Genetics: Analysis and Principles
6th Edition
ISBN: 9781259616020
Author: Robert J. Brooker Professor Dr.
Publisher: McGraw-Hill Education
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Chapter 10, Problem 5EQ
When chromatin is treated with a salt solution of moderate concentration, the linker histone H1 is removed (see Figure 10.12a). A higher salt concentration removes the rest of the histone proteins (see Figure 10.18b). If the experiment of Figure 10.11 was carried out after the DNA was treated with moderately or highly concentrated salt solution, what would be the expected results?
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Students have asked these similar questions
You are studying a large eukaryotic gene that is 439,515 base pairs long. You find the polypeptide that this gene produces in liver cells is 46,771 amino acids long. Your colleague studies the function of this gene in brain cells, and finds the polypeptide produced in the brain is much larger – 61,438 amino acids long. How do you explain this difference?
Possible Answers:
A. The cell cycle of liver cells is much longer than that of brain cells.
B. This is due to alternative splicing. in the brain
C. There was a different complement of sequence-specific transcription factor binding sites in the CRM of the brain cells.
D. There is no 5' cap added to the gene product from the liver cells.
Cell (A) contains 3.1 billion (3.1 x 109) base pairs of DNA. Each nucleosome has about 200 bp of DNAwrapped around the histone core.a. What is the maximum number of nucleosomes that can be present in the cell?b. What is the maximum number of H2A histone protein molecules that can be present in the cell?
In the human gene for the beta chain of haemoglobin (the oxygen-carrying protein in the red blood cells), the first 30 nucleotides in the amino-acid-coding region is represented by the sequence:
3'-TACCACGTGGACTGAGGACTCCTCTTCAGA-5'.
What is the sequence of the partner strand?
4B. If the DNA duplex for the beta chain of haemoglobin above were transcribed from left to right, deduce the base sequence of the RNA in this coding region.
4C. In NOT more than 200 words, explain how eukaryotic RNA synthesized by RNA polymerase II is modified before leaving the nucleus?
Chapter 10 Solutions
Genetics: Analysis and Principles
Ch. 10.1 - 1. A bacterial chromosome typically contains
a. a...Ch. 10.2 - Mechanisms that make the bacterial chromosome more...Ch. 10.2 - 2. Negative supercoiling may enhance activities...Ch. 10.2 - 3. DNA gyrase
a. promotes negative supercoiling....Ch. 10.3 - 1. The chromosomes of eukaryotes typically contain...Ch. 10.4 - Which of the following is an example of a...Ch. 10.5 - What are the components of a single nucleosome? a....Ch. 10.5 - 2. In Noll’s experiment to test the...Ch. 10.5 - Prob. 3COMQCh. 10.5 - Prob. 4COMQ
Ch. 10.6 - Prob. 1COMQCh. 10.6 - 2. The role of cohesin is to
a. make chromosomes...Ch. 10 - Prob. 1CONQCh. 10 - Prob. 2CONQCh. 10 - 3. Describe the mechanisms by which bacterial DNA...Ch. 10 - Why is DNA supercoiling called supercoiling rather...Ch. 10 - Prob. 5CONQCh. 10 - Prob. 6CONQCh. 10 - Prob. 7CONQCh. 10 - Prob. 8CONQCh. 10 - Prob. 9CONQCh. 10 - 10. What is the function of a centromere? At what...Ch. 10 - Prob. 11CONQCh. 10 - 12. Describe the structures of a nucleosome and a...Ch. 10 - Beginning with the G1 phase of the cell cycle,...Ch. 10 - Draw a picture depicting the binding between the...Ch. 10 - 15. Compare heterochromatin and euchromatin. What...Ch. 10 - 16. Compare the structure and cell localization of...Ch. 10 - 17. What types of genetic activities occur during...Ch. 10 - Lets assume the linker region of DNA averages 54bp...Ch. 10 - 19. In Figure 10.12, what are we looking at in...Ch. 10 - 20. What are the roles of the core histone...Ch. 10 - A typical eukaryotic chromosome found in humans...Ch. 10 - Which of the following terms should not be used to...Ch. 10 - Discuss the differences between the compaction...Ch. 10 - 24. What is an SMC complex? Describe two...Ch. 10 - Two circular DNA molecules, which we can call...Ch. 10 - 2. Let’s suppose you have isolated DNA from a cell...Ch. 10 - 3. We seem to know more about the structure of...Ch. 10 - In Nolls experiment of Figure 10.11, explain where...Ch. 10 - When chromatin is treated with a salt solution of...Ch. 10 - 6. Let’s suppose you have isolated chromatin from...Ch. 10 - If you were given a sample of chromosomal DNA and...Ch. 10 - Consider how histone proteins bind to DNA and then...Ch. 10 - In Chapter 23, the technique of fluorescence in...Ch. 10 - Bacterial and eukaryotic chromosomes are very...Ch. 10 - The prevalence of highly repetitive sequences...Ch. 10 - Discuss and make a list of the similarities and...
