By definition, a sympatric speciation is a biological process wherein the evolution of a certain species most likely comes from their single ancestral species. Its main difference from the allopatric speciation is that the evolution does not necessary require to be physical in nature, therefore, it is most likely to occur less frequent.
In some capacity, earth scientists study all of the above listed. They focus on ecology (how different organisms function within ecosystems), food chains, climates and weather patterns, aspects of the atmosphere, soil and the Earth's mineral components, farming, and how man's machines both help and hurt the environment.
Answer:
Suppose that at a given point along a capillary, the following forces exist: Capillary hydrostatic pressure (HPc) = 30 mmHg Interstitial fluid hydrostatic pressure (HPif) = 0 mmHg Capillary colloid osmotic pressure (OPc) = 25 mmHg Interstitial fluid colloid osmotic pressure (OPif) = 2 mmHg. The net filtration pressure at this point in the capillary is <u>7mmHg.</u>
Explanation:
Capillary hydrostatic pressure (HPc) = 30 mmHg
Interstitial fluid hydrostatic pressure (HPif) = 0 mmHg
Capillary colloid osmotic pressure (OPc) = 25 mmHg
Interstitial fluid colloid osmotic pressure (OPif) = 2 mmHg
Net filtration pressure= hydrostatic pressure gradient - Oncotic pressure gradient
Hydrostatic pressure gradient = Capillary hydrostatic pressure - Interstitial hydrostatic pressure = 30mmHg - 0 mmHg = 30 mmHg
Oncotic pressure gradient = Capillary colloid osmotic pressure - Interstitial fluid colloid osmotic pressure =25 - 2 = 23 mmHg
Net filtration pressure= hydrostatic pressure gradient - Oncotic pressure gradient = 30 mmHg - 23 mmHg = 7 mmHg.
Hence, The net filtration pressure at this point in the capillary is <u>7mmHg.</u>
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.
