Lesson 3 practice← Back to the lesson

Practice: The central dogma

Test yourself on the math from the lesson. There are 35 kinds of problem, from warm-ups to challenges, and each one gives you new numbers whenever you ask.

Practice: How it works

Type your answer and press Check (or Enter). If you've made a common slip, such as using the diameter instead of the radius, the feedback says which one.
Stuck? Press Hint. Every problem has hints, then a full worked solution.
Press New numbers for a fresh version of the same problem.
Type big and small numbers any of these ways: 4.6e6, 4.6 × 10^6,4.6x10^6, or 4600000. Negative exponents work the same way: 3e-5or 3 × 10^-5. You can leave off the units.
For a DNA or RNA sequence, type the letters, such as ATGC. Small letters work too, and spaces, hyphens, and the 5′ and 3′ labels are ignored.
For a protein, type one letter per amino acid, such as MKV, or the three-letter names, such as Met-Lys-Val. Leave out the stop codon: it doesn't add an amino acid.
When a problem asks what kind of mutation it is, type silent, missense, ornonsense.
You don't have to be exact. Where an answer is a measurement you get a few percent of leeway, so rounding in the middle steps is fine.

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

1. Positions and strands

RememberGenBank counts bases from 1 and includes both ends, so a gene at start..end is end − start + 1 bases long. complement(a..b) means the gene is on the other strand and reads from b down to a. Python counts from 0 and stops before the end, so start..end is the slice genome[start − 1 : end]. A gene for a protein of n amino acids has n + 1 codons, counting the stop codon.

Review this in the lesson

GenBank lists syn3A's dnaK gene (JCVISYN3A_0542, a chaperone that helps proteins fold) at complement(352102..353877). How many bases long is the gene, from the first letter of its start codon to the last letter of its stop codon?

syn3A's ftsZ protein, which forms the ring that divides the cell, is 385 amino acids long. How many bases long is its gene, counting the start codon and the stop codon?

A program finds a gene at genome[297889:299317] in Python. Python counts from 0, and a slice genome[a:b] stops just before index b. How many bases does the slice hold?

syn3A's gyrB gene (JCVISYN3A_0006, part of DNA gyrase, which relieves twists in DNA) sits on the forward strand at 5515..7419, so base 5,515 is the first letter of its start codon. Base 6,738 of the genome is inside the gene. Which codon is it in? Number the codons from 1, so the start codon is codon 1.

syn3A's infA gene (JCVISYN3A_0649, translation start factor IF-1) is at complement(407104..407328), on the other strand. Which genome coordinate lines up with the first letter of its start codon?

2. Transcription

RememberRNA polymerase reads the template strand and builds the mRNA 5′ to 3′, pairing A–U, T–A, G–C and C–G. So the mRNA matches the coding strand, with U for T. From the template written 5′ to 3′, the mRNA is the reverse complement, with U for T. In syn3A, a polymerase copies about 25 letters per second (rRNA genes about 90), and about 29 of its 187 polymerases are busy at a time.

Review this in the lesson

These are the first 12 letters of the coding strand of syn3A's ptsH gene (a sugar-uptake protein). This is real syn3A DNA: $5'\text{-}\texttt{ATG}$ $\texttt{GCA}$ $\texttt{AAG}$ $\texttt{TTT}\text{-}3'$ . Write the mRNA made from this stretch, from 5′ to 3′.

RNA polymerase reads this template strand, written 5′ to 3′ as sequences always are: $5'\text{-}\texttt{CGG}$ $\texttt{ACG}$ $\texttt{TGT}$ $\texttt{TGG}\text{-}3'$ . The sequence is made up. Write the mRNA it makes, from 5′ to 3′.

In syn3A's model, RNA polymerase copies about 25 letters per second. How many seconds does one polymerase take to copy a typical syn3A gene, 1,050 letters long?

Suppose a cell starts 20 new transcripts per minute, and each one is 1,000 letters long. RNA polymerase copies 25 letters per second. On average, how many polymerases are busy copying at any moment?

