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6 Feb 2018

I am thinking I need to use the formula V=4/3 3.14 r^3 so r=2.5 cause d=5 so (4/3)(3.14)(15.625)= 65.42 um^3 to convert to L 65.42 x 1x10^-15 = 6.54x10^-14L. for part A

Part B i know to do the same think for volume and divide the two than times it by 100 to get a percent but i am uncertain on how to find the length

Part C dont know where to start.

The nucleus of a typical human cell is roughly spherical with a diameter of about 5 Î¼m. The nucleus contains, among other things, the DNA making up the genome, which for humans consists of 46 chromosomes containing a total of about 3Ã109 basepairs (bp). The DNA molecules are roughly cylindrical (but flexible) with a diameter of approximately 2 nm and a length of 0.34 nm per base pair.

(a) What is the volume of a typical nucleus? Express your answer in both Î¼m3 and liters.

(b) What is the total length of the DNA making up the human genome? What volume

does the DNA occupy? What fraction of the cell nucleus is occupied by DNA?

(c) An E. coli bacterium is roughly a cylinder 1Î¼m in diameter and 2Î¼m long. Its genome contains approximately 5Ã106 bp of DNA. What is the volume occupied by the E. coli genome? What fraction of the cell volume does the DNA occupy?

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Patrina SchowalterLv2

7 Feb 2018

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A typical eukaryotic chromosome found in humans contains about 100 million bp of DNA. 1 bp of DNA has a linear length of 0.34 nm.

A. What is the linear length of the DNA for a typical human chromosome in micrometers?

B. What is the linear length of a 30-nm fiber of a typical human chromosome?

C. Based on your calculation of part B, would a typical human chromosome fit inside the nucleus (with a diameter of 5 Î¼m) if the 30-nm fiber were stretched out in a linear manner? If not, explain how a typical human chromosome fits inside the nucleus during interphase.

raspberrycamel466

The bacterium E. coli, which is a major model system for the study of genes and DNA, has a singular circular chromosome of about 4x106 bp. In contrast, human diploid cells have __1__ chromosomes, of __2__ different types (donât forget that the X Â¹ Y!). The total amount of DNA found in a haploid cell such as a sperm or egg is called the organismâs âC-valueâ; the human C-value is __3__.

Also in contrast to the E. coli circular chromosome is the fact that eukaryotic chromosomes are not circular, but instead are __4__. Eukaryotic chromosomes are complexed heavily with protein to form a DNA-protein complex generically called __5__. The proteins found can be considered to fall into two general classes: __6__ and __7__. The first class is made up of 5 major proteins called __8__, __9__, __10__, __11__, and __12__. On a weight to weight basis, these proteins make up about __13__% of the total mass of a chromosome. Since these proteins must associate relatively non-specifically with DNA, they are rich in the basic amino acids __13__ and __14__, which gives the proteins a net __15__ charge.

One goal of packaging the DNA into chromatin is to compact the genome so it will fit into a nucleus. The basic unit of chromatin structure is the __16__, which has __17__ base pairs of DNA wrapped around a âcore particleâ composed of __18__ copies each of the core histones. Each nucleosome is separated by a stretch of âlinkerâ DNA so that the repeating unit is approximately 200 bp. Lets do a little arithmetic: If we consider that the nucleosome is approximately __19__ nm in diameter, that there is about 200 bp in this structure, and that 200 bp of DNA would normally occupy __20__ nm if stretched out straight (recall that each base is spaced about 0.34 nm from the next), we can see that the nucleosome alone allows for a __21__-fold compaction ratio.

The nucleosome structure is further compacted by histone __22__, which is not a part of the core particle. Instead, this histone __23__ (does what?) to cause the nucleosomal 10 nm fiber to condense into a higher-order structure referred to as the __24__ nm fiber. In the nucleus, chromosomes do not just float around freely: they are tethered to a proteinaceous structure called the __25__ at specific DNA regions called âSARsâ, which stands for __26__. SARs are rich in topoisomerases, to allow the winding and unwinding of the DNA that accompanies replication and transcription to occur, despite this constraint.

The 30 nm fiber discussed above is the form in which transcriptionally active DNA is believed to exist in the nucleus; however, further condensation leads to a highly condensed inactive form called __27__; in this terminology, the active form is referred to as __28__. The inactive condensed form can be divided into two types: one type is essentially invariant between cell types, and since it is âon all t timeâ, it is referred to as __29__. The other type comes and goes, depending on the expression of the genes present in that region of the chromosome; it is referred to as __30__.

Eukaryotic chromosomes carry the genes, of course, but also have other functional regions. One such region is the place where the spindle apparatus attaches during mitosis and meiosis; this region of the chromosome is called the __31__. Another important part of the linear eukaryotic chromosome is the ends; these are called __32__. In humans, the end of each chromosome is composed of many repeats of the short sequence __33__. One more type of functional region on each chromosome is the place where DNA replication starts; in humans, each chromosome has several of these __34__. Thus, eukaryotic chromosomes carry several different types of functional sequences, which together allow for their __35__ during S phase, __36__ during mitosis and meiosis, and the __37__ of their genes

In addition to identifying the genetic material, the experiments of Avery, MacLeod, and McCarty with different strains of Streptococcus pneumoniae demonstrated that

	DNA is a double helix.
	DNA may be taken up by bacterial cells and be active.
	DNA is present in bacteriophage T2.
	eukaryotic cells may be genetically transformed.
	DNA binds to the cell wall of the bacterium.

