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BC3009: Biophysical & Biochemical Methods > Quantities > Flashcards

Quantities Flashcards

1
Q

What is Avogadro’s number?

A

The number of molecules in one mole of a substance

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2
Q

How many molecules are in a mole of a substance?

A

6.022 x 10^23

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3
Q

How many molecules of NaCl in 10g?

A
  1. 10g NaCl/58.44g/mol = 0.171moles NaCl
  2. 0.171 moles NaCl x 6.022 x 10^23 =
    1.03 x 10^23 molecules
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4
Q

How many molecules in 10µL of a 1M NaCl solution?

A
  1. 10µL = 10 x 10^-6L
  2. n = MV
  3. n = 1mol/L x 10 x 10^-6L = 1 x 10^-5 mol
  4. (1 x 10^-5 mol) x (6.022 x 10^23 molecules/mol) =
    6.022 x 10^18 molecules
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5
Q

How many molecules in 44g of CO2?

A
  1. 44g CO2/44g/mol = 1 mole CO2
  2. 1 mole CO2 x 6.022 x 10^23 =
    6.022 x 10^23 molecules
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6
Q

What is the average weight of an amino acid?

A

110 Da

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7
Q

What is the average weight of a dNTP?

A

327.0 Da
(single strand DNA)

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8
Q

What is a Dalton?

A

Atomic mass unit

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9
Q

What is the approximate weight of the following peptide:
M-A-R-S-H?

A

M = 149 Da
A = 89 Da
R = 174 Da
S = 105 Da
H = 155 Da
TOTAL = 672 Da

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10
Q

How many molecules of M-A-R-S-H in 1 pmol?

A
  1. 1pmol = 1 x 10^-12 moles
  2. Molecules = moles x Avogadro’s number
  3. (1 x 10^-12 moles) x (6.022 x 10^23 molecules/mole) =
    6.022 x 10^11 molecules
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11
Q

What is the weight of 1 pmol of M-A-R-S-H?

A
  1. 1pmol = 1 x 10^-12 moles
  2. Weight (g) = Moles x Molar Mass (g/mol)
  3. (1 x 10^-12) x 672 g/mol =
    6.72 x 10^-10 g
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12
Q

What is the approximate weight of the the following oligonucleotide?
A-A-A-C-C-C-G-G-G-T-T-T

A

(3 x 331.2) + (3 x 307.2) + (3 x 347.2) + (3 x 322.2) = 3923.4 Da (or g/mol)

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13
Q

How many molecules of A-A-A-C-C-C-G-G-G-T-T-T in 1 nmol?

A
  1. 1nmol = 1 x 10^-9 mol
  2. Molecules = moles x Avogadro’s number
  3. 1 x 10^-9mol x 6.022 x 10^23 = 6.022 x 10^14 molecules
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14
Q

What is the weight of 1 nmol of A-A-A-C-C-C-G-G-G-T-T-T?

A
  1. 1nmol = 1 x 10^-9 moles
  2. Weight (g) = Moles x Molar Mass (g/mol)
  3. (1 x 10^-9) x 3923.4 g/mol =
    3.92 x 10^-6g
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15
Q

How would you make up 50ml of a 5M NaCl solution (MW NaCl = 58.44)?

A
  1. n = MV
  2. n = 5M x 0.05L = 0.25 mol
  3. Mass = moles x molar mass
  4. Mass = 0.25 mol x 58.44 g/mol = 14.61 g
  5. Add 14.61 g NaCl to 50ml solution
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16
Q

How would you make up a 100µM solution of the following oligonucleotide A-A-A-A-A-C-C-C-C-C-G-G-G-G-G-T-T-T-T-T if you were provided with 25 nmol of the oligo as a dried powder?

A
  1. 100µM = 1 x 10^-4M (mol/L)
  2. 25nmol = 2.5 x 10^-8 mol
  3. V = n/M
  4. V = (2.5 x 10^-8 mol) / (1 x 10^-4 mol/L) = 2.5 x 10^-4 L = 0.25ml
  5. Add 25 nmol dried powder to 0.25ml solution
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17
Q

How would you make up a 100µM solution of the following oligonucleotide A-A-A-A-A-C-C-C-C-C-G-G-G-G-G-T-T-T-T-T if you were provided with 10µg of the oligo as a dried powder?

