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IMPORTANTÉ > Mathematics, Practicals, > Flashcards

Mathematics, Practicals, Flashcards

1
Q

what is ACCURACY

A

a measure of the closeness of agreement between an individual test result and the true value

if accurate, it is close in agreement with the true value- can be an accepted reference value, though usually not known

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2
Q

what is PRECISION

A

the closement of agreement between a number of independent measurements obtained under the same conditions in the experiment

it depends only on the distribution of random errors- i.e. the spread of measurements- and does not relate to the true value

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3
Q

what is ERROR of measurement

A

the difference between an individual measurement and the true value or accepted reference value of the quantity being measured

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4
Q

what are the two types of error and what is the difference between them

A
  1. RANDOM ERROR- errors in measurement caused by factors which vary from one measurement to another
  2. SYSTEMATIC ERROR- errors whose effects are not reduced when observations are averaged/the mean
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5
Q

what is the formula for the volume of a sphere

A

V = 4/3 π r3

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6
Q

which is RQ

(give formula)

A

respiratory quotient is the ratio of carbon dioxide produced and oxygen consumed per unit time by an organism

RQ = volume of CO2 produced/ volume of oxygen used

per unit of time

eg. if glucose, 6 CO2 molecules are produced and 6 O2 molecules are consumed; 6/6 = 1.0

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7
Q

what are the RQ values of different respiratory substrates

A

glucose = 1.0

triglycerides = 0.7

protein = 0.9

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8
Q

what is the equation for the Simpson’s Index of Diverisity (D)

A

D = 1 - Σ (n/N)²

n = number of individuals in one population

N = number of individuals in the whole community

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9
Q

what does the result of the Simpson’s Index of Diversity equate to

A

value of D is 0-1

0 = N0 diversity (where only 1 species exist)

near 1 = high diversity

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10
Q

how can you measure these abiotic factors:

  • Temperature
  • pH
  • Light intensity
  • Depth of water
  • Flow rate
  • O2 concentration
  • Turbidity
  • Type of substrate
  • Dissolved solids (conc. of ions eg. Ca2+)
A

Thermometer or temperature probe

Universal indicator or pH probe

Light meter

Metre ruler

Flow meter or timing how long a floating object takes to travel a set distance

O2 probe

Colorimeter or arbitrary scale (0-10)

Visual description eg. rock, mud, sand

Conductivity metre

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11
Q

give the formulae for NITRIFICATION

A

Oxidation of ammonia to nitrite

2 NH3 + 3 O2 → 2 NO2- + 2H+ + 4 H2O + release of energy*

  • ammonia+oxygen → nitrite+protons+water (*energy released used by bacteria for chemosynthesis)*
  • Oxidation of nitrite to nitrate*

2 NO2- + O2 → 2 NO3- + release of energy*

nitrite+oxygen → nitrate+energy

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12
Q

when should you use the STUDENT’s T-TEST

how does it work

A

> when there are two strata eg. a woodland habitat and grassland habitat and how many bluebells in each

  • State the null hypothesis
  • Calculate the t value, using the difference between the means and SD of the two sets of data.
  • Compare t with the critical value for the appropriate degrees of freedom at a p value of 0.05
  • If t > critical value, the difference is significant
  • Reject the null hypothesis
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13
Q

what are the types of line transect

A

CONTINUOUS LT = count all species that touch the transect

I N T E R R U P T E D LT = count species that touch the line at fixed intervals

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14
Q

how is productivity measured

A

kJ/m2/yr

due to seasons

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15
Q

what is standard deviation

how is it better than range

A

Spread from the mean i.e. distribution

The wider the spread, the larger the SD.

Range can include outliers.

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16
Q

how is % efficiency of biomass transfer between trophic levels measured

A
  • biomass at higher trophic level*
  • ——————————————– x100*
  • biomass at lower trophic* level
    eg. 250/1250 x 100 = 20%
17
Q

how to make sampling more accurate

A

USE MORE QUADRATS/LARGER SAMPLE SIZE, to

  • reduce effect of outliers
  • find a running mean/one closer to true value/find more consistent measure of mean

PLACE QUADRATS RANDOMLY eg. using random number generator/coordinates

or randomly within stratified sampling PROPORTIONALLY TO AREA SIZE

18
Q

when should you use chi-squared test

A

Finding if there is a difference between observed and expected values

eg. dihybrid crosses & ratios of expected vs. studied

IF x2 value calculated IS GREATER THAN THE CRITICAL VALUE, ACCEPT/SUPPORT NULL HYPOTHESIS

19
Q

what is a plan diagram

A
20
Q

when is the CHI-SQUARED stats test used

A

To assess the significance of differences between expected and observed results.

