Answer:
Explanation:
Bacterial count in stock- 1.85x10^6 cfu/ml
Dilution methods
Take 100 uL or (0.1ml) from stock and add to 900ul (0.9ml) saline and mixed it- this makes 10^1dilution.
Now take 100ul from 10^1 dilution and add to next 900ul saline this is 10^2 dilution, similarly do upto 10^5 dilution.
Then take 100ul from 10^ 4 and 10^5 dilution seperately and plate on LB agar plate seperetely and count the colonies.
Cfu/ml formula= (No.of colonies x dilution factor)/0.1 ml
So suppose, 18 colonies formed on 10^4 dilution then total no. Of cells in stock will be 18x10^4/ 0.1= 18x10^5 cfu/ml.
If we dilute 10^4 or 10^5 that's leads to colony count of 18-19 colonies on 10^4 dilution while 2 colonies should come on plate of 10^5 dilution.
The correct answer is stabilising selection. This is selection that favours an average trait value. It is thought that this sort of selection is very common as traits among most animals do not appear to change drastically over time. Stabilising selection selects against the extreme traits, and leads to a decrease in genetic diversity by favouring the average phenotypes.
<span>A soil horizon is a layer parallel to the soil surface</span><span>, whose physical characteristics differ from the layers above and beneath</span>
Anticline I believe would be the answer but in not the smartest so yeah