A
  1. 100 µM = 1 x 10^-4M (mol/L)
  2. 10 µg = 1 x 10^-5 g
  3. (5 x 331.2) + (5 x 307.2) + (5 x 347.2) + (5 x 322.2) = 6539 Da (or g/mol)
  4. Moles = mass / molar mass
  5. Moles = (1 x 10^-5 g) / (6539 g/mol) = 1.53 x 10^-9 moles
  6. Volume = moles / concentration
  7. Volumes = 1.53 x 10^-9 moles / 1 x 10^-4 mol/L = 1.53 x 10^-5 L = 0.0153ml
  8. Add 10µg oligo to 0.0153ml solution
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18
Q

How many µm^3 in 1cm^3?

A

1 µm^3 = 1 x 10^-6 m^3
1 cm^3 = 1 x 10^-2 m^3
(1 x 10^-2)^3 m^3 / (1 x 10^-6)^3 m^3 =
1, 000, 000, 000, 000 µm^3 in 1 cm^3

19
Q

How many nm^3 in 1cm^3?

A

1 nm^3 = 1 x 10^-9 m^3
1 cm^3 = 1 x 10^-2 m^3
(1 x 10^-2)^3 m^3 / (1 x 10^-9)^3 m^3 =
1 x 10^21 nm^3 in 1 cm^3

20
Q

What is the approximate diameter of a human hair?

A

0.1 mm

21
Q

What is the approximate diameter of a human cell?

A

10 - 100 µm

22
Q

What is the approximate diameter of a bacterium?

A

0.5 µm

23
Q

What is the approximate diameter of a virus?

A

10 - 100 nm

24
Q

What is the approximate diameter of DNA?

A

10 nm

25
Q

What is the approximate diameter of a protein?

A

1 - 100 nm

26
Q

What is the approximate volume of an oocyte?

A

4,000,000 µm^3

27
Q

What is the approximate volume of a fibroblast cell?

A

2,000 µm^3

28
Q

What is the approximate volume of a red blood cell?

A

100 µm^3

29
Q

What is the approximate volume of an E.coli?

A

1 µm^3

30
Q

How many E.coli would fit in 1cm^3?

A

10^12

31
Q

How many E.coli would you expect in 1ml of an overnight culture grown in a rich media?

A

10^9 to 10^10

32
Q

How many E.coli would fit in a fibroblast?

A

2000

33
Q

How many antibodies in IgM?

A

5

34
Q

What is the approximate weight of an IgM molecule?

A

850 kDa

35
Q

What exactly does it mean to say the affinity of a protein receptor for its ligand is 1 picomolar?

A
  • Receptor has dissociation constant (Kd) of 10^-12 mol/L
  • If a ligand is present at 10^-12 mol/L then 50% of receptors bound
36
Q

Approximately how many protein molecules in a yeast cell?

A

42 million

37
Q

How many proteins encoded by the yeast genome?

A

6,000

38
Q

What is the abundance range of proteins in a yeast cell?

A

1000 - 10,000

39
Q

About how many mRNA are there in a yeast cell?

A

60,000

40
Q

What is the approximate weight of now human genome?

A

3.2 pg

41
Q

Explain Dynamic Range

A

Dynamic range is the SIZE of the concentration range which a sensor or receptor can detect a molecule

42
Q

Explain the Law of Mass Action

A

The rate at which a chemical reaction takes place is proportional to the product of the active masses of the reactants

43
Q

Explain Kd

A
  • Equilibrium Dissociation Constant
  • When ligand concentration is equal to Kd, half of the receptors are occupied by a ligand at equilibrium
44
Q

Explain the relationship between Kd and affinity

A

HIGHER the Kd, the LOWER the Affinity of receptor for ligand

BC3009: Biophysical & Biochemical Methods (12 decks)

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