Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Chapter 1, Problem 13P
Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book.
An Estimation of Minimal Genome Size for a Living Cell Studies of existing cells to determine the minimum number of genes for a living cell have suggested that 206 genes are sufficient. If the ratio of protein-coding genes to non-protein-coding genes is the same in this minimal organism as the genes of Mycoplasma genitulium, how many proteins are represented in these 206 genes. How many base pairs would be required to form the genome of this minimal organism if the genes are the same size as M genilalium genes? (Section 1.5)
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Chapter 1 Solutions
Biochemistry
Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...
Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...
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- Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Preparing cDNA Libraries from Different Cells Describe an experimental protocol for the preparation of to cDNA libraries, one from anaerobically grown yeast cells and the second from aerobically grown yeast cell.arrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. A Simple Genome and Its Protein-Encoding Capacity The genome of the Mycoplasma genitalium consists of 523 genes, encoding 484 proteins, in just 5S0.U74 base pairs (see Table 1.6). What fraction of the M. genitalium genes encode proteins? What do you think the other genes encode? If the fraction of base pairs devoted to protein-coding genes is the same as the fraction of the total genes that they represent, what is tlie average number of base pairs per protein-coding gene? If it takes three base pairs to specify an amino acid in a protein, how many amino acids are found in the average M. genitalium protein? If each amino acid contributes, on average, 12U daltons to the mass of a protein, what is the mass of an average M. genitalium protein? (Section 1.5)arrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Abundance of the Different Bases in the Human Genome Results on the human genome published in Science (Science 291 :1304—1350 [2001]) indicate that the haploid human genome consists of 2.91 gigabase pairs (2.91 X ]09 base pairs} and that 27% of the bases in human DNA are A. Calculate the number of A. T, G, and C residues in a typical human cell.arrow_forward
- Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Calculating Tms and Separating DNA Molecules That Differ in G:C Content At 0.2 M Na+, the melting temperature of double-stranded DNA is given by the formula, Tm = 69.3 + 0 41 (% G + C). The DNAs from mice and rats have (G + C) contents of 44% and 40%, respectively. Calculate the Tms for these DNAs in 0.2 M NaCl. If samples of these DNAs were inadvertently mixed, how might they be separated from one another?arrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Deducing DNA Sequence from Sanger Sequencing Results The output of an automated DNA sequence determination by the Sanger dideoxy chain termination method, performed as illustrated in Figure 11.3, is disp1ayed at right. What is the sequence of the original oligonucleotide?arrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. An Estimation of the Number of Cenes in a Virus Virus genomesrange in size from approximately nucleotides to approximately 280,000 base pairs. If viral genes are about the same size as M. genilalium genes, what is the minimum and maximum number of genes in viruses? (Section 1.5)arrow_forward
- Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. The Sequence Relationship Between an Antisense RNA Strand and Its Template DNA Strand The DNA strand that is complementary to the template strand copied by RNA polymerase during transcription has a nucleotide sequence identical to that of the RNA being synthesized (except T residues are found in the DNA strand at sites where U residues occur in the RNA). An RNA transcribed from this nontem-plate DNA strand would be complementary to the mRNA synthesized by RNA polymerase. Such an RNA is called antisense RNA because its base sequence is complementary to the “sense mRNA. One strategy to thwart the deleterious effects of genes activated in disease slates (such as cancer) is to generate antisense RNAs in affected cells. These antisense RNAs would form double-stranded hybrids with mRNAs transcribed from the activated genes and prevent their translation into protein. Suppose transcription of a cancer-activated gene yielded an mRNA whose sequence included the segment 5’-UACGUCUAAGCUGA. What is the corresponding nucleotide sequence (5’ The template strand in a DNA duplex that might be introduced into these cells so that an untisense RNA could be transcribed from it?arrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Chargaff’s Rules for the Base Composition of DNA Chargaff’s results (Table 10.1) yielded a molar ratio of 1 1.29 for A to G in ox DNA, 1.43 for T to C, 1.04 for A to T, and 1.00 for G to C. Given these values, what are the approximate mole fractions of A. C, G. and T in ox DNA?arrow_forwardAnswers to all problems are at the end οΓthis book. Detailed solutions are available in the Student Solutions Manual. Study Guide, and Problems Book. Superbug infections are becoming more common around the world. Many of these infections arise from the action of -lactamases, of which there are several types with different mechanisms of action. Consult the end-of-chapter reference by von Nussbaum and Schiffer and write detailed mechanisms for the serine -lactamases and metallo- -lactamases.arrow_forward
- Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Calculate the Number of Nucleosomes in a Human Diploid Cell Assume that there is one nucleosome for every 200 bp of eukaryotic DNA. How many nucleosomes are there in a diploid human cell? Nucleosomes in be approximated a disks 11 nm in diameter and 6 nm long. If all the DNA molecules in a diploid human cell are in the B-conformation, what is the sum of their lengths? If this DNA is now arrayed on nucleosomes in the beads-on-a-string motif, what would be the approximate total height of the nucleosome column if these disks were stacked atop one another?arrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Structural complementarity is the key to molecular recognition, a lesson learned in Chapter 1. The principle of structural complementarity is relevant to answering problems 5, 6, 7,11, 12, and 19. The quintessential example of structural complementarity in all of biology is the DNA double helix. What features of the DNA double helix exemplify structural complementarity?arrow_forward
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