3. Codons and reading frames

RememberWith 4 letters, codons of 3 give $4^3 = 64$ kinds, enough for 20 amino acids plus stop (2-letter codons give only 16). A strand has 3 reading frames, starting at index 0, 1 or 2, and double-stranded DNA has 6. The standard code has 61 codons for amino acids and 3 stops; syn3A's code 4 reads UGA as tryptophan (W), leaving 62 and 2. The first codon of a gene is always read as methionine (M).

Review this in the lesson

DNA has 4 letters: A, C, G and T. How many different codons 1 letter long can be made from them?

syn3A's 455 protein-coding genes hold 159,283 codons in all. The most common codon, AAA (lysine), appears 15,126 times. What percent of all codons is that?

These are the first 7 codons of the mRNA of syn3A's ftsZ gene (the cell-division ring protein). This is a real syn3A sequence: $5'\text{-}\texttt{AUG}$ $\texttt{ACA}$ $\texttt{AAC}$ $\texttt{GAA}$ $\texttt{UUU}$ $\texttt{AAA}$ $\texttt{CAA}\text{-}3'$ . Translate it with syn3A's genetic code. Codons you need: $\texttt{AAA}$ K (Lys), $\texttt{AAC}$ N (Asn), $\texttt{ACA}$ T (Thr), $\texttt{AUG}$ M (Met), $\texttt{CAA}$ Q (Gln), $\texttt{GAA}$ E (Glu), $\texttt{UUU}$ F (Phe).

A stretch of double-stranded DNA is 39 base pairs long. Each strand can be read in 3 frames (starting at index 0, 1 or 2 of that strand, read 5′ to 3′), so there are 6 frames in all. How many complete codons do the 6 frames hold together?

Here is a made-up stretch of DNA: $5'\text{-}\texttt{GCTGCGAGCATCTC}\text{-}3'$ . Reading frames are numbered by the Python index of their first letter, so frame 0 starts at the first letter. Write codon 2 of frame 2.

4. Chance and stop codons

RememberIn random DNA with an even mix of letters, a codon is a stop with chance $p = 3/64 \approx 0.047$ in the standard code and $2/64 \approx 0.031$ in syn3A's code 4. Stops are then about $1/p$ codons apart. The chance of no stop in $n$ codons is $(1 - p)^n$ , and with logs, $\log_{10}\left((1 - p)^n\right) = n \log_{10}(1 - p)$ . With an uneven mix, multiply the letter chances: $P(\texttt{TAA}) = p_T \, p_A \, p_A$ .

Review this in the lesson

In random DNA with the four letters equally likely, each of the 64 codons is equally likely. syn3A uses genetic code 4. What is the chance that a random codon is a stop codon in syn3A's code? Give a decimal.

In random DNA with the four letters equally likely, read with the standard code, each codon is a stop with chance $p = 3/64 \approx 0.0469$ . In a stretch of 349 codons, the average length of syn3A's proteins, read in one frame, how many stop codons would you expect?

In a made-up A–T-rich genome, the letters are A 35%, T 35%, G 15% and C 15%. What is the chance that a random codon is a stop in syn3A's code 4? Treat each letter as independent.

In random DNA with syn3A's A–T-rich mix of letters, read with the standard code, each codon is a stop with chance $p = 0.088$ . What is the chance that 50 codons in a row contain no stop at all? Give a decimal.

5. Half-lives and steady state

RememberAfter time $t$ , the fraction left is $(1/2)^{t / t_{1/2}}$ . The average lifetime is $\tau = t_{1/2} / \ln 2 \approx 1.44\, t_{1/2}$ . At steady state, the number of copies = production rate × average lifetime. In syn3A, mRNA has a half-life of about 4 minutes (an average life of 5.8), proteins about 25 hours, and the cell cycle is 105 minutes.