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- In the human gene for the beta chain of haemoglobin (the oxygen-carrying protein in the red blood cells), the first 30 nucleotides in the amino-acid-coding region is represented by the sequence: 3'-TACCACGTGGACTGAGGACTCCTCTTCAGA-5'. What is the sequence of the partner strand? 4B. If the DNA duplex for the beta chain of haemoglobin above were transcribed from left to right, deduce the base sequence of the RNA in this coding region.arrow_forwardWhich of the following set(s) of primers a-d could you use to amplify the following target DNA sequence, which is part of the last protein-coding exon of the CFTR gene? Explain briefly. (Note: The three dots represent the body of the region to be amplified, whose beginning and end are only being shown.) 5' GGCTAAGATCTGAATTTTCCGAG . TTGGGCAATAATGTAGCGCCTT 3' 3' CCGATTCTAGACTTAAAAGGCTC . AACCCGTTATTACATCGCGGAA 5' a. 5' GGAAAATTCAGATCTTAG 3'; 5' TGGGCAATAATGTAGCGC 3' b. 5' GCTAAGATCTGAATTTTC 3'; 3' ACCCGTTATTACATCGCG 5' c. 3' GATTCTAGACTTAAAGGC 5'; 3' АССCGTTATTАСАТСGCG 5 d. 5' GCTAAGATCTGAATTTTC 3'; 5' TGGGCAATAATGTAGCGC 3'arrow_forwardA diploid human cell contains approximately 6.4 billion base pairs of DNA. Assuming that the linker DNA encompasses 35 bp, how many nucleosomes are present in such a cell? Use two significant figures. How many histone proteins are complexed with this DNA? use two significant figures.arrow_forward
- The nucleosome assembly factor exclusively functions in DNA-dependent nucleosome assembly and binds to the PCNA (clamp) of DNA polymerase. (Capitalization and dashes/spaces are not critical.)arrow_forwarda) Complete the table below. Assume that reading is from left to right and that the columns represent transcriptional and translational alignments. Label the 5’ and 3’ ends of DNA and RNA and carboxy and amino acid ends of protein. (You may fill in this chart by hand writing- no typing necessary here.) 2. b) Is the top or bottom DNA strand the template strand?arrow_forwardIn the first figure of “DNMT3L Connects Unmethylated Lysine 4 of Histone H3 to de novo Methylation of DNA,” the authors determined that the DNMT3L protein interacts with several histone proteins. Using Figure a and b (from that paper), explain how each of the 4 histones interact with DNMT3L.arrow_forward
- In the catalytic center of DNA polymerases, the presence of Mg2+ ions and aspartate residues close to these ions is required. What is the role of these aspartate residues that are found close to the magnesium ions? What is the role of the magnesium ions in the polymerization of nucleotides?arrow_forwardThe D1S80 locus is located on human chromosome 1 and is characterized by a repeating 16 base pair (bp) sequence. Alleles for this locus vary depending on the number of repeats present, thus affecting the size of the locus. The D1S80 locus also contains two conserved sequences, a 32bp sequence at one end and a 113bp sequence at the other end. If the DNA of an individual is targeted for D1S80 amplification, and one of the resulting amplicons is approximately 785bp in size, how many repeats would be present in this D1S80 allele? The amplicon of interest is indicated by a red arrow in the diagram below.arrow_forwardWhat is the length in AA’s of the LilP protein? Assume fMet is NOT CLEAVED. Enter just the number, nothing else! Write out the sequence of the polypeptide in AA: use the three letter notation, e.g. Met-Ser-Pro- A lilP mutant called lilPXS is isolated that produces a truncated polypeptide of only 6 AA in length. Describe a single basepair DNA change that would lead to this truncated version of the protein. Multiple options are possible (100 words max.)arrow_forward
- In the table below, there are four versions of gene A, one of which is normal, and the other three which contain mutations that make the gene product nonfunctional. Focus on the shaded region of the sequence. Use the genetic code table to answer the question. How would you describe Mutation #2? Partial DNA sequence for gene A ("..." indicates many nucleotides of sequence not shown) 5' ... ATG GTG AGC AAG GAG GAG CTG TTC ACC TGT AAA TAG ... Normal Mutation #1 5' ... ATG GTG AGC AAG GAG AAG CTG TTC ACC TGT AAA TAG ... Mutation #2 5' ... ATG GTG AGC AAG TAG GAG CTG TTC ACC TGT AAA TAG ... Mutation #3 5' ... ATG GTG AGC AAG GAG CTG TTC ACC TGT AAA TAG ... Silent mutation Nonsense mutation Frameshift mutations Missense mútationarrow_forwardAssuming that the histone octamer forms a cylinder 9 nm in diameter and 5 nm in height and that the human genome forms 32 million nucleosomes, what volume of the nucleus (6 μm in diameter) is occupied by histone octamers?arrow_forwardDraw the structure of the double Holliday junctionthat would result from strand invasion by both ends of thebroken duplex into the intact homologous duplex shownin Figure Q5–3. Label the left end of each strand in the Hol-liday junction 5ʹ or 3ʹ so that the relationship to the paren-tal and recombinant duplexes is clear. Indicate how DNAsynthesis would be used to fill in any single-strand gaps inyour double Holliday junction.arrow_forward
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