HW 1- MBB347

Determine if the following features characterize eukaryotes, prokaryotes, both groups, or neither group. 1. They use a deoxyribonucleoside triphosphate as their "energy currency". 2. They chiefly use deoxyribonucleic acid molecules to store genetic information. 3. RNA molecules are their main catalyst. 4. Their DNA is typically found within sub-cellular organelles such as mitochondria, chloroplasts but mainly in the nucleus. 5. Their proteins consist of a common set of 20 different amino acids 6. Most of their genomes comprises protein-coding genes. 7. Their cell membranes chiefly comprise of phospholipid-based bilayer.

1. A. Neither prokaryotes nor eukaryote 2. B. Both prokaryotes and eukaryotes 3. A. Neither prokaryotes nor eukaryotes 4. C. eukaryotes 5. B. Both prokaryotes and eukaryotes 6. D. prokaryotes 7. B. Both prokaryotes and eukaryotes

Viruses are often not considered " living" because they have the following characteristics (True/False) : 1. They are acellular. 2. They are not made of organic molecules. 3. They lack membranes. 4. They lack nuclei. 5. They do not evolve. 6. They don't use DNA as their information storage molecule.

1. A. True 2. B. False 3. B. False 4. B. False 5. B. False 6. B. False

What is the molecular weight of glucose?

180

A 70-kg adult person could meet all of his/her entire energy needs for one day by consuming 3 moles of glucose (540 g - you are strongly advised, naturally, against such a diet!). Each molecule of glucose generates 30 ATP molecules when completely oxidized to CO2. The intracellular concentration of ATP is maintained constant in the cells at about 2 mM, and a 70-kg adult has about 25 liters of intracellular fluid. How many times per day, on average, does an ATP molecule turnover (= hydrolyzed and resynthesized)?

1800 1 mole of glucose yields 30 moles of ATP 3 moles of glucose yield 90 moles of ATP. If there were no turnover (that is if ATP were not used by hydrolysis into AMP), then the concentration of ATP would rise to 90 moles/25 liters = 3.6 M = 3,600 mM. Since the intracellular ATP concentration remains constant at 2 mM that means that it continuously turning over: Turnover = 3,600 mM/ 2 mM = 1,800 times

A person whose blood glucose levels drop to below 70 mg/dL enters a serious hypoglycemic state. Hypoglycemia is potentially fatal if the glucose levels are not rapidly restored by consuming something loaded with sugar, like soda (for example sprite has 38 g sugar per a 12-oz can). How much Sprite should an 80-kg hypoglycemic man drink to raise his blood sugar level to normal? Express your answer in mL (rounded to the nearest mL). Please do not enter the units, though. You will need to make some simplifying assumptions. Assume that 45% of the sugar in sprite (infamous high fructose corn syrup) is glucose. Assume complete absorption, uniform distribution (in all aqueous compartments) and water content of 60% of body weight. Similarly ignore, elimination, metabolism etc. Remember: This is a very crude approximation and is not to be taken as a clinical recommendation.

202 1. First let's calculate by how much the blood glucose concentrations need to be raised to bring the levels to normal. [glucose]blood = 90 mg/mL -70 mg/mL = 20 mg/mL = 200 mg/L = 0.2 g/L 2. To translate this raise in concentration we need to know the volume. Our assumption treats the person as a barrel 60% of whose weight is water. Volume = 80 kg × 60% = 48 kg ⇒The volume of this amount of water is 48 L. 3. The mass of glucose we need is therefore Needed Glucose = 0.2 g/L × 48 L= 9.6 g 4. Next let's calculate what is the concentration of glucose in Sprite. [Glucose]Sprite = 0.45 × 38 g/can = 17.1 g/can = 17.1 g/12 oz = 1.425 g/oz [Glucose]Sprite = 1.425 g/oz = 1.425 g/30 mL = 0.0475 g/mL 5. Finally let's calculate what is volume of Sprite needed to give 9.6 g. Sprite = Needed Glucose / [Glucose]Sprite = 9.6 g / 0.0475 g/mL = 202 mL (~⅔ can)

In the US, the concentration of blood glucose is usually reported in milligrams (mg) per deciliter (dL = 100 mL = 0.1 L). A concentration of 90 mg/dL is considered the mean normal value. What is this value expressed in milimolar (mM) units (rounded to the nearest milimolar)?

5 Because 1 M = 1 mol/L, let us first calculate the normal blood sugar value in mg (or grams) per 1 liter: 90 mg/dL = 90 mg/0.1 L = 900 mg/L = 0.9 g/L Now, let us convert the mass (in g) to moles: If 180 g is 1 mol Then 0.9 g is x mole = 0.9 g • 1 mol/180 g = 0.005 mol Finally, therefore: [glucose] = 90 mg/dL = 0.005 mol/L = 0.005 M = 5 mM

Organism Comments Influenza virus Genome consists of 8 segments of single-stranded RNA. Genome's over all G+C content is about 45%. [A]%[T]% [G]%[C]%

A: N/A T: 0% G: N/A C: N/A

All of the following statements are true EXCEPT _______. -Differences in ionic composition across membranes can be generated through the action of ATP driven pumps. -Differences in ionic composition across membranes can be used to drive molecular motor. -Membrane potential associated with differences in ionic composition across membranes can be quickly and temporarily dissipated by transiently opening ion channels to allow rapid flow of certain ions according to their concentration gradient. -Differences in ionic composition may exist between cells in different environments, but the external and internal ionic compositions of a cell within a certain environment are always identical to each other.

Differences in ionic composition may exist between cells in different environments, but the external and internal ionic compositions of a cell within a certain environment are always identical to each other.

