Practicals

Question 1

What method is used to extract plasmid DNA from bacterial cells?

Answer 1

Miniprep

Question 2

What is the definition of transforming bacteria?

Answer 2

The process of inserting plasmids into the bacterial cells.

Question 3

What does transforming bacteria allow?

Answer 3

This allows genes of interest to be inserted and replicated within the cells through the growth of bacterial colonies.

Question 4

Why are genes inserted into plasmids for antibiotic resistance?

Answer 4

Allows bacterial cells which have been transformed to be selected for, by growing them on a media containing an antibiotic.

Question 5

Each colony on a plate containing the selective growth medium will…

Answer 5

have grown from a single transformed bacteria.

Question 6

What will happen when single colony is removed and allowed to grow into a liquid cell culture?

Answer 6

Replicate the inserted plasmid and gene of interest.

Question 7

In liquid cell culture what happens to the bacteria?

Answer 7

Bacteria is added to a tube containing growth medium, rather than being spread onto an agar plate.

Question 8

Why do you need to count the number of bacterial cell colonies on the plates?

Answer 8

To estimate the frequency with which the bacteria develop a mutation that makes them antibiotic-resistant.

Question 9

Why do we count colonies?

Answer 9

Each colony will have grown from a single bacterium and gives an estimate of the concentration of cells added to a plate.

Question 10

Why is it important to know the volume of bacterial culture and the number of dilutions?

Answer 10

It will be possible to calculate the concentration of cells in the original culture.

Question 11

What is meant by logarithmic dilutions?

Answer 11

Serial dilutions which decrease in concentration by the same factor.

Question 12

Why are undiluted suspensions or solutions used?

Answer 12

used to produce a range of dilutions, usually so the original concentration can be calculated.

Question 13

Give an example of a diluent?

Answer 13

Distilled water or a buffer solution (added to to make each dilution)

Question 14

Why is it beneficial that dilutions increase by the same factor at each step?

Answer 14

Creates a broad range of concentrations.

Question 15

What does miniprep involve?

Answer 15

This involves breaking open the bacterial cells, and then separating the plasmid DNA from the proteins, chromosomal DNA and other cellular material.

Question 16

What is the difference in the pellet and the supernatant after centrifugation?

Answer 16

Pellet is the material that collects at the bottom of the tube, and the supernatant is the liquid that remains above the pellet in the tube.

Question 17

Why is the bacteria culture centrifuged?

Answer 17

To collect the bacterial cells at the bottom of the tube.

Question 18

What is Lysis?

Answer 18

The growth media (supernatant) is removed, and a detergent is added to the bacteria. This dissolves the cell membranes, breaking open the bacteria and releasing the cellular contents.

Question 19

Why is a high salt concentration added to the extract?

Answer 19

causes the chromosomal DNA to become insoluble, so it precipitates. The plasmid DNA does not precipitate. Centrifugation of the extract collects the chromosomal DNA at the bottom of the tube as a pellet. The supernatant containing the plasmid is removed.

Question 20

Why is an alcohol (isopropanol) added to the plasmid solution?

Answer 20

Causes the plasmid to precipitate.

Question 21

Why is agarose gel electrophoresis used?

Answer 21

The plasmids inserted into transformed bacteria can be separated from other nucleic acids

Question 22

What buffer is used for running the gel?

Answer 22

Tris-acetate-EDTA or TAE, with a pH of ~8.5. It is a common buffer for agarose gel electrophoresis and is suitable for short gel run times (<4 hours) and large pieces of DNA (>20 kb).

Question 23

How are the DNA molecules separated in the gel?

Answer 23

applying an electrical field across the gel, which causes DNA (naturally negatively charged) to move towards the positively-charged anode. Smaller DNA molecules move through the gel faster, thus the DNA is separated according to size. Plasmids are smaller than the bacterial chromosomal DNA, for example, so move further within the gel.

Question 24

How does voltage effect the run time of the gel electrophoresis?

Answer 24

A low voltage will mean the gel takes too long to run. We are separating large pieces of DNA (~4kb), therefore we need a relatively high voltage to move them. Gel density also affects run time.

Question 25

Why is the DNA in the gel stained with ethidium bromide?

Answer 25

binds to DNA and fluoresces brightly under UV light. This makes it possible to visualise the nucleic acids within the gel, so it can be analysed.

Question 26

In your experiment last week, you transformed E. coli cells with a plasmid containing the bla gene.

How will this change the phenotype of the bacterial cells?

Answer 26

The cells became resistant to ampicillin.
Transformed cells became resistant to ampicillin because they contained the bla gene.

This means ampicillin resistance can be used as a selective marker for transformed cells.

Question 27

It may be necessary to measure the size of samples viewed using a microscope. Why does this create a challenge?

Answer 27

because it is not possible to directly measure microscopic samples.

Question 28

How do you make it possible to measure samples?

Answer 28

n eyepiece graticule can be used, which is simply a scale with numbered divisions which is situated within the eyepiece of the microscope.

Question 29

There are a number of steps to measuring using a graticule:

Answer 29

1. Measure the total length of the graticule using a stage micrometer, which is a scale of known length which is placed on the stage and aligned with the graticule. At this stage the total length of the graticule and the number of divisions needs to be recorded, e.g. 500 µm graticule length, 50 graticule divisions.

Note: The total length of the graticule will be different for each objective lens.

2. Determine the distance represented by each graticule division - Simply use the equation
   Total graticule length / Number of graticule divisions = Division length
3. Count the number of divisions for the structure being measured - The graticule can be rotated and the specimen moved around to align the scale with the structure being measured.
4. Determine length of structure - Simply multiply the length of 1 division (step 2) by the number of divisions counted (step 3)
   Division length x Number of divisions = Measured length

Question 30

Which environmental factors can influence the aperture of stomata?

Answer 30

Light

Water availability in the soil

Carbon dioxide concentration in the air

Oxygen concentration in the air

Light usually leads to an opening of stomata, as photosynthesis takes place in the light and requires CO2 (substrate for photosynthesis) needs to enter into the leaf. If the CO2 concentration in the air is high, the stomata do not need to open widely to allow diffusion of CO2 into the leaf. Conversely, in environments with low CO2, stomata will open widely to improve diffusion into the leaf. If plants grow in well-watered soil, they can open stomata widely as any water lost through transpiration can be replenished from the soil. However, if the soil is dry, plants will close their stomata to reduce water loss through transpiration. The oxygen concentration of the air and the nutrient content of the soil do not have an influence on the opening and closing of stomata.

Question 31

Which of the following statements about abscisic acid (ABA) are correct?

Answer 31

ABA is a plant hormone
Plants increase the synthesis of ABA synthesis when they are under drought stress.
Stomata close in response to ABA.

Question 32

Which statement(s) about stomatal density are true?

Answer 32

Stomatal density is the number of stomata per leaf surface area.

Changes in stomatal density are a long-term response to changes in environmental conditions.

Stomatal density increases in response to low CO2 concentrations.

Question 33

During the practical the cockroach leg will be pinned to the corkboard with two pin electrodes. These will be placed through the

Answer 33

Femur

Question 34

Spontaneous action potentials will be recorded from (choose an option) leg

Answer 34

Spontaneous action potentials will be recorded from an unstimulated leg.

Question 35

To calculate spontaneous action potential frequency the ______ will be recorded.

Answer 35

Number of spikes