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Lab 2 MICROBIOLOGY Background  Bacterial species can be distinguished from one another by looking at a variety of characteristics: PowerPoint PPT Presentation

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Lab 2 MICROBIOLOGY Background  Bacterial species can be distinguished from one another by looking at a variety of characteristics:  Morphology of individual bacterium of a species:- cocci - rods / bacilli - spirilla  Colonial appearance of a bacterial species: - size

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Lab 2 MICROBIOLOGY Background  Bacterial species can be distinguished from one another by looking at a variety of characteristics:

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 Bacterial species can be distinguished from one another by looking at a variety of


 Morphology of individual bacterium of a species:- cocci

- rods / bacilli

- spirilla

 Colonial appearance of a bacterial species: - size

- shape

- colour

 Staining properties: Gram staining - due to differences in cell wall:

- Gram positive bac purple

- Gram negative bac pink

 Bacterial cultures

 Culture medium provides nutrients to bac

Can be liquid (test tube cultures) or solid (petri plate cultures)

 Colony consists of identical cells. A culture made from a colony = pure culture

 Bacterial growth

 Most species divide by binary fission: one cell divides into two

Generation time = time it takes for bac population to double

Bacterial growth = determined by counting # cells over time


 Sterile techniques:- Inoculating loop

- Spreader

What you did:

1) Observation of three different bac species: - Colony appearance (size and shape)

- Cell appearance following Gram staining

- Gram +/-

2) Bacterial growth

- Plating technique p.2-9

- Calculate generation time p.2-11

2 # cell generations= # cells at 120 min

# cells at 0 min

Generation time = 120 min = about 40 min for S. marcescens

# cell generations




A reducing substance gives up electrons or Hatom

A substance that gains electrons or H atom is reduced

A substance that loses electrons or H atom is oxidized

What you did:

1) Benedict's test: demonstrates the presence of a reducing sugar

- Add Benedict's solution to sugar. If it is a reducing sugar color change

Glucose = reducing

Sucrose (glucose + fructose) is not reducing because reducing ends are attached


2) Iodine test: demonstrates the presence of starch

- Add iodine to sugar. If it is composed of starch purplish color

Cotton is not made of starch but of cellulose (polymer of glucose)

3) Starch and cellulose (macromolecules)

- Both are polymers of glc but the bonds between the glc units are different:

Starch = -linked Cellulose = -linked

- Observed outcome of 2 chemical treatments on both starch and cellulose:

i) acid (HCl) + high heat

ii) enzyme (amylase) + body temperature


- Both macromolecules are not reducing however their subunits (glc) are

- Both kinds of treatment cut starch

- However only HCl treatment cut cellulose

amylase is specific for cutting -linkages


succinic dehydrogenase

Succinate + coenzyme

fumarate + coenzyme - H2

(natural H2 acceptor)

(found in MT)


1) Artificial H2 acceptor: DCIP

DCIP gets blue when oxidized

becomes decolorized when reduced

succinic dehydrogenase

Succinate + DCIP

fumarate + DCIP-H2


(reduced and colorless)

2) Competitive inhibitor: Malonate

Succinic dehydrogenase will complexe with malonate but cannot dehydrogenate it

it is inhibited

A competitive inhibitor binds to the same site (active site ) as the natural substrate

What you did:

- Mixed MT and DCIP with either succinate or malonate

- Assessed enzyme activity by looking at color change over time using spectrophotometry

Absorbance = direct measure of solute concentration (non-reduced DCIP) and thus

= direct measure of enzyme activity


 When added succinate: the more enzyme (MT) added, the faster the initial reaction

(blue color lost rapidly).

Adding tons of enzyme will not speed up the reaction indefinitely as the substrate

concentration becomes the limiting factor.

