGC AAT ACC GAC GAA 5’ ‘Mutation Effect on amino acid sequence i. Substitution of T for G at _—_| This changes the codon! in mRNA to a stop| position 8. codon; translation stops at this point. A shorter (truncated) polypeptide is pro duced and this shortened polypeptide is likely to be nonfunctionél. ii, Addition of T between Serine is still incorporated as the third positions 8 and 9. amino acid, but the amino acids that follow all differ from the sequence in part b above. This is a frameshift mutation, iii, Deletion of C at position 15. | ‘The first four amino acids in the chain are not affected. The fifth amino acid becomes cysteine and the subsequent amino acids are also changed from part b. iv. Substitution of T forC at | The original mRNA'codon, CUG, and the position 18. ‘one resulting from the substitution, CUA, both code for leucine, $8 no change occurs, in the polypeptide sequence, ¥. Deletion of C at position 18. | Leucine is still inserted as the sixth amino acid in the polypeptide. However, since we're given only a part of the sequence, it is uncertain what the next amino acid in the chain will be, vi. Which of the mutations produces the greatest change in the amino acid sequence of the polypeptide coded for by this 21-base-pair gene? ‘The addition of T between positions 8 and 9 still leaves the third amino acid intact; however, all amino acids after that are different. In the substitution of T for G at position 8, a stop codon is inserted and only the first two amino acids are unaltered. As atesult, this mutation produtes the greatest change.4. Given your understanding of transcription and translation, fill inthe blanks below and indicate the 5’ and 3’ ends of each nucleotide sequence. Again, assume no RNA processing occurs Nontemplate strand of DNA: 5’ ATGTATGCCAATGCA3’ ‘Template strand of DNA: mRNA: Anticodons on complementary tRNA: Template strand of DNA: 3’ TACATACGGTTACGTS’ mRNA: "A UGUAUGCCAAUGCAY’ (RNA: 3’ UAC/AUA/CGG/UUA/CGU/S' Q 5. Scientists struggled to understand how four bases could code for 20 different amino acids. If one base coded for one amino acid, the cell could produce only four different kinds of amino acids (4). If two bases coded for each amino acid, there ‘would be four possible choices (of nucleotides) for the first base and four possible choices for the second base. This would produce 4” or 16 possible amino acids. a, Whatis the maximum number of three-letter codons that can be produced using only four different nucleotide bases in DNA? #, or 64 b, How many different codons could be produced if the codons were four bases long? 4, or 256 ‘Mathematical logic indicates that at least three bases must code for each amino acid. ‘This led scientists to ask: + How can we determine whether this is true? + Which combinations of bases code for each of the amino acids? ‘To.answer these questions, scientists manufactured different artificial mRNA strands. When placed in appropriate conditions, the strands could be used to produce polypeptides.‘Assume a scientist makes three artificial mRNA strands: (@) 5’ AAAAAAAAAAAAAAAAAAAAAAAAAA 3° (9) 5’ AAACCCAAACCCAAACCCAAACCCAAA 3 @ 5’ AUAUAUAUAUAUAUAUAUAUAUAUAU 3’ ‘When he analyzes the polypeptides produced, he finds that: x produces a polypeptide composed entirely of lysine. produces a polypeptide that is 50% phenylalanine and 50% proline. produces a polypeptide that is 50% isoleucine and 50% tyrosine. c. Do these results support the three-bases-per-codon or the four-bases-per-codon hypothesis? Explain. Only if there were three bases per codon would both y and z produce only two different kinds of amino acids in equal proportions. In fact, each strand would produce the two in alternating order; for example, the z strand would produce a polypeptide chain of isoleucine followed by tyrosine followed by isoleucine, then tyrosine, and so on, d. This type of experiment was used to discover the mRNA nucleotide codons for each of the 20 amino acids. If you were doing these experiments, what sequences would you try next? Explain your logic. ‘There are many possible ways to answer this question. One possibility follows: ‘Continue as above and make the remaining three types of mRNA made up of only one type of nucleotide—that is, poly G, poly U, or poly C. Then make all possible