Review this in the lesson

syn3A's proteins have a half-life of about 25 hours. Of the protein molecules present at one moment, what fraction are still intact 75 hours later?

A made-up protein lasts 50 hours on average. What is its half-life? Use: average lifetime = half-life ÷ ln 2.

A syn3A cell has 100 copies of an mRNA when it stops making new ones. The mRNA's half-life is 4 minutes. On average, how many copies are left 6 minutes later?

A gene in syn3A makes new mRNAs at an average of 0.3 per minute, and each mRNA has a half-life of 3 minutes. Once things settle (steady state), how many copies of this mRNA does the cell hold on average?

Suppose a new ribosome starts on an mRNA every 15 seconds, and each start makes one protein. The mRNA's half-life is 4 minutes. On average, how many proteins does one mRNA make in its life?

Suppose a protein needs 1,000 new copies per cell cycle of 105 minutes. Each mRNA makes about 3 proteins per minute, and its half-life is 4 minutes. How many new mRNAs per minute must the gene make?

A cell starts its cycle with 50 copies of a protein. At division, each of the two daughter cells should get as many as the cell started with. Ignoring breakdown, how many new copies must the cell make during the cycle?

6. Speed and cost

RememberTime = length ÷ speed. In syn3A's model a ribosome adds about 5 amino acids per second (E. coli 10 to 20). Ribosomes ride an mRNA about 300 letters apart, so a new one starts about every 20 seconds: 3 proteins per minute from each mRNA. Each amino acid costs about 4 ATP. syn3A has about 500 ribosomes, and its proteins average 349 amino acids.

Review this in the lesson

In E. coli, a ribosome adds 10 to 20 amino acids per second. Take 15. A syn3A gene is 852 letters long, counting its stop codon. How many seconds does one ribosome take to make its protein?

Making a protein costs about 4 ATP per amino acid: about 2 at the ribosome, plus 2 to load each amino acid onto its tRNA. How many ATP does it take to make one copy of JCVISYN3A_0034, syn3A's longest protein, 1,795 amino acids long?

Ribosomes ride an mRNA about 300 letters apart, all moving at the same speed. About how many ribosomes ride at once on a typical syn3A mRNA, with 1,050 letters of code?

syn3A has about 500 ribosomes, and each adds about 5 amino acids per second. Suppose they all work nonstop. syn3A's proteins are 349 amino acids long on average, and its cell cycle lasts 105 minutes. About how many average proteins could the cell make in one cycle?

syn3A has about 500 ribosomes, each adding about 5 amino acids per second, and each amino acid costs about 4 ATP. If every ribosome worked nonstop through the 105-minute cell cycle, how many ATP would translation use?

7. Mutations

RememberA one-letter swap is silent (same amino acid), missense (a different amino acid) or nonsense (a new stop codon, so the protein ends early). Each letter can change in 3 ways. Inserting or deleting letters that aren't a multiple of 3 shifts the reading frame: every codon after it is regrouped, usually until a stop turns up. In syn3A, TGA means tryptophan, so TGG → TGA is silent there but nonsense in E. coli.

Review this in the lesson

Codon 6 of syn3A's dnaA gene (which starts each round of DNA copying) is ATT. A mutation changes its first letter to C. In syn3A's code: $\texttt{ATT}$ I (Ile), $\texttt{CTT}$ L (Leu), $\texttt{GTT}$ V (Val), $\texttt{TTT}$ F (Phe). What kind of mutation is this?

These are the first 7 codons of the coding strand of syn3A's rpmJ gene (ribosomal protein L36). This is real syn3A DNA: $5'\text{-}\texttt{ATG}$ $\texttt{AAA}$ $\texttt{GTT}$ $\texttt{AGA}$ $\texttt{TCA}$ $\texttt{TCA}$ $\texttt{GTC}\text{-}3'$ . A mutation changes letter 7 (counting from 1 at the first A) from G to C. Using syn3A's code, write the amino acids this stretch now codes for, stopping if you reach a stop codon. Codons you need, in DNA letters: $\texttt{AAA}$ K (Lys), $\texttt{AAC}$ N (Asn), $\texttt{AGA}$ R (Arg), $\texttt{ATG}$ M (Met), $\texttt{CTT}$ L (Leu), $\texttt{GCT}$ A (Ala), $\texttt{GTC}$ V (Val), $\texttt{GTT}$ V (Val), $\texttt{TCA}$ S (Ser).