 When added malonate: no change in blue color over time since malonate inhibited the



Meaning: energy conversion

Organic molecules such as carbohydrates are a form of long-term storage for energy

Problem: cells cannot use carbohydrates directly

Solution: convert it to ATP

How? Respiration

Aerobic vs anaerobic respiration - in yeast


hydrolytic enz


glycolytic enz

2 pyruvic acids

fermentation enz

citric acid cycle enz + 6 O2

2 CO2 + 2 ethyl alcohol + E

6 CO2 + 6 H2O + E

Anaerobic (fermentation)


Gas exchange in aerobic respiration (peas)

- How do you measure respiration? By measuring gas exchange

- Problem: in aerobic resp, O2 consumption is couterbalanced by CO2 release

won't see difference in gas volume over time

- Solution: use an alkaline solution (NaOH) which will remove CO2

Gas exchange in anaerobic respiration (yeast)

- As long as O2 is present, yeast cells respire aerobically and anaerobically

- To look at fermentation only, O2 was eliminated in the experiment

- Glc and galactose were tested as well as the effect of sodium fluoride (NaF)

Results:- Yeast cells only respire when glc was present but not galactose

This is because they lack an enzyme that converts gal to glc (an isomerase).

Therefore, glycolytic enz are specific for breaking down glc but not gal

- Also, NaF inhibited fermentation because it combined with Mg2+ and removed

it from important enzymes of respiratory pathway

Lab 8 Sex-linked cross - week 1

Genetic problems

Dihybrid cross: a mating in which the parents differ in two traits

As long as the 2 genes are on different chromosomes, the way one pair

of alleles segregates has no effect on how the other pair of alleles

segregates, i.e. the alleles of maternal (or paternal) origin won't

necessarily end up in the same gamete independent assortment

Sex-linked inheritance

- Males are hemizygous for genes on X chr

- One mutant recessive allele is sufficient to produce a mutant phenotype in males

This is because there are no normal copy of this gene present (only one X).

- However in females, 2 mutant copies = needed for expression of the mutant phenotype

Incomplete dominance

- Some genes have alleles that are not dominant or recessive to each other.

Instead, the htz phenotype is intermediate

These genes = governed by incomplete dominance

Test cross

- Way to test whether a given individual showing a dominant trait is hmz or htz

- Cross this individual with a hmz recessive mutant and do progeny testing:

- if all progeny shows a dom pheno parent was hmz dom

- if about half the progeny shows a rec pheno parent was htz


- To be considered a true-breeding, the observed trait must be the only form present

in many generations

- True-breeding lineages are therefore hmz for the trait of interest

Lab 10 Microbiology

Part I Growth cycle of bacteriophage


- Viruses cannot replicate outside of living cells

- If plate a suspension of bacteria (E. coli) in which a couple cells are infected by phage (T4)

and incubate, you will observe a dense lawn of bateria with clear areas (plaques).

- Plaques = areas of dead cells due to phage infection (phage lyse cell after replication

and infect neighbouring cells)

- Counting # plaques tells you about the # of infected bac in the original suspension and

thus about the density of infective phage particles

What you did:

- Prepared T4-infected E.coli suspension

- Every 10 minutes, plated one drop of this infected culture with one drop of uninfected

culture + incubate. Count # plaques

- Calculated: - length of growth cycle (middle of rise period)

- burst size = # progeny / bac = # plaques at 40 min

# plaques at 10 min

Part II Effects of UV radiation on bac

What you did:

- Looked at the effect of varying doses (0, 5, 10, 15 sec) of radiation on survival rate of

S. marcescens

- The longer the exposure, the less colony you get because UV light causes DNA damage


- Recombination = genetic exchange between two homologous chromosomes

- Occurs during prophase of meiosis I, when the homologous chromosomes synapse

- The closer together are two genes on a chromosome, the less chance they have of being

separated by recombination (recombination is a chance event)

- Cis arrangement: all wild-type alleles = on same homologue

all mutants are on the other

Trans arrangement: each homologue has both mutant and normal alleles

- The percentage of recombination between two genes on a chr can be calculated as follows:

# of recombinants X 100 = % recombination

total # progeny

- Remember that 1% = 1 map unit distance

A couple of genetic problems...

1. If you had a fruit fly of phenotype A, what test would you make to determine if it is

AA or Aa?

