Lesson 2 practice← Back to the lesson

Practice: Molecules of life

Test yourself on the math from the lesson. There are 28 kinds of problem, from warm-ups to challenges, and each one gives you new numbers whenever you ask.

28 kinds of problem. Solve as many as you like.

1. Atoms and bonds

RememberEach atom makes a set number of covalent bonds: H 1, O 2, N 3, C 4, S 2, and P 5 in a phosphate group. A double bond counts as 2. In a formula such as C6H12O6\mathrm{C_6H_{12}O_6}, the number after each symbol counts its atoms, and no number means 1.

Review this in the lesson
1.1

Bonds per atom

Warm-up

How many covalent bonds does a sulfur atom usually make? Count a double bond as 2.

1.2

Reading a formula

Warm-up

Glycine has the formula C2H5NO2\mathrm{C_{2}H_{5}NO_{2}}. How many nitrogen atoms are in one molecule?

1.3

Filling in hydrogens

Practice

Sulfur makes 2 bonds and hydrogen makes 1. A sulfur atom is joined to one carbon atom by a single bond. Its other bond goes to hydrogen. How many hydrogen atoms are attached to the sulfur?

2. Formulas and molecular mass

RememberMolecular mass = the sum of (number of atoms × atomic mass). Atomic masses in daltons: H 1.008, C 12.011, N 14.007, O 15.999, P 30.974, S 32.06. A mass of M Da means M grams per mole, and a mole is 6.022×10236.022 \times 10^{23} molecules.

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2.1

Molecular mass from a formula

Warm-up

What is the molecular mass of ATP, C10H16N5O13P3\mathrm{C_{10}H_{16}N_{5}O_{13}P_{3}}? Use these atomic masses, in daltons (Da): C 12.011, H 1.008, N 14.007, O 15.999, P 30.974.

Da
2.2

Mass percent of an element

Practice

Glycine, C2H5NO2\mathrm{C_{2}H_{5}NO_{2}}, has a molecular mass of 75.067 Da. Nitrogen has an atomic mass of 14.007 Da. What percent of the mass of glycine is nitrogen?

2.3

Grams to molecules

Practice

One milliliter of water weighs about 1 g. Water has a molecular mass of 18.015 Da, so a mole of it weighs 18.015 g. How many water molecules is that? One mole is 6.022×10236.022 \times 10^{23} molecules.

molecules
2.4

How concentrated is water?

Practice

A liter of pure water weighs 1,000 g. Water, H2O\mathrm{H_2O}, is made of hydrogen (1.008 Da) and oxygen (15.999 Da). What is the concentration of water, in moles per liter (M)?

M
2.5

Water molecules in a cell

Challenge

A yeast cell has a volume of about 40 µm³ and about the density of water, so it weighs about 40 pg (picograms). About 70% of a cell's mass is water, and water has a molecular mass of 18.015 Da. About how many water molecules are in the cell? One mole is 6.022×10236.022 \times 10^{23} molecules.

molecules

3. Energy and kT

RememberkT=R×TkT = R \times T, with R = 0.008314 kJ/(mol·K) and T in kelvin (°C + 273.15). At 37 °C, kT is about 2.58 kJ/mol. A state that costs E more energy is rarer by the Boltzmann factor e−E/kTe^{-E/kT}, and e−x=10−x/2.303e^{-x} = 10^{-x/2.303}.

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3.1

kT and energies in kT

Warm-up

Hydrogen bonds take about 5 to 20 kJ/mol to break. Suppose one takes 20 kJ/mol. At 37 °C, kT is 2.58 kJ/mol. How many kT is that?

kT
3.2

The Boltzmann factor

Practice

A change in a molecule's shape costs 5 kJ/mol of energy. At 37 °C, kT is 2.58 kJ/mol. What is the Boltzmann factor, e−E/kTe^{-E/kT}, for this change?

3.3

How many times more likely?

Practice

Two shapes of a molecule differ in energy by 10 kJ/mol. At 37 °C, kT is 2.58 kJ/mol. How many times more likely is the lower-energy shape? Use ratio =eΔE/kT= e^{\Delta E / kT}, where ΔE\Delta E is the energy gap.

times
3.4

Turning up the heat

Challenge

Breaking a bond costs 10 kJ/mol. kT is 2.58 kJ/mol at 37 °C and 3.06 kJ/mol at 95 °C. How many times bigger is the bond's Boltzmann factor, e−E/kTe^{-E/kT}, at 95 °C than at 37 °C?

times

4. DNA and RNA

RememberA pairs with T (with U in RNA), and G pairs with C. A–T pairs have 2 hydrogen bonds and G–C pairs have 3. The two strands run opposite ways, so to write a partner strand 5′ to 3′, pair each letter and then read backward. Each base pair adds 0.34 nm, and the helix turns once every 10.5 base pairs.

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4.1

Pairing letters

Warm-up

A DNA strand reads 5′-TACGATA-3′5'\text{-}\texttt{TACGATA}\text{-}3'. Write the RNA strand that pairs with it, letter by letter from left to right. In RNA, U takes the place of T: DNA's A pairs with U, T with A, G with C, and C with G.

4.2

G–C content

Warm-up

What percent of the bases in 5′-TTAGTCACATTACCTATCTG-3′5'\text{-}\texttt{TTAGTCACATTACCTATCTG}\text{-}3' are G or C? This is the strand's G–C content.

4.3

The partner strand, 5′ to 3′

Practice

An RNA strand, 5′-CAGUUACGUUU-3′5'\text{-}\texttt{CAGUUACGUUU}\text{-}3', was made by pairing with a DNA strand. Write that DNA strand from 5′ to 3′. RNA's A pairs with DNA's T, U pairs with A, G with C, and C with G.