combinations of the nucleotides taken two at a time—for example, GCGC, CGCG, AGAG, and so on. Next, make other combinations of nucleotides taken three at a time—for example, AAAGGGAAAGGG and so on. Continue with combinations of nucleotides taken four at a time—for example, AAAAUUUUAAAAUUUUAAAAUUUU and so on. In this last example, if the codon for each amino acid is three bases long, these’ combinations of nucleotides should give you a maximum of three different types of amino acids in equal proportions or percentages. . Now that the complete genetic code has been determined, you can use the strand of DNA shown here and the codon chart in Figure 17.4 on page 329 in Campbell Biology, 9th edition, to answer the next questions, “Original template strand of DNA: 3’ TAC GCA AGC AAT ACC GAC GAA 3’ a. Ifthis DNA strand produces an mRNA, what does the sequence of the mRNA " read from 5’ to 3'2 Z mRNA = 5’ AUG CGU UCG UUA UGG CUG CUU 3’b. For what sequence of amino acids does this mRNA. code? (Assume it does not contain introns.) ‘Sequence of amino acids: methionine-arginine-serine-leucine-tryptophan- leucine-leucine ‘c. The chart lists five point mutations that may occur in the original strand of DNA. ‘What happens to the amino acid sequence or protein produced as a result of each mutation? (Nofe: Position 1 refers to the first base at the 3’ end of the transcribed strand. The last base in the DNA strand, at the 5’ end, is at position 21.) Original template strand: 3! ‘TAC GCA AGC AAT ACC GAC GAA 5’ ‘Mutation Effect on amino acid sequence i. Substitution of T for Gat * | This changes the codon in mRNA to a stop| position 8. codon; translation stops at this point. A. shorter (truncated) polypeptide is pro- : duced and this shortened polypeptide is likely to be nonfunctional, Addition of T between Serine is still incorporated as the third positions 8 and 9. amino acid, but the amino acids that follow all differ from the sequence in part b above. This is a frameshift mutation, iii, Deletion of C at position 15. | The first four amino acids in the chain are not affected. The fifth amino acid becomes cysteine and the subsequent amino acids ate also changed from part b, iv. Substitution of Tfor C at —_| The original mRNA codon, CUG, and the position 18, cone resulting from the substitution, CUA, both code for Jeivcine, s0 no change occurs in the polypeptide sequence. ¥.- Deletion of C at position 18. | Leucine is still inserted as the sixth amino acid in the polypeptide. However, since we're given only a part of the sequence, it is uncertain what the next amino acid in the chain will be, vi, Which of the mutations produces the greatest change in the amino acid sequence of the polypeptide coded for by this 21-base-pair gene? ‘The addition of T between positions 8 and 9 still leaves the third amino acid intact; however, all amino acids after that are different. In the substitution of T for G at position 8, a stop codon is inserted and only, the first two amino acids are unaltered. As a result, this mutation produées the greatest change.7. Sickle-cell disease is caused by a single base substitution in the gene for the beta subunit of hemoglobin. This base substitution changes one of the amino acids in the hemoglobin molecule from glutamic acid to valine. Look up the stiuctures of glutamic acid (glu) and yaline (val) on page 79 of Campbell Biology, 9th edition. ‘What kinds of changes in protein structure might result from this substitution? Explain, Glutamic acid is polar, and valine is nonpolar. Being polar, the glutamic acid molecule would have been able to interact with water and other polar molecules, but the valine molecule cannot.’As a result, unlike glutamic acid, valine is more likely to have an interior position in the hemoglobin molecule. 8. Why do dentists and physicians cover patients with lead aprons when they take mouth or other X-rays? ‘As noted in this and other chapters, X-rays, UV light, and many chemicals can damage DNA. Such damage can result in point mutations such as base substitutions, deletions, and insertions. These mutations can cause cancer. If they occur in the cells that will produce the gametes, the mutations can be passed on to offspring. As a result, lead aprons are used 10 shield the rest of your body from any stray radiation.