2. Two black guinea pigs were mated and over several years produced 29 black and

9 white offspring. Explain these results, giving the genotypes of parents and progeny.

3. Can it ever be proven that an animal is not a carrier of a recessive allele

(that is, not a heterozygote for a given gene)? Explain.

4. In dogs, dark coat color is dominant over albino and short hair is dominant over

long hair. Assume te two genes are on different chromosomes and write the

genotypes of the parents in each of the crosses shown below.

Parental Number of progeny

phenotypes D,S D,s d,S d,s

a. D,S x D,S89 31 29 11

b. D,S x D,s 18 19 0 0

c. D,S x d,S 20 0 21 0

d. d,S x d,S 0 0 28 9

e. D,s x D,s 0 32 0 10

5. Consider the cross: A/a;B/b;C/c;D/d;E/e x a/a;B/b;c/c;D/d;e/e

a. What proportion of progeny will phenotypically resemble (1) the first parent,

(2) the second parent, (3) either parent, and (4) neither parent?

b. What proportion of progeny will genotypically resemble (1) the first parent,

(2) the second parent, (3) either parent, and (4) neither parent?

Assume independent assortment.

6. When a cell of genotype A/a;B/b;C/c having all genes on separate chromosome

pairs divides mitotically, what are the genotypes of the daughter cells?

7. The recessive allele s causes Drosophila to have small wings and the s+allele

causes normal wings. This gene is X-linked.

a. If a small winged male is crossed with a homozygous wild-type female,

what ratio of normal to small-winged flies can be expected in each sex in the F1?

b. If F1 flies are intercrossed, what F2 progeny ratios are expected?

c. What progeny ratios are predicted if F1 females are backcrossed to their father?

8. Human cells have 46 chromosomes. For each of the following stages, state the

number of chromosomes present in a human cell (here, 1 chromatid = 1 chromosome):

a. metaphase of mitosis

b. metaphase I of meiosis

c. telophase of mitosis

d. telophase I of meiosis

e. telophase II of meiosis

9. Four of the following events are part of both meiosis and mitosis, but only one is

only meiotic. Which one?

(a) chromatid formation, (b) spindle formation, (c) chromosome condensation,

(d) chromosome movement to poles, (e) chromosome pairing.

10. In Drosophila, there is a dominant allele for grey body color and a dominant allele

of another gene for normal wings. The recessive alleles of these two genes result in

black body color and vestigial wings respectively. Flies hmz for grey body and

normal wings were crossed with flies that had black bodies and vestigial wings.

The F1 progeny were then test-crossed, with the following results:

Grey body, normal wings236

Black body, vestigial wings253

Grey body, vestigial wings 50

Black body, normal wings 61

Would you say that these two genes are linked? If so, how many units apart

are they?

Answers to genetic problems

1. Test cross

2. B/b x B/b

3. When you do a test cross, you assume that if the animal tested is homozygous

dominant, all the progeny should have a dominant phenotype. However if

looking at a small sample size, it cannot be excluded that the animal was indeed

heterozygous and that just by chance, its recessive allele was not passed on to

its progeny. Therefore, in order to be able to say that an animal is not a carrier

for a recessive allele, one has to look at a large number of progeny.

4. a. C/c;S/s x C/c;S/s

b. C/C;S/s x C/C;s/s

c. C/cS/S x c/c;S/S

d. c/c;S/s x c/c;S/s

e. C/c;s/s x C/c;s/s

5. a. (1) 9/128 (2) 9/128 (3) 9/64 (4) 55/64

b. (1) 1/32 (2) 1/32 (3) 1/16 (4) 15/16

6. Both will be A/a;B/b;C/c

7. a. All F1 progeny will be normal for wing size

b. For males: half will be be normal, the other half will be small-winged

For females: all will be normal

c. Samething for both males and females: half will be be normal, the other half

will be small-winged

8. a. 92 (the chromosomes are duplicated)

b. 92

c. 46

d. 46

e. 23

9. e

10. The two genes are linked and are 18.5 m.u. apart.

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