Organism Comments Herpes simplex virus type 1 (HSV-1) HSV-1 has a large double-stranded DNA genome with a very high G+C content (68%) [A]%[T]% [G]%[C]%

Specified Answer for: A 16 Specified Answer for: T 16 Specified Answer for: G 34 Specified Answer for: C 34

The rule of complementarity of DNA originated in studies made by Erwin Chargaff during the 1940s. He demonstrated that the DNA obtained from many sources was composed of equal amounts of As and Ts and equal amounts of Gs and Cs. Consider the following information and fill in the blanks with the missing values in the next 5 questions. Round your answer to the nearest percent. If the values cannot be calculated based on the given information fill in with "N/A" (without the quotation marks). Organism Comments Yersinia pestis A Gram-negative bacterium that causes plague. [A]% [T]% 24% [C]%

Specified Answer for: A 26 Specified Answer for: T 26 Specified Answer for: C 24

Coltivirus Coltivirus, the causative agent of Colorado tick fever, has a genome composed of 11 segments of double-stranded RNA with an overall G+C content of 46% [A]% [T]% [G]% [C]%

Specified Answer for: A 27 Specified Answer for: T 0 Specified Answer for: G 23 Specified Answer for: C 23

Organism Comments Bacillus anthracis A Gram-positive bacterium, the causative agent of anthrax. The (G+C) content of the B. anthracis genome is 32%. [A]% [T]% [G]% [C]%

Specified Answer for: A 34 Specified Answer for: T 34 Specified Answer for: G 16 Specified Answer for: C 16

The tubulin polymerization reaction is temperature dependent, occurring spontaneously at 37 ºC but not at lower temperatures. In fact microtubules will spontaneously depolymerize (fall apart) when cells are cooled down to 0 ºC and this phenomenon can be replicated also in vitro (in a test tube, see Fig. 2). This process is governed by the basic thermodynamic function of free energy according to the equation: ΔG = ΔH - TΔS. Which of the following statements is correct? -The polymerization reaction (left to right in Fig. 2) has ΔG < 0 at 37 ºC and at 0 ºC. -The polymerization reaction has ΔG > 0 at 37 ºC and ΔG < 0 at 0 ºC. -The polymerization reaction has ΔG < 0 at 37 ºC and ΔG > 0 at 0 ºC. -The polymerization reaction has ΔG > 0 at 37 ºC and at 0 ºC. .

The polymerization reaction has ΔG < 0 at 37 ºC and ΔG > 0 at 0 ºC.

What is the maximal number of double covalent bonds that a carbon atom can make? -None -Two -One -Three -Four

Two

During mitosis, chromosomes are pulled to the two cell poles by the spindle. The spindle consists of filamentous "microtubules" which are made up of polymerized monomers of the protein tubulin. Tubulin has many hydrophobic amino acid side-chains on its surface (giving these surfaces overall hydrophobic nature) and the hydrophobic effect (so called "hydrophobic interactions") is the major force that holds the tubulin subunits together during polymerization and assembly of microtubules. Explain in your own words what is the physical-chemical basis for hydrophobic interaction.

When hydrophobic molecules are dispersed in aqueous solutions, water molecules interfacing with the hydrophobic surface of the solute become ordered. This is an entropically unfavored state that is the driving force for the aggregation of the hydrophobic solute molecules thereby reducing surface area available for interaction of the water molecules, more water molecules are released from this highly ordered state and the entropy increases.

The diameter of a typical bacteria cell can be all of the following except -between 1.0 µm and 2.0 µm. -between 1,000 nM and 2,000 nM. -between 104 Å and 2x104 Å. -between 0.001 mm and 0.002 mm. -between 10-6 m and 2x10-6 m.

between 1,000 nM and 2,000 nM.

According to the endosymbiosis hypothesis, which of the following eukaryotic organelles evolved from originally independent prokaryotic ancestors? -chloroplasts -ribosomes -endoplasmic reticulum -chromoplasts -lysosomes -mitochondria

chloroplasts chromoplasts mitochondria

The constituents of biological membranes that are most important in facilitation of movement of hydrophilic compounds across the membrane are ____________. -peripheral membrane proteins. -nucleic acids. -carbohydrates. -phospholipids. -integral membrane proteins.

integral membrane proteins.

Which of the following sub-cellular structures contain DNA? -ribosomes -mitochondria -lysosomes -chromoplasts -endoplasmic reticulum -chloroplasts

mitochondria chromoplasts chloroplasts

What is the configuration around carbon atoms covalently binding four other partners? -linear -trigonal bipyramidal -trigonal planar -tetrahedral

tetrahedral

The synthesis of the three fundamental polymers of life, DNA, RNA and protein, is template-directed. This means that _____. -the monomers are assembled one by one in a sequence prescribed by an existing polymer. -DNA is copied from DNA, RNA is copied from RNA and protein is copied from protein. -all of them utilize a semi-conservative mode of synthesis. -the system evolved in direction of more complexity from a 4-letter code of nucleotides to a 20-letter code of amino acids.

the monomers are assembled one by one in a sequence prescribed by an existing polymer.

Living cells require input of free energy in order to be able to perform all of the following except _______. -to synthesize ATP from ADP and inorganic phosphate. -to breakdown macromolecules. -transport ions across the membrane against their -concentration gradient. -to accumulate nutrients in the cell.

to breakdown macromolecules.

Assuming that ΔH and ΔS do not change with the temperature, are the values of ΔH and ΔS negative or positive? ΔH > 0 and ΔS > 0 ΔH < 0 and ΔS < 0 ΔH > 0 and ΔS < 0 ΔH < 0 and ΔS > 0

ΔH > 0 and ΔS > 0