Answer is B. You can tell that from the table.
Answer:
E. glucosuria (glucose in the urine)
Explanation:
Generally, glycosuria occurs in patients with kidney changes due to diseases such as Wilson's disease or cystinosis, can also be a hereditary problem, but is not expected in patients with kidney damage caused by prolonged lithium use.
Normally, the kidneys filter the blood, eliminating all substances that are not necessary for the body to function, while glucose is reabsorbed in the blood because of its importance in energy production, but people with renal glycosuria do not reabsorb glucose. , which causes it to be eliminated in the urine, occurring glucosuria.
The answer for this question is D.Mechanical weathering of very fing grained blue grey clays
Hope i helped please brainliest answer if possible
Answer:
The correct answer is a. absent spinal reflexes below the level of injury.
Explanation:
Spinal shock strictly refers to the neurological condition that occurs immediately after a spinal cord injury, in which the loss of not only motor and sensory functions occurs, but also the abolition of all reflexes below the injury (reflexes of muscular or myotatic stretching and cutaneous reflexes). There is also flaccidity, loss of reflexes. It is characterized by hypotension associated with cervical or upper thoracic spinal injuries. This characteristic shock results from the lesion of the descending sympathetic pathway in the spinal cord, producing a loss of vasomotor tone and sympathetic innervation of the heart. This causes vasodilation of the affected area with accumulation of blood and a decrease in venous return to the heart as well as cardiac output.
Answer:
4 percent (4%)
Explanation:
A single crossover occurs between two (non-sister) chromatids belonging to homologous chromosomes. In this case, 16 percent of the meioses have a single crossover, thereby it will produce 8 percent of the chromosomes with the original (parental) combination in the progeny and the remaining 8 percent should be recombinants. From this result, it is reasonable to conclude that half of these recombinants should be 'Br' (and the other remaining 4 percent should be recombinants 'bR'), and therefore the answer is 4 percent (4%).
