Hemophilia is an X-linked recessive trait. If a non-hemophiliac man mates with a non-hemophiliac woman and they have a hemophiliac son, (a) what is the probability that their next son will have hemophilia? (b) What phenotypes and genotypes are possible for their daughters?

(a) 1/2

(b) Possible Genotype =

    Possible phenotype = non-haemophilic

Explanation:

Let the allele for haemophilia be h, the alternate form would be H.

Genotype of non-haemophilic man would be

Genotype of non haemophilic woman could be either

If they have an haemophilic son, it means the genotype of the woman is

Crossing the two:

  x    =

(a) One out of the two sons is affected. Hence the probability of their next son having haemophilia is 1/2.

(b) the possible genotypes for their daughters are and both of them have no haemophilia.

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.

the answer is c. it will give up it's valence electron to the other atom.