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17.117-18 · Selection, Evolution, Classification and Conservation

Variation — Cambridge AS & A Level Biology 9700 Revision Notes & Topical Questions

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17.1 Variation — CAMBRIDGE AS & A LEVEL BIOLOGY 9700 — REVISION NOTES notes: Explain phenotypic and genetic variation, continuous and discontinuous variation, and use the t-test.Phenotype — the observable characteristics of an organism.Phenotypic variation — differences in phenot…

  • Genetic causes
  • Environmental causes
  • Definition
  • Inheritance
  • Independent assortment
  • Crossing over

Section 1 of 20Phenotypic variation

Phenotypic variation
Phenotype
  • Phenotype the observable characteristics of an organism.
Phenotypic variation
  • Phenotypic variation differences in phenotype between organisms of the same species.
Genetic causes
  • Genetic causes for example the four ABO blood groups from three alleles of one gene.
Environmental causes
  • Environmental causes for example cloned plants growing to different heights in different conditions.
Both
  • Both some variation has genetic and environmental causes, such as the sickle cell allele being common where malaria is prevalent.
Equation
  • Equation phenotype = genotype + environment.

Section 2 of 20Genetic variation

Definition
  • Definition small differences in DNA base sequences between individuals of a species.
Inheritance
  • Inheritance passed from one generation to the next and generates phenotypic variation.
Independent assortment
  • Independent assortment random alignment of homologous chromosomes at metaphase I.
Crossing over
  • Crossing over exchange between non-sister chromatids in prophase I.
Random fusion of gametes
  • Random fusion of gametes any male gamete can fuse with any female gamete.
Mutation
  • Mutation a random change in the base sequence producing new alleles, which must be in gametes to be passed on.
New alleles
  • New alleles may be advantageous, disadvantageous or neutral, and may stay hidden for several generations.

Section 3 of 20Environmental variation

Factors
  • Factors length of sunlight hours, supply of nutrients, temperature range and oxygen levels.
Effect
  • Effect a tall genotype in poor conditions cannot reach its full height.
Inheritance
  • Inheritance variation caused solely by the environment is not inherited.
Reason
  • Reason only changes to the genetic component of gametes are inherited.

Section 4 of 20Discontinuous variation

Cause
  • Cause qualitative differences in phenotype.
Categories
  • Categories discrete categories with no intermediates, such as ABO blood groups.
Genetic basis
  • Genetic basis caused solely by genetic factors; different alleles at a single locus have a large effect.
Example
  • Example alleles of the F8 gene decide whether factor VIII is made and whether haemophilia occurs.
Display
  • Display distinct categories in a table or graph.

Section 5 of 20Continuous variation

Cause
  • Cause quantitative differences with a range of values between two extremes, such as height and mass.
Genetic basis
  • Genetic basis an interaction between genetics and the environment.
Alleles
  • Alleles different alleles at one locus have a small effect.
Polygenes
  • Polygenes many genes with additive effects.
Additive effect
  • Additive effect genotypes with more dominant alleles contribute more to the phenotype.
Display
  • Display no categories and a range of values.

Section 6 of 20The t-test

Purpose
  • Purpose compares the means of two sets of continuous data to see if they differ significantly.
Conditions
  • Conditions data roughly normally distributed and standard deviations about equal.
Formula
  • Formula given in the exam, but the degrees of freedom formula must be learnt.
Null hypothesis
  • Null hypothesis no significant difference between the two means.

Section 7 of 20Carrying out the t-test

Carrying out the t-test
Calculate the mean of each data set.
Calculate the standard deviation of each set.
Square each standard deviation and divide by the sample size n.
Add the two values and take the square root.
Divide the difference between the means by this value to give t.
Calculate degrees of freedom v
(n1 − 1) + (n2 − 1).
Compare t with the critical value at p
0.05.

Section 8 of 20Interpreting the t value

1t greater than critical value
  • t greater than critical value a significant difference with a probability below 5 % that chance caused it;
    • reject the null hypothesis.
2t smaller than critical value
  • t smaller than critical value no significant difference; accept the null hypothesis.
3Larger t
  • Larger t a lower probability that chance causes the difference.
4Conclusion
  • Conclusion say that the difference between the data sets is significant.

Section 9 of 20More from the specification

More from the specification
Variation
  • Differences between individuals in a species population; it can be continuous or discontinuous.
Phenotype
  • Variation refers to differences caused by genetic factors, environment, or both.
Populations
  • Different populations of a species may show different mean values for a feature such as height.
Statistics
  • The t-test method compares the means of two samples to determine if the difference is statistically significant.
Significance
  • If probability is below 0.05, the difference is unlikely to be due to chance.
Population
  • A group of organisms of one species in the same area at the same time.
Statistically significant
  • A difference with a probability of less than 5% of being due to chance.

