Question

patient has a mutation that removes the negatively charged region on the sodiu lucose cotransporter located on proximal tubule cells. as a result, na+ and glucose cannot bind properly to the transport protein. which explanation best accounts for the patient developing glucose in the urine (glucosuria) despite normal blood glucose levels? without the transporters negatively charged region, the positively charged na+ cannot interact with the protein, so neither na+ nor polar glucose can enter the cell; both remain in the filtrate, and glucose is lost in the urine. glucose freely diffuses across the lipid membrane without transport proteins, so the mutation has no effect on reabsorption. the mