Biology 2581B Lecture Notes - Lecture 5: Chloroplast, Hepatocyte, Species Richness

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jadedolphin26

10 May 2016

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Genome diversity: size: recall: a genome is a set of genetic instructions within a biological compartment. Genome size is the length of one set of all the genetic instructions within a genetic compartment. If all the genomes are identical within a compartment, then you only need to count one set since they are all identical. 1 000 000 000 base pair (gb) = 1pg (picograms for mass) Nucleotide (used for single stranded dna) & base pair (used for double stranded dna) Haploid human nuclear genome: mass: 3 pg, length: 3 billion bp = 3 gb, how to estimate genome size, rst sequence, then assemble sequencing. Extract dna from the polytomella sample, isolate the total dna via palleting the sample. Obtain sequence (for polytomella 2 genomes) assembling. Assemble into contigs by looking for overlaps, create different sets of contigs that corresponds to different genomes (for polytomella, a nuclear genome set and a mitochondrial genome set: large genomes could lead to mistakes.

Related Questions

Question 1

1 pts

Cystic fibrosis is caused by nonsense and missense mutations in the CFTR gene, which encodes for a chloride channel. You are studying cystic fibrosis patients to determine what mutation they possess in the CFTR gene. The difference between the mutant and wild type CFTR genes can be uncovered by examining the CFTR:

	DNA
	mRNA
	protein
	tRNA

Question 2

1 pts

You decide to identify the CFTR mutation by analyzing the genomic DNA of your patients compared to healthy individuals. You specifically are looking to see whether a specific 3' gene truncation has occurred in the patients. You will determine this using hybridization techniques with samples from healthy and CF patients. Which of the following will allow you to accomplish this?

	Using an RNA probe complementary to the region not removed by the truncation.
	Using an RNA probe complementary to the region removed by the truncation.
	Using an DNA probe complementary to the region not removed by the truncation.
	Using an DNA probe complementary to the region removed by the truncation.

Question 3

1 pts

You would like to ensure that this experiment (to determine whether patients have a specific CFTR gene truncation using hybridization) is properly controlled. Which of the following samples must you test?

	The genomic DNA of a healthy individual who does not have cystic fibrosis.
	The genome of a CFTR patient known to have the specific truncation you are trying to identify.
	The genome of a CFTR patient with a missense mutation but full length gene.
	The genome of a healthy individual married to a CFTR patient with the specific truncation you are trying to identify.
	The genome of a patient with muscular dystrophy, which can be due to a trucation in the dystrophin gene.

Question 4

1 pts

To conduct the hybridization experiment, you are trying to decide between using a DNA or RNA probe. Which would be ideal to use and why?

	As both are composed of nucleic acids, using either would result in identical results.
	An RNA probe because RNA has uracil bases.
	An RNA probe because it could also be used in a translation experiment.
	A DNA probe because it is more stable than RNA.
	A DNA probe because RNA cannot bind to DNA.

Question 5

1 pts

Which of the following will lower the Tm of a given DNA strand?

	Increasing the percentage of GC base pairs.
	Raising the pH of the solution from neutral to basic.
	Decreasing the buffer concentration from 50mM NaCl to 5mM NaCl.
	None of the above.

Question 6

1 pts

One step of the Hershey/Chase experiment involved blending the virus/cell mixture before centrifugation and probing the pellet for radioactivity. Why was the blending step necessary?

	To collect the bacteria at the bottom of the tube.
	To break open the bacteria to release the genome.
	To separate the bacteria from the bacteriophages.
	To be able to detect the radioactivity.

Question 7

1 pts

Imagine Hershey/Chase had used an RNA virus (genome composed of RNA) instead of a DNA virus in their experiment. Would radioactivity still have been found in the pellet?

	No, because only DNA can be labeled with radioactivity.
	No, because the RNA genome would not enter the bacteria upon infection.
	No, because while DNA and RNA nucleotides are similar, they are not identical.
	Yes, because DNA and RNA nucleotides are similar.
	Yes, because genome in any form (DNA, RNA, protein) would be labeled similarly.

Question 8

1 pts

Griffith and Avery's transformation experiments allowed us to identify that DNA is our genetic information. Which of the following scenarios would result in bacterial cells that are capable of killing mice upon injection?

