The day trip What it is for To apply the key concepts from the lectures to a real- life example To produce your own experimental design To develop your group-working skills (and yes, this includes dealing with group members not pulling their weight - welcome to the real world)
Evolutionary Genetics Day Trip (Note the date is Tues 8th Nov) You are unlikely to pass Evolutionary Genetics at all if you do not attend this trip. Where is it? At Pulpit Hill reserve near Monk's Riseborough What should I do if I miss the coach ? Ring the department on 7882 3058 to let use know, and to get instructions for how to get to the site. Catch the train from Marylebone to Monk's Riseborough at your own expense and join up with the party by foot.
Left outside station Right at T junction, continue over roudabout & Ikm Left through gate onto footpath & up hill before the steep hill
Its best to avoid this! So don’t be late. You should meet outside Baker St Tube (ie. on the north side of Marylebone Rd) at 9.15 am Bring: a packed lunch･ sturdy boots/shoes･ rainwear (we will not stop for rain) and writing materials.
Think about the number and type of comparison Why do mulitple comparisons, on what scale?
How about studying boundaries? Environment Snail genotypes What sort of environmental transition? What sort of scale relative to the snail’s gene flow
You will find live and dead snails What use could you make of them? Remember all snails eventually die! Remember the environment may change over time.
A comparison of two sites Dead & Red Live & Red Dead & Blue Live & Blue Site 1 30 40 10 1 Site 2 50 10 4 2 Our null hypothesis was that the two sites have the same frequencies of phenotypes. As the expected values for Live and Blue are less than 3, we merged the live and dead categories to give the following contingency table.
Dead & Red Live & Red Blue Totals Site 1 30 40 11 81 Site 2 50 10 6 66 80 50 17 147 Expected values Dead & Red Live & Red Blue Site 1 80*81/147 50*81/147 17*81/147 Site 2 80*66/147 50*66/147 17*66/147
The 2 value is (Observed - Expected)2/Expected = 23.2 The table had 2 rows & 3 columns so there are (2-1)*(3-1)=2 degrees of freedom From the table of 2 values we can read of that this value exceeds the critical value for p<0.01 (23.2 > 9.21). This means that 2 values greater than 0.21 would be seen in less than 1% of tests if the null hypothesis were true. We can therefore reject the null hypothesis at the p=0.01 level. (NB we have not disproved the null hypothesis, it just appears highly unlikely on the basis of this test).
We therefore have strong evidence that the difference in phenotype frequencies between Site 1 and Site 2 cannot be attributed to sampling error. They might be due to genetic drift and/or selection. Nb. You do not prove “its random and hence due to drift!” Conversely there appears to be insufficient gene flow between the two sites to homogenise the shell colour allele frequencies. In the discussion you should expand on this “..the alternative interpretations are..“ (Remember that if your test shows that the frequencies are not significantly different, this might be because the shell phenotypes are in the same frequencies, but it might also indicate that your sample size was too small to detect the difference in frequency between the sites, or you were unlucky)