Answer:
C-able to relieve haemophilia
Explanation:
Factor V111 is a blood clotting factor;produced in the liver and body endothelial cells. It is usually exist as inactive form beacuse it is binded to glycoprotein (glycogen +protein) called Von Willebrand factor when released into the blood stream.
If absent in the gene of certain individuals due to mutations, it result in a sex-linked conndtion called haemophilia, where blood failed to clot when blood vessels are damaged leading to profuse bleeding/ hemorrhage.
Ideally in a normal individuals the inactive from of factor VIII are continuously circulated in inactive forms in the blood stream, until a damaged blood vessels occurs. At the injured site they are converted to active from called coagulating factor VIIIa,,by separation from Willebrand factor, react with coagulating factor IX to initiate clotting. The factor VIII is a co- factor for factor IX, and together with calcium ion,phospholipids converts t to factor IXa,
The latter converts prothrombin to thrombin. The thrombin convert fibrinogen to fibrin, which form a stable fibrin clot at the injured site to arrest bleeding.<u> </u><u>The cascade of events is lacking individual without factor VIII. Therefore bleeding occurs.
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Consequently, symptoms of hemophilia can be relieved by a functional factor VIII, because it will prevent the deficiency in factor VIII in the affected individuals by acting as co-factors for the conversion of factor IX to IXa, to initiate clotting.
C. Single events or several causes working together to produce extinction in a short period is referred to as mass extinction.
<span>Example:
</span><span>Destruction of forests, mountains, and bodies of water to satisfy human needs result to the imbalance of nature which also give rise to global warming.
</span><span>Animals and plants have lost their habitats. They lost their means to live. They lost their food sources and they suffer and die because of humans irresponsible acts.</span>
Answer:
the genotype of the parent guinea pigs are as follows:
RrBb (rough, black) and Rrbb (rough, white)
Explanation:
Since the phenotype of smooth coat is present in the offspring, the parent genotype cannot have RR alleles for their coat. Similarly, the rough, black parent cannot have BB alleles in its genotype because that will not yield a white coat color in its offspring.
After making the dihybdrid cross, the probability obtained for each phenotype is given below:
rough black: 
rough white:
smooth black: 
smooth white:
Hope that answers the question, have a great day!
