Question

Name: PRACTICE EXAM 1 - Cell Biology Spring 2020 MULTIPLE CHOICE (5 points each, 100 points total). Choose the one alternat. completes the statement or answers the question. 1) Leeuwenhoek's simple home-made microscope had a numerical aperture (NA) of anproximately 0.6. Using light with a wavelength of 500 nm, about what was in resolution limit (R) of this microscope? a) 500 pm b) 5 pm c) 0.5 pm d) 5 nm e) 0.6 pm CH,DH 2) This polymeric macromolecule is: a) glycogen b) polylactose c) cellulose d) glucose e) chitin 3) Two major cytosolic proteins that form filaments of the eukaryotic cytoskeleton: a) actin, myosin b) kinesin, dynein c) tubulin, actin d) microtubules, myosin 4) About how large (atomic mass) would you expect for a protein that is 329 amino acids in length (DA = Daltons)? a) 329 KDA b) 329 DA c) 3,290 DA d) 32,900 DA e) it is not possible to estimate this based on the information given Cell Bio Practice Exam 1 Page 1 ria hrea

Answer 1

Ans-1- resolution (d)= 0.61 T / NA

T = wavelength = 500 nm

Numerical Apperature (NA) = 0.6

d = 0.61 * 500 / 0.6 = 508 nm = 500 nm = 0.5 um

option C is correct.

Ans-2- in the given molecule there are two types of bond alpha1-4 glycosidic and alpha1-6 glycosidic bond and all the subunits are glucose. glycogen is the polymer of glucose and contains the same type of bonding pattern. It is not cellulose because it is branched. option A (glycogen) is right.

Ans-3- two major cytoskeleton proteins are actin and tubulin that make microfilament and microtubules respectively. option C is correct.

Ans-4- average mass of 1 amino acid = 110 Da

so the molecular mass of 329 amino acids = 329 * 110 = 36190 Da

So the option E is correct