	Heat killed non-virulent bacteria is added to a live virulent bacteria strain.
	Heat killed virulent bacteria is added to a heat killed non-virulent bacteria strain.
	A heat killed virulent bacteria that is treated with a nuclease, is then added to a non-virulent bacteria strain.
	Heat killed mouse cells are added to a non-virulent bacteria.

Question 9

1 pts

The human genome consists mostly of non-coding DNA. Which of the following are benefits of this?

	Random DNA mutations generally won't affect RNA and protein function.
	It is faster to duplicate the genome when these are present.
	The existence of introns can lead to multiple variations of proteins encoded by a single gene.
	It is unlikely transposons would exist in the genome if there was too much protein coding DNA.

Question 10

1 pts

Andrew Murray's sister, Andrea, is adding to her brother's work on chromosomes. She is using cells that are unable to synthesize adenine (âade) and histidine (âhis). The plasmid she is currently working with consists of an origin of replication and the Ade gene.

Following her transformation of the plasmid into her yeast, what media will the cells be plated on to select for cells that have picked up the plasmid?

	Media containing histidine
	Media containing adenine
	Media lacking adenine
	Media lacking histidine

tealcockroach755

1. What recent research has indicated that studying Down Syndrome may hold a key to cancer growth and slowing tumor growth? In your answer be sure to mention the DSCR1 gene, angiogenesis suppression, and why individuals with Downâs Syndrome rarely have solid tumor growth. 2. Select your favorite chromosome and list the following: Three disorders that are found on this chromosome, Indicate the type of disorder (dominant, recessive) and the type of numerical or structural aberration. 3. Assume that crossing over does not occur in the region of each chromosome that is associated with the centromere. You have 46 centromeric regions in each cell, 23 that are copies of the DNA in your motherâs egg and 23 that are copies of the DNA fatherâs sperm. a. For any given gamete, what is the probability that all the centromeric regions will be copies of the same ones that you received from your mother? b. It has been estimated that the average person is heterozygous for at least 2,000 protein forming genes. How many different kinds of gametes is it theoretically possible for each person to produce? c. Considering the maternal and paternal origin of the centromeric regions alone, how many different kinds of gametes can you produce? 4. Find a current article on Telomerase (within the last year). Summarize the article and answer the following: What is telomerase? Why is it viewed as the fountain of youth? Attach the abstract or the first page. 5. Promiscuous Sun Flower has five lovers, but she has decided that the only true father of her child can have her hand in marriage. Sunâs Blood type is A- her five suitors blood types are as follows: a. Maury AB+ b. Elmer O+ c. Willie A- d. Jiggs B- e. Cousin it B+ 6. What is the epigenome. Explain why the study of the epigenome may not follow normal Mendelian genetics 7. Explain what is meant by the statement âWomen are just as hemizygous as menâ. 8. After having 10 children you remarryâ¦â¦ What is the probability of you now having a boy and then a girl, and then a girl, and then a boy, and then a boy or a girl, and then a girl? You now have 16 childrenâ¦â¦ 9. The genetic code has four codons for the amino acid Valine (GUG, GUC, GUU, GUA). Organisms have evolved to use some codes more than other codes. Which of the valine codes would you expect to be preferentially used in the genome of an algae growing in a hot spring at Yellowstone Park? EXPLAIN your answer 10. You discovered a new virus that needs to be studied. You begin by digesting the virus with protease (an enzyme that completely digests all proteins) until only amino acids remain. You then determine that the base composition of the viral nucleic acid is 38% Cytosine, 35% Glycine, 15% Thymine, and 12 % Adenine. To which class could this virus belong? EXPLAIN YOUR ANSWER a. Single-stranded RNA Virus b. Double-stranded RNA Virus c. Single-stranded DNA Virus d. Double=stranded DNA Virus e. None of the above because there is not enough information given to make a guess. 11. Adenine is found in numerous biological molecules (for example, DNA, RNA, NADH, ATP). You are now asked to explain the evolution of adenine and at least 5 of these biological (Adenine containing) molecules and how they may have evolved (the order).

amethysthippopotamus941

__________ are fairly small organelles that provide a safe place within the cell to carry out certain biochemical reactions that generate harmful, highly reactive oxygen species. These chemicals are both generated and broken down in the same location.

