A plant breeder is trying to increase the medicinal quality of belladonna (Atropa belladonna), which is derived from atropine, an alkaloid in the roots. Her previous research indicates that about 55% of atropine variation is due either to environmental factors (e. g., soil conditions or rainfall), or to genetic dominance and genetic interactions. If the average plant has 10 units of atropine and our breeder selects for plants with an average of 50 units of atropine, what will be the average amount of atropine in the new population

The average amount of atropine in the new population will be 32 units of atropine.

Explanation:

To answer this question, we need to remember how can we calculate the selection differential and the heritability in the narrow sense.

• We can get the selection differential, SD, by getting the difference between the mean value of the population´s units of atropine, PoUA, and the mean value of the parents´ units of atropine, PaUA. So, in this example the selection differential is:

SD = PoUA - PaUA    

SD = 10 - 50  

SD = - 40    

•The heritability in the narrow sense, h², for units of atropine in the population is

h² = units of atropine of the offspring average/selection differential

But we already know the value of h² = 55% = 0.55. And we want to know the units of atropine of the offspring. So we just need to clear the following equation.

OUA = PoUA + h² ( PaUA - PoUA)

where,  

•offspring average units of atropine = OUA

•population average units of atropine = PoUA = 10 units

•parents average units of atropine = PaUA = 50 units

•Selection diferential = SD = -40 units

•narrow sense heritability = h² = 0.55

OUA = PoUA + h² ( PaUA - PoUA)

OUA = PoUA + h² (SD)

OUA = 10 + 0.55 x ( 50 - 10)

OUA = 10 + 22

OUA = 32 units.  

the bald eagle got 50 because everything gets 10% of what they consume and since 50 is ten % of 500 hundred that means the   grasshopper had to have consumed a plant with 5,000 calories of energy

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