In order to show that DNA in cell extracts is responsible for genetic transformation in Streptococcus pneumoniae, important corroborating evidence should indicate that _______ also destroy transforming activity.

	temperatures high enough to kill cells
	enzymes that hydrolyze proteins
	enzymes that hydrolyze DNA
	enzymes that hydrolyze polysaccharides
	enzymes that hydrolyze RNA

Based on what you have learned about the experiments conducted by Griffith and Avery and colleagues with bacteria, which of the following would result in transformation of living R cells?

	Pure RNA from S cells
	Cell-membrane extracts from S cells
	A mixture of pure RNA and protein from S cells
	Cell-free extracts from S cells
	Pure protein from S cells

A-T base pairs in a DNA double helix

	form three hydrogen bonds with each other.
	are more heat-stable than G-C base pairs.
	are of a substantially different length than G-C base pairs.
	are not accessible to DNA binding proteins.
	are chemically distinct from G-C base pairs.

If 23 percent of the bases in a sample of double-stranded DNA are adenine, what percentage of the bases are uracil?

	27
	50
	0
	73
	23

The uniform diameter of the DNA structure provides evidence for

	antiparallel DNA strands.
	a right-handed helix.
	a double-stranded helix.
	3â² end phosphorylation.
	purineâpyrimidine pairing.

If a sequence of one strand of DNA is 5â²-TGACTATC-3â², what is the complementary strand?

	5â²-ACTGATAG-3â²
	3â²-TGACTATC-5â²
	3â²-ACTGATAG-5â²
	3â²-ACTGATAG-3â²
	5â²-TGACTATC-3â²

What structural aspect of the DNA facilitates dissociation of the two DNA strands for replication?

	3â² end phosphorylation
	Purineâpyrimidine pairing
	Antiparallel DNA strands
	Covalent bonds between sugars and phosphate groups
	Hydrogen bonding between paired bases

If the MeselsonâStahl density gradient experiment had resulted in two bands of DNA molecules after only one round of replication, one containing only ¹⁵N and the second only ¹⁴N, this result would have indicated that replication was

	unidirectional.
	dispersive.
	conservative.
	bidirectional.
	semiconservative.

10.

The nucleoside analogue acyclovir, which is used to treat herpes simplex virus (HSV) infections, lacks a 3â² hydroxyl group (âOH). Predict what will happen if the host cell DNA polymerase incorporates a molecule of acyclovir into an elongating strand of HSV DNA.

	The replication complex will no longer bind to origins of replication.
	The DNA polymerase will only incorporate acyclovir molecules.
	Acyclovir will bind single-strand binding proteins.
	DNA polymerase will not be able to link a successive nucleotide.
	Acyclovir will block the activity of the DNA helicase.

11.

Which of the following does not demonstrate the stability of the DNA double helix?

	Single-strand binding proteins are required to keep the strands apart.
	Paired bases form hydrogen bonds.
	High temperatures are required to denature it.
	DNA synthesis requires energy.
	DNA helicases require ATP.

12.

What effect would a primase inhibitor have on DNA replication?

	DNA polymerase would not be able to add bases to the DNA.
	Primase would be blocked from joining the DNA replication complex.
	Primase would not be able to provide primers for DNA polymerases.
	DNA polymerase would not be able to use DNA as a template.
	There would be no effect on DNA replication.

13.

To replicate their DNA in a timely manner, most eukaryotic chromosomes

	normally have uncoiled DNA.
	have multiple origins of replication.
	contain linear molecules of single-stranded DNA.
	have two replication forks that move in the same direction.
	are composed entirely of DNA.

14.

Which statement about DNA replication is false?

	Okazaki fragments are synthesized as parts of the leading strand.
	Error rates for DNA replication are reduced by proofreading of the DNA polymerase.
	The sliding clamp increases the rate of DNA synthesis.
	Replication forks represent areas of active DNA synthesis on the chromosomes.
	Ligases and polymerases function in the vicinity of replication forks.

15.

In many eukaryotes, there are repetitive sequences called telomeres at the ends of chromosomes. After successive rounds of DNA replication, the _______ strand becomes shorter. In some cells, an enzyme called _______ repairs the shortened strand.

	leading; telomerase
	lagging; telomerase
	lagging; DNA polymerase I
	leading; DNA polymerase I
	leading; replicase

16.

A researcher studies normal human fibroblast cells. They can be maintained in culture but die off after about 30 cell generations. Unexpectedly, a colony of cells continues to survive and divide past 30 generations. Which scenario is most likely true for these cells?

	DNA polymerase is constantly active.
	Primase is able to extend the telomeres.
	The mismatch repair system is extra efficient.
	The telomerase gene is being expressed.
	The cells do not need the ends of their chromosomes.

17.

If DNA polymerase III introduces an incorrect nucleotide, what is the first corrective action made by the DNA repair system?

	The replication complex excises the incorrect nucleotide.
	The excision repair proteins remove the incorrect nucleotide.
	The mismatch repair system scans for base-pairing mismatches.
	There is no correction, because nucleotide errors by DNA polymerases are not repaired.
	DNA ligase connects the correct nucleotide.

18.

Choose the correct order of the following four events in the excision repair of DNA:
(1) Base-paired DNA is made complementary to the template.
(2) Damaged bases are recognized.
(3) DNA ligase seals the new strand to existing DNA.
(4) Part of a single strand is excised.

	1, 2, 3, 4
	2, 1, 3, 4
	3, 4, 2, 1
	2, 4, 1, 3
	4, 2, 3, 1

19.

Six complete cycles of PCR should result in a _______-fold increase in the amount of DNA.

	64
	32
	12
	128
	6

20.

When double-stranded DNA is heated to temperatures above 90Â°C, it denatures. Denaturation is a process that

	breaks covalent bonds.
	hydrolyzes DNA.
	separates double-stranded DNA into two single strands.
	causes new hydrogen bonds to form.
	irreversibly destroys the double helical structure.