4.4

Hydrogen bonds in a double helix

Practice

A short piece of double-stranded DNA has this sequence on one strand: 5′-ATAAGCAGTACGTA-3′5'\text{-}\texttt{ATAAGCAGTACGTA}\text{-}3'. A–T pairs are held together by 2 hydrogen bonds, and G–C pairs by 3. How many hydrogen bonds hold the two strands together?

4.5

Chargaff's rule

Practice

In a sample of double-stranded DNA, 23% of the bases are T. What percent are G or C (the G–C content)?

4.6

How long is the DNA?

Practice

E. coli's genome is 4,641,652 base pairs long. Each base pair adds 0.34 nm to the length of the double helix. How long is it, stretched out straight, in micrometers?

µm
4.7

DNA in a small box

Challenge

E. coli's genome is 4,641,652 base pairs long, and each base pair adds 0.34 nm of length. An E. coli cell is about 2 µm long. How many times longer than the cell is its DNA, stretched out straight?

times
4.8

A genome as a file

Challenge

Each DNA base is one of 4 letters, so it carries 2 bits of information. A gene is 1,500 base pairs long. How many bytes does it take to store its sequence? Use 8 bits = 1 byte.

bytes

5. Counting possibilities

RememberWith 4 choices at each of nn positions there are 4n4^n DNA sequences, and with 20 amino acids there are 20n20^n protein chains. As powers of ten: 4n=100.602n4^n = 10^{0.602 n} and 20n=101.301n20^n = 10^{1.301 n}.

Review this in the lesson
5.1

Counting sequences

Warm-up

How many different DNA sequences are exactly 2 bases long? Each position can be A, C, G, or T.

sequences
5.2

Huge counts as powers of ten

Practice

How many different DNA sequences are 50 bases long? The answer is 4504^{50}. Write it as a power of ten, 10?10^{?}, using log⁡10(4)=0.602\log_{10}(4) = 0.602, and type just the exponent.

5.3

Four letters, two bits

Practice

Fill in the exponent: 410=2?4^{10} = 2^{?}.

5.4

Long enough to outnumber

Challenge

What is the shortest length nn for which there are more possible DNA sequences of length nn, 4n4^n, than Avogadro's number, 6.022×10236.022 \times 10^{23}? Use log⁡10(4)=0.602\log_{10}(4) = 0.602.

6. Proteins and membranes

RememberAn amino acid in a protein weighs about 110 Da. A sphere has surface area 4πr24\pi r^2. A lipid bilayer is about 4 to 5 nm thick and holds about 5 million lipids per µm², counting both layers (Alberts, Molecular Biology of the Cell).

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6.1

Protein mass and length

Warm-up

A protein is 150 amino acids long. At about 110 Da per amino acid, what is its mass in kDa (kilodaltons)?

kDa
6.2

How thin is a membrane?

Warm-up

A lipid bilayer is about 5 nm thick. An E. coli cell is about 1 µm wide. How many membrane thicknesses would fit across the width of the cell?

6.3

The area of a cell membrane

Practice

Treat a human cell as a sphere 20 µm across. What is the surface area of its membrane, in µm²? Use A=4πr2A = 4\pi r^2.

µm²
6.4

Lipids in a membrane

Challenge

A lipid bilayer holds about 5 million lipid molecules per µm², counting both layers. A round bacterium is 1 µm across. About how many lipids are in its membrane? Use A=4πr2A = 4\pi r^2.

lipids

Mixed review

8 problems from across the lesson, in random order. Here the topic is hidden until you solve each one, so part of the job is working out which tool you need.

M1

Problem

Practice

Fill in the exponent: 412=2?4^{12} = 2^{?}.

M2

Problem

Practice

How many different DNA sequences are 50 bases long? The answer is 4504^{50}. Write it as a power of ten, 10?10^{?}, using log⁡10(4)=0.602\log_{10}(4) = 0.602, and type just the exponent.

M3

Problem

Practice

A short piece of double-stranded DNA has this sequence on one strand: 5′-TGGATCAGGTCGG-3′5'\text{-}\texttt{TGGATCAGGTCGG}\text{-}3'. A–T pairs are held together by 2 hydrogen bonds, and G–C pairs by 3. How many hydrogen bonds hold the two strands together?

M4

Problem

Practice

Adenine, C5H5N5\mathrm{C_{5}H_{5}N_{5}}, has a molecular mass of 135.13 Da. Nitrogen has an atomic mass of 14.007 Da. What percent of the mass of adenine is nitrogen?

M5

Problem

Practice

The DNA double helix makes one full turn every 10.5 base pairs. How many turns are there in a stretch of DNA of 1,050 base pairs?

turns
M6

Problem

Challenge

Breaking a bond costs 50 kJ/mol. kT is 2.58 kJ/mol at 37 °C and 3.06 kJ/mol at 95 °C. How many times bigger is the bond's Boltzmann factor, e−E/kTe^{-E/kT}, at 95 °C than at 37 °C?

times
M7

Problem

Challenge

What is the shortest length nn for which there are more possible DNA sequences of length nn, 4n4^n, than a billion, 10910^9? Use log⁡10(4)=0.602\log_{10}(4) = 0.602.

M8

Problem

Challenge

A syn3A cell has a volume of about 0.0335 µm³ and about the density of water, so it weighs about 0.0335 pg (picograms). About 70% of a cell's mass is water, and water has a molecular mass of 18.015 Da. About how many water molecules are in the cell? One mole is 6.022×10236.022 \times 10^{23} molecules.

molecules