packback 10-12 Flashcards

Background (Quick + Helpful for Understanding):

λ phage chooses between:

• Lytic cycle = kill host, make new phage
• Lysogenic cycle = integrate into host genome, stay dormant

Key regulatory proteins:

- Cro: Lytic Blocks CI expression

- CI (λ repressor): Lysogenic Represses lytic genes & maintains lysogeny

- CII: Lysogenic Turns on CI (and thus lysogeny) early after infection

- CIII: Lysogenic Protects CII from degradation

- N & Q: Lytic (anti-termination)Allow expression of lytic regulatory genes

a) Nonfunctional Cro Cro normally inhibits CI.
If Cro doesn’t work → CI wins → lysogeny established.

How the Dideoxy (Sanger) Method Works Key Idea

DNA polymerase builds DNA by adding nucleotides to a growing chain.
Normal nucleotide: dNTP → has a 3’ OH → chain continues
Dideoxynucleotide: ddNTP → no 3’ OH → chain stops when added

Step-by-Step Mechanism

1. Denature template DNA into single strands
2. Add:
   • DNA primer (binds to known sequence)
   • DNA polymerase
   • All 4 dNTPs
   • Small amount of ddNTPs (each labeled, historically radioactive or fluorescent)
3. DNA polymerase copies template
4. Whenever a ddNTP is incorporated → chain terminates
5. This produces many DNA fragments of different lengths, each ending at a specific base

How We Read it

The fragments are separated by size using gel electrophoresis

• Shorter fragments travel further down
• Each lane represents one nucleotide (A, T, G, C) if using a 4-lane gel (modern methods combine lanes with fluorescent bases, but same logic applies)

You read the gel:

• Bottom → Top
• because smallest fragment (lowest band) = earliest termination = first base after primer

So you read which lane each band is in:

Bottom → Top might read:
G → A → C → T → A → T → G …

Step 0 – Make up a DNA sequence (first 5 bases known)

Let’s say the strand we want to know is:

5'-ATGCC TAGACTGT-3'

We are told we already know the first 5 nucleotides:

Known: 5'-ATGCC-3'
Unknown (what we’ll sequence): TAGACTGT

So the unknown part we’ll “discover” by Sanger is:

5'-TAGACTGT-3'

a) Design the sequencing primer

We design a primer that binds just before the unknown region, i.e., to the known sequence ATGCC.

To keep it simple, we’ll just use those 5 known bases as the primer:

Primer (5'→3') = 5'-ATGCC-3'

Conceptually:

• The template strand in the tube is the complement of our target strand.
• Our primer binds to this complement and DNA polymerase extends it into the unknown region.

You don’t have to stress the template strand sequence for this question; the key is:
primer = known sequence at the 5' end of what we’re sequencing.

b) Perform a “virtual” Sanger sequencing experiment

We set up 4 reactions (classic version), each with:

• Template DNA
• Primer: 5'-ATGCC-3'
• DNA polymerase
• All 4 normal dNTPs
• A small amount of one ddNTP in each tube:
  • Tube 1: ddATP
  • Tube 2: ddTTP
  • Tube 3: ddGTP
  • Tube 4: ddCTP

As DNA polymerase extends from the primer into the unknown region (TAGACTGT), sometimes it adds a normal dNTP and keeps going, and sometimes it accidentally adds a ddNTP, which stops the chain at that base.

Our newly synthesized unknown region is:

5'-TAGACTGT-3'
index the bases:
1: T
2: A
3: G
4: A
5: C
6: T
7: G
8: T

Where can termination occur?

• ddTTP (T): positions 1, 6, 8 → fragments of length 1, 6, 8
• ddATP (A): positions 2, 4 → fragments of length 2, 4
• ddGTP (G): positions 3, 7 → fragments of length 3, 7
• ddCTP (C): position 5 → fragment of length 5

(Remember: these lengths are for the newly added region after the primer.)

c) Run the DNA products “virtually” on a gel

We run the 4 reactions in 4 lanes on a polyacrylamide gel.

Let’s order the lanes (left → right) as:

Lane G | Lane A | Lane T | Lane C

Shorter fragments run farther (toward the bottom), longer fragments stay higher.

Using the fragment lengths we listed, here’s what the gel would look like conceptually:

• Lane G (ddGTP): bands at lengths 3, 7
• Lane A (ddATP): bands at lengths 2, 4
• Lane T (ddTTP): bands at lengths 1, 6, 8
• Lane C (ddCTP): band at length 5

Now, imagine the gel:

(top, largest fragments)

len 8: G | A | T● | C (T at position 8)
len 7: G●| A | T | C (G at position 7)
len 6: G | A | T● | C (T at position 6)
len 5: G | A | T | C● (C at position 5)
len 4: G | A●| T | C (A at position 4)
len 3: G●| A | T | C (G at position 3)
len 2: G | A●| T | C (A at position 2)
len 1: G | A | T● | C (T at position 1)

(bottom, smallest fragments)

You don’t need the exact picture on the exam, but you need to know:

• Bottom = smallest fragment = first base after the primer
• Each band tells you: “at this length, the chain ends in the base in this lane.”

d) Read the DNA sequence from the gel

To read the sequence:

1. Look at the bottom band first (shortest fragment = first base added after the primer).
2. Move upwards, noting which lane each band is in.
3. Write down the corresponding base.

From the “gel” above, going bottom → top:

• len1 → T lane → T
• len2 → A lane → A
• len3 → G lane → G
• len4 → A lane → A
• len5 → C lane → C
• len6 → T lane → T
• len7 → G lane → G
• len8 → T lane → T

So the newly synthesized unknown sequence is:

5'-TAGACTGT-3'

Put it together with the known first 5 nucleotides (ATGCC):

Full sequenced strand: 5'-ATGCCTAGACTGT-3'

That matches the “random” sequence we started with.