Variation (Cambridge (CIE) A Level Biology): Flashcards
Exam code: 9700
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- FrontVariation: Phenotype
Define phenotype.
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Cards in this collection (27)
Define phenotype.
The phenotype is the observable characteristics of an organism, resulting from the interaction of its genotype with the environment.
What are the three possible causes of phenotypic variation?
Genetic factors alone.
Environmental factors alone.
A combination of genetic and environmental factors.
Give an example of variation caused only by genetic factors.
ABO blood group in humans.
It is determined solely by the alleles inherited, with no environmental influence.
Give an example of variation caused only by the environment.
A scar, or a plant becoming tall and spindly (etiolated) when grown in the shade.
These are not caused by differences in alleles.
Give an example of variation caused by both genetic and environmental factors.
Height or body mass in humans.
Alleles set the potential, but diet and nutrition affect the value actually reached.
Variation caused by both alleles and the surroundings is due to a of genetic and environmental factors.
Variation caused by both alleles and the surroundings is due to a combination of genetic and environmental factors.
True or False?
Identical twins raised apart can differ in phenotype.
True.
They share the same genotype, so any differences must be caused by the environment.
Define discontinuous variation.
Discontinuous variation is variation showing distinct categories with no intermediates (e.g. blood group).
Define continuous variation.
Continuous variation is variation showing a full range of values between two extremes (e.g. height).
How is each type of variation usually displayed as a graph?
Discontinuous data is shown as a bar chart.
Continuous data is shown as a histogram, often giving a normal distribution curve.
Explain the genetic basis of discontinuous variation.
It is controlled by one or two genes (monogenic).
Different alleles have a large effect, and there is little or no environmental influence.
Explain the genetic basis of continuous variation.
It is controlled by many genes (polygenic).
Each gene has a small additive effect, and it is strongly influenced by the environment.
Variation controlled by many genes is described as .
Variation controlled by many genes is described as polygenic.
Blood group is an example of variation.
Blood group is an example of discontinuous variation.
True or False?
Continuous variation is usually controlled by a single gene.
False.
Continuous variation is polygenic — controlled by many genes with additive effects.
When is the t-test used?
To compare the means of two samples of continuous, normally distributed data, to see whether the difference between them is significant.
Define null hypothesis.
The null hypothesis states that there is no significant difference between the two samples, any difference being due to chance.
What conditions must the data meet to use a t-test?
The data must be continuous and approximately normally distributed.
The two samples should have similar variance.
How are the degrees of freedom calculated for a t-test?
Degrees of freedom = n~1~ + n~2~ − 2
(the two sample sizes added together, minus 2).
How do you interpret the calculated t against the critical value (at p = 0.05)?
If calculated t ≥ critical value: reject the null hypothesis — the difference is significant.
If calculated t < critical value: accept the null hypothesis — no significant difference.
The t-test compares the of two samples.
The t-test compares the means of two samples.
Why is the probability level p = 0.05 used?
It is the conventional 5% threshold.
Below it, you are at least 95% confident the difference is not due to chance.
Outline the steps to carry out a t-test.
Step 1: State the null hypothesis.
Step 2: Calculate the mean and standard deviation of each sample.
Step 3: Substitute into the formula to find t.
Step 4: Find the degrees of freedom and read off the critical value at p = 0.05.
Step 5: Compare and accept or reject the null hypothesis.
Two samples each have n = 12. What are the degrees of freedom?
Degrees of freedom = n1 + n2 − 2
= 12 + 12 − 2 = 22.
Calculated t = 2.85; critical value at p = 0.05 is 2.10. What do you conclude?
Calculated t (2.85) is greater than the critical value (2.10).
So reject the null hypothesis: there is a significant difference between the means.
Calculated t = 1.4; critical value at p = 0.05 is 2.09. What do you conclude?
Calculated t (1.4) is less than the critical value (2.09).
So accept the null hypothesis: there is no significant difference between the means.
If the calculated t value is greater than the critical value, you the null hypothesis.
If the calculated t value is greater than the critical value, you reject the null hypothesis.
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