NULL HYPOTHESIS = no significant deviation between E and O results.

The larger the deviation, the larger the x2

IF x2 EXCEEDS p=0.05, THE O RESULTS DIFFER SIGNIFICANTLY

21
Q

when sampling, what is important to keep in mind for the method

A
  • REPEATS! in each area, with the same technique in each area
  • Use number of quadrats/sampling sites proportional to area’s size
  • Classify + count species, using identification key
  • Tape measure, quadrat, sweep net/pitfall traps,
22
Q

what is the null hypothesis in a Spearman’s Rank stats test

and what can we say about the null hypothesis if rs = 0.075 and the critical value at p = 0.05 is 0.564

A

There is no correlation between (factor A) and (factor B).

The critical value is larger than R. The correlation between factor A and factor B is not significant at the 95% confidence level.

Therefore there is a greater than 5% probability that the relationship is due to chance, so we accept the null hypothesis: no correlation.

23
Q

why can the STUDENT’s TEST not be used to analyse a set of observed + expected results

A
  • t-test compares two or more means
  • no mean in this data/cannot calculate mean/standard deviation from it
24
Q

why might observed results differ from expected results

A

RANDOM FERTILISATION

25
Q

Calculate rs for the following values

SOIL MOISTURE 0% 20% 40% 60% 80%

RANK (R1)

NUMBER OF SEEDLINGS GERMINATED 0 8 15 42 33

RANK (R2)

d (R1-R2)

critical value at p = 0.05 is 1.0

A

(R1) 1 2 3 4 5

(R2) 1 2 3 5 4

d (R1—R2)

0 0 0 1 1 d = 2 (no negatives)

d2 = 2 (i.e. 12 + 12)

rs = 1 - (6 x 2 / 5(25-1))

rs = 0.9

Critical value > rs , the correlation between soil moisture (%) and number of seedlings germinated is not significant.

26
Q

A farmer crossed an ash red male (XX) with a brown female (XY).

Ash red XCr is dominant to brown XCbr is dominant to blue XCbl

Pigeons are kept in small mixed flocks but tend to be monogamous- each season the female produces two eggs.

The offspring of the same pair of birds:

Year 1 : brown male, blue female

Year 2 : ash red male, ash red female

Year 3 : ash red male, blue female (etc. etc.)

x2 in Chi-squared test = 1.6 critical value at p=0.05 is 7.81

(a) What can be concluded about the observed results collected by the breeder? 1 mark

(b) Explain why the observed results did not exactly match the predicted results. 1 mark

(c) In Year 4 two brown females were produced. The breeder had not expected any, so had left the category out of the results table. What effect would adding this unexpected results have on the value of chi-squared? 1 mark

(d) Assuming the breeder had made an accurate prediction about the ratio of of offspring, what might the breeder have concluded about the parents of the chicks in Year 4? 1 mark

(e) Explain how you have reached this conclusion. 2 marks

A

(a) There is no significant difference between the expected & observed results.

(b) RANDOM FERTILISATION (not random mating)

(c) The value would increase to infinity.

(d) They were not monogamous- there was another bird involved.

(e) In female offspring the allele for feather colour comes from male parent. Original male bird did not have allele for brown feathers. Brown feathers allele in female would not produce brown female offspring.

27
Q

Heterozygous male x heterozygous female

TtDd x TtDd

offspring phenotypes: curly/pink, curly/black, straight/pink, straight/black

  1. What is the phenotypic ratio?
  2. Describe in words how this phenotypic ratio might be different if the two genes were autosomally linked.
A
  1. 9 : 3 : 3 : 1
  2. There would be a higher proportion of heterozygous (curly/pink) / more phenotypes like parents.

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