Section 10 of 20Causes of variation

Variation
  • Variation the differences between individuals.
Discontinuous variation
  • Discontinuous variation distinct categories, e.g. blood group.
Continuous variation
  • Continuous variation a range of values, e.g. height and mass.
It may be caused by genes, the environment or both.
  • Height depends on genes and on nutrition.
  • The environment has no direct effect on some characters, such as blood group.

Section 11 of 20Genes and phenotype

1
Different genes have different effects on the phenotype.
2
Different alleles at a single locus can have a large effect on the phenotype.
3
In polygenic inheritance many genes each contribute a small, additive amount.
4
Check the information given for polygenes, since each gene may not have the same effect.

Section 12 of 20The t-test

The t-test
  • The t-test compares the means of two sets of data to see if they differ significantly.
The formula for t is given in the exam, but the formula for degrees of freedom must be learnt.
The data must be continuous, roughly normally distributed and have approximately equal standard deviations.
The standard deviation (s) must be calculated for each data set first.

Section 13 of 20Steps of the t-test

1
State the null hypothesis: no significant difference between the means.
2
Calculate the mean and standard deviation of each sample (s1 and s2).
3
Square each standard deviation and divide by n for each sample.
4
Add the two values and take the square root.
5
Divide the difference between the means by this value to give t.
6
Calculate the degrees of freedom: (n1 − 1) + (n2 − 1).
7
Compare t with the critical value at p = 0.05.
8
State the conclusion.

Section 14 of 20Interpreting t

1
The table gives critical values at p = 0.10, 0.05, 0.01 and 0.001 for each degree of freedom.
  • For 2 degrees of freedom the values are 2.92, 4.30, 9.93 and 31.6.
2
If t is less than the critical value at p = 0.05, the null hypothesis is accepted and there is no significant difference.
3
If t is greater, the null hypothesis is rejected and the difference is significant.
4
Questions usually require you to say why a t-test is used, state the null hypothesis, give how degrees of freedom are found and state the conclusion.
5
Say the difference between the data sets is significant, not just that the data is significant.

Section 15 of 20Choosing the test and stating the hypothesis

The t-test is used because the data are continuous, roughly normally distributed and two means are compared.
Record the sample sizes n1 and n2.
Null hypothesis
  • Null hypothesis there is no significant difference between the mean values of the two populations.

Section 16 of 20Calculating the means and standard deviations

1
Add the values in each sample and divide by the number of values to give each mean.
2
For each value find x − mean and square it.
3
Add all squares for each sample to give Σ(x − x̄)2.
4
Divide by n − 1 and take the square root to give s = √[Σ(x − x̄)2 ÷ (n − 1)].

Section 17 of 20Calculating t and degrees of freedom

1
Square each standard deviation and divide by its sample size.
2
Add the two values and take the square root.
3
Divide the difference between the means by this value to give t.
4
Degrees of freedom = (n1 − 1) + (n2 − 1).

Section 18 of 20Interpreting t

1
Look up the critical value for the degrees of freedom at p = 0.05 in the table.
2
If t is less than the critical value, the null hypothesis is accepted.
  • There is no significant difference and any difference between the means is due to chance.
3
If t is greater than the critical value, the null hypothesis is rejected.
  • The probability that the difference is due to chance is less than 5%, so the difference is significant.
4
State the conclusion in terms of the characteristic measured.

Section 19 of 20Key terms

TermMeaning
PhenotypeObservable characteristics of an organism.
Genetic variationDifferences in DNA base sequences within a species.
Discontinuous variationQualitative variation in distinct categories.
Continuous variationQuantitative variation across a range.
PolygeneOne of many genes with additive effects.
t-testTest comparing the means of two data sets.
Null hypothesisStatement of no significant difference.
Standard deviationMeasure of spread around the mean.
TermMeaning
VariationDifferences between individuals.
Continuous variationA range of values between two extremes.
Discontinuous variationDistinct categories with no intermediates.
PolygeneOne of many genes with additive effect on one phenotype.
t-testTest comparing the means of two continuous data sets.
Null hypothesisNo significant difference between the means.
Degrees of freedom(n1 − 1) + (n2 − 1).
TermMeaning
Null hypothesisStatement that there is no significant difference.
Standard deviationSpread of data about the mean, using n − 1.
Critical valueTable value that t must reach for significance.
Degrees of freedom(n1 − 1) + (n2 − 1).
SignificantLess than 5% probability that the difference is due to chance.

Section 20 of 20Exam technique

1
Saying environmental variation is inherited — only genetic variation is inherited.
2
Saying discontinuous variation is partly environmental — it is solely genetic.
3
Saying the t-test compares categories — it compares means of continuous data.
4
Saying the data are significant — say the difference between the means is significant.
1
Saying the data are significant — the difference between the means is significant.
2
Forgetting the degrees of freedom formula — it is not given in the exam and must be learnt.
3
Rejecting the null hypothesis when t is below the critical value.
4
Using the t-test on categorical data.
1
Dividing the sum of squares by n instead of n − 1.
2
Forgetting to compare t with the critical value before concluding.
3
Accepting the null hypothesis when t is above the critical value.
4
Not stating the conclusion in context.

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