Suppose 64% of a remote mountain village can taste phenylthiocarbamide (ptc) and must, therefore, have at least one copy of the dominant ptc taster allele. if this population conforms to hardy-weinberg equilibrium for this gene, what percentage of the population must be heterozygous for this trait?

The answer is; 48%

This is how to calculate;

Let's assign the dominant allele letter ‘T’ and the recessive letter ‘t’

Using the hardy weignburg equation; p2 + 2pq + q2  = 1

P2 + 2pq = 0.64

q2 = 1 – 0.64 = 0.36

q = 0.6

Remember that; p + q = 1

P = 1 – 0.6 = 0.4

Now to calculate population with herozygour trait = 2pq

2 * 0.4 * 0.6 = 0.48

This means 48% of the population is heterozygous for the trait that enable tasting of phenylthiocarbamide

it has to be understood that the sea water contains a huge amount of sodium and chlorine ions and if they enter the cell of the atlantic cod fish, the cell will shrivel and the fish will die. to avoid such a condition, the fish uses active transport for the purpose of removing the ions against the gradient.

explanation:

the atlantic cod is not a maritime osmoconformer. its native application of salts is less than that of the seawater it lives in. it is continually falling water from the gills to the ocean via osmosis.

countering that, the fish is going to be swallowing seawater. but since its internal collection of salts is lower than seawater, the fish will want to go free of excess salt, and since that's going upon a gradient, then active transport, requiring energy, must be used to pump out the salt. much marine fish use the organs to filter out the salt and these fish discharge very straight urine.