The codon CAA means Q (glutamine). Each of its 3 letters can change to 3 others, so there are 9 single-letter changes. In syn3A's code, the codons for glutamine are CAA and CAG, and the stop codons are TAA and TAG. How many of the 9 changes are nonsense (a stop codon)?

syn3A's rpmH protein (ribosomal protein L34) is 44 amino acids long. Codon 5 of its gene is TGA, which means tryptophan (W) in syn3A, and there is no other TGA before it. If a cell that uses the standard code, such as E. coli, read this gene, how many amino acids long would the protein be?

This made-up gene reads, in codons: $\texttt{ATG}$ $\texttt{ATT}$ $\texttt{AAT}$ $\texttt{AAA}$ $\texttt{TTT}$ $\texttt{ATA}$ $\texttt{CAT}$ $\texttt{TAT}$ $\texttt{CCT}$ $\texttt{TTA}$ $\texttt{TAG}$ . Its protein is 10 amino acids long. A mutation inserts an extra A after letter 7. Using syn3A's code, where the stop codons are TAA and TAG, how many amino acids long is the new protein?

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.

Suppose a protein needs 1,000 new copies per cell cycle of 105 minutes. Each mRNA makes about 3 proteins per minute, and its half-life is 4 minutes. How many new mRNAs per minute must the gene make?

Here is a made-up mRNA, starting at its start codon: $5'\text{-}\texttt{AUG}$ $\texttt{UGU}$ $\texttt{AAA}$ $\texttt{ACA}$ $\texttt{ACU}$ $\texttt{UAG}$ $\texttt{AGG}$ $\texttt{GAA}\text{-}3'$ . Translate it with syn3A's genetic code. Codons you need: $\texttt{AAA}$ K (Lys), $\texttt{ACA}$ T (Thr), $\texttt{ACU}$ T (Thr), $\texttt{AGG}$ R (Arg), $\texttt{AUG}$ M (Met), $\texttt{GAA}$ E (Glu), $\texttt{UAG}$ stop, $\texttt{UGU}$ C (Cys).

syn3A's rpoB gene (JCVISYN3A_0804, the beta part of RNA polymerase) is at complement(479651..483526), on the other strand. Number its codons from 1, starting with the start codon. Which genome coordinate lines up with the first letter of codon 282?

A syn3A cell has 100 copies of an mRNA when it stops making new ones. The mRNA's half-life is 4 minutes. On average, how many copies are left 10 minutes later?

Codon 3 of syn3A's rpmH gene (ribosomal protein L34) is AGA. A mutation changes its second letter to A. In syn3A's code: $\texttt{AAA}$ K (Lys), $\texttt{ACA}$ T (Thr), $\texttt{AGA}$ R (Arg), $\texttt{ATA}$ I (Ile). What kind of mutation is this?

syn3A has about 500 ribosomes, and each adds about 5 amino acids per second. Suppose they all work nonstop. syn3A's proteins are 349 amino acids long on average. About how many average proteins does the cell finish per second?

Number the letters of a gene from 1, starting with the first letter of its start codon. Think about letter 398. Is it the 1st, 2nd or 3rd letter of its codon? Type 1, 2 or 3.

In random DNA with syn3A's A–T-rich mix of letters, read with the standard code, each codon is a stop with chance $p = 0.088$ . What is the chance that 100 codons in a row contain no stop? Write it as a power of ten, $10^{?}$ , and type just the exponent. Use $\log_{10}(1 - p) = -0.04001$ .