The _____________ cells in the _________________ of a nephron can cause and increase or decrease in body pH. A. intercalated; co
llecting duct B. principal; proximal convoluted tubule C. macula densa; distal convoluted tubule D. glomerular; nephron loop
2 answers:
Answer:
The intercalated cells in the distal convoluted tubule of a nephron can cause and increase or decrease in body pH.
Explanation:
The renal collecting duct is the nephron segment where the final urine content of acid equivalents and inorganic ions are determined.
Two types of cells regulate the acid-base and volume homeostasis.
Intercalated cells, which express acid-base transporters and vacuolar H+-ATPase, maintain an apropriate acid-base balance and principal cells that express the epithelial sodium channel ENaC and aquaporin 2, regulate electrolyte reabsorption.
Filtered bicarbonate is predominantly reabsorbed from the proximal convoluted tubule about 80% and the remaining bicarbonate is reabsorbed from the thick ascending limb of the loop of Henle, the distal nephron, specifically the distal convoluted tubule & the connecting tubule, and finally the cortical and both the outer and inner medullary collecting duct.
Intercalated cells are enriched in mitochondria, and express proteins involved in transport of proton equivalents such as vacuolar H+-ATPase, carbonic anhydrase II and bicarbonate transporters. There are two types of intercalated cells Type-A and Type-B.
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Answer:
50%
Explanation:
This question involves a gene coding for eye color in humans. The allele for brown eyes (B) are dominant to blue eyes (b). This means that blue-eyed individual will possess genotype: bb.
According to this question, a blue-eyed man (bb) has children with a brown-eyed woman (B_) whose mother has blue eyes (bb). Since the mother of the brown-eyed woman has a blue eye, this means that the woman will be heterozygous for brown eye (Bb).
Hence, the parents in this question will cross as follows: blue eyed man (bb) × brown-eyed woman (Bb). The offspring/children will have the following genotypes (see attached punnet square); Bb, Bb, bb, bb.
Based on the question, 2/4 = 1/2 of the children will be heterozygous for the eye color trait. That is, ½ × 100 = 50%.
