Answer: Phototherapy
Explanation:
As the infant is only 2 weeks old, the most appropriate technique used is the Phototherapy. As the condition is not that severe it can be cured by phototherapy.
Photo therapy is usually given to the children suffering from jaundice when they are born. This is a special type of light that is not sunlight.
It is used in case of high bilirubin level.This light lowers the level of bilirubin levels in the baby's blood.
This process is basically photo oxidation in which the oxygen gets added to the bilirubin which helps easy dissolution of the bilirubin in the water.
This lowers down its level in the blood.
Answer:
Frameshift mutation
Explanation:
The first nucleotide is removed or deleted but it's not up to three so it can't code for any amino acid.
The coding shifts from one point to the next.
Answer:
That the adult has more cells than the baby?
Explanation:
Cells are the basic unit of structure and function in all living organisms. Which describes the GREATEST difference between the cells of a baby gorilla and the cells of an adult gorilla? The adult has more cells than the baby.
Sorry if its wrong
Answer:
a) Astrocytes
Explanation:
The astrocytes are components of the central nervous system that provide the necessary supports that the brain needs to perform vital tasks such as learning and other daily activities. They are typical cell types in the central nervous system that aid blood flow in the brain as well as the migration of neurons.
Answer:
option C: B = 0.48; b = 0.52
Explanation:
Of the 50 mice, 12 were homozygous black (BB), 24 were heterozygous black (Bb), and 14 were brown (bb).
BB Bb bb Total
12 + 24 + 14 = 50
the formular for alleleic frequency (p) is (BB+ Bb/2) / total,
q = (bb + Bb/2) / total
Allelic frequecy of B (p) is (12 + 24/2) /50 = 12 + 12 = 24/50 = 0.48
Allelic frequency of b (q) is (14 + 24/2) / 50= 14 + 12 = 26/50 = 0.52
check p + q = 1; 0.48 + 0.52 = 1