Lysosomes

Endosomes

Peroxisomes

Nucleosomes

Biologists cannot possibly study all living species. Instead, they try to understand cell behavior by studying a select subset of them. Which of the following characteristics are useful in an organism chosen for use as a model in laboratory studies?

rapid rate of reproduction

all of the above

ability to grow under controlled conditions

amenability to genetic manipulation

By definition, prokaryotic cells do not possess __________.

ribosomes.

a nucleus.

membrane bilayers.

replication machinery.

Cell biologists employ targeted fluorescent dyes or modified fluorescent proteins in both standard fluorescence microscopy and confocal microscopy to observe specific details in the cell. Even though fluorescence permits better visualization, the resolving power is essentially the same as that of a standard light microscope because the resolving power of a microscope is limited by the __________ of light.

intensity

absorption

wavelength

filtering

Choose the phrase that best completes this sentence: Microtubules ____________ and are required to pull duplicated chromosomes to opposite poles of dividing cells.

generate contractile forces

can rapidly reorganize

are found in especially large numbers in muscle cells

are intermediate in thickness

Despite the differences between eukaryotic and prokaryotic cells, prokaryotes have proteins that are distantly related to eukaryotic actin filaments and microtubules. What is likely to be the most ancient function of the cytoskeleton?

cell division

cell motility

vesicle transport

membrane support

DNA and RNA are different types of nucleic acid polymer. Which of the following is true of DNA but not true of RNA?

It contains thymine.

It has 5'-to-3' directionality.

It is single-stranded.

It contains uracil.

Double covalent bonds are both shorter and stronger than single covalent bonds, but they also limit the geometry of the molecule because they ____________________.

prevent additional bonds from being formed with the bonded atoms.

limit the rotation of the bonded atoms.

create a new arrangement of electron shells.

change the reactivity of the bonded atoms.

Drosophila melanogaster is a/an __________. This type of animal is the most abundant of all animal species, making it an appropriate choice as an experimental model.

mammal

amphibian

bird

insect

10.

If the isotope 32S has 16 protons and 16 neutrons, how many protons, neutrons, and electrons will the isotope 35S have, respectively?

16, 19, 17

16, 19, 15

16, 19, 16

16, 20, 15

11.

Living systems are incredibly diverse in size, shape, environment, and behavior. It is estimated that there are between 10 million and 100 million different species. Despite this wide variety of organisms, it remains difficult to define what it means to say something is alive. Which of the following can be described as the smallest living unit?

cell

organelle

protein

DNA

12.

Macromolecules in the cell can often interact transiently as a result of noncovalent interactions. These weak interactions also produce stable, highly specific interactions between molecules. Which of the factors below is the most significant in determining whether the interaction will be transient or stable?

the concentration of each molecule

the size of each molecule

surface complementarity between molecules

the rate of synthesis

13.

Many of the mechanisms that cells use for maintenance and reproduction were first studied at the molecular level in bacteria. Which bacterial species had a central role in advancing the field of molecular biology?

E. coli

S. pombe

C. elegans

D. melanogaster

14.

Many types of cells have stores of lipids in their cytoplasm, usually seen as fat droplets. What is the lipid most commonly found in these droplets?

cholesterol

isoprene

triacylglycerol

palmitic acid

15.

Mitochondria contain their own genome, are able to duplicate, and actually divide on a different time line from the rest of the cell. Nevertheless, mitochondria cannot function for long when isolated from the cell because they are __________________.

anaerobes.

viruses.

parasites.

endosymbionts.

16.

Mitochondria perform cellular respiration, a process that uses oxygen, generates carbon dioxide, and produces chemical energy for the cell. Which answer below indicates a correct pairing of material âburnedâ and the form of energy produced during cellular respiration?

17.

Polar covalent bonds are formed when the electrons in the bond are not shared equally between the two nuclei. Which one of these molecules contains polar bonds?

Biology 2581B Lecture Notes - Lecture 5: Chloroplast, Hepatocyte, Species Richness

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