It's the first one where it resist change in pH solutions by neutralizing :)
Answer:
Indivisuals with producing an abnormal form of extracellular protein fibrillin are suffering from Marfan syndrome ehich is caused by genetic mutation in the FBN1 gene.
Explanation:
Gene mutations in FBN1 gene results in the production of an abnormal extracellular matrix fibrillin-1 protein that cannot function properly. These gene mutations basically reduce the amount of fibrillin-1 produced by the cell, alter the structure of fibrillin-1, or causes the impairment of the transport of fibrillin-1 out of the cell.
As a result, protein is poorly incorporated into extracellular matrix. Hence, indivisuals with Marfan syndrome present following symptoms
Tall stature.
Disproportionately long arms, legs and fingers.
Sternum either protrudes outward or dips inward.
Arched palate and crowded teeth.
Heart murmurs.
Extreme nearsightedness.
Question : T<span>he humpback whale can be seen off the coast of Alaska during the summer months, but is found as far south as California and Hawaii during the winter. Why would the humpback whale most likely spend the summer months off the coast of Alaska?
Answer : The humpback whale likes being in the cold like all other fishes.
The humpback whale is usually found around the coast of Alaska because it is a little colder over there. Then in winter it is found around Hawaii and California because the pacific ocean cools down and is not so hot anymore!
-procklown
</span><span>
</span>
Heat from the Sun causes water<span> to evaporate from the surface of lakes and oceans. This turns the liquid </span>water<span> into</span>water<span> vapor in the </span>atmosphere<span>. Plants, too, help </span>water<span> get into the </span>atmosphere<span> through a </span>process<span> called transpiration</span>
Let us assign the
recessive allele for freckle to be f and
the dominant F. According to the statement, we deduce that ff = 0.04.
The frequency of the recessive f allele is, therefore, is
0.2 (square root of 0.04 ).
Hardy-Weinburg p + q = 1
P + 0.2 = 1; p = 1- 0.2 = 0.8
Therefore using the Hardy-Weinburg equation of a population
in equilibrium
P2+2pq+q2=1
Heterozygous individuals are;
2pq = 2*0.8*0.2 = 0.32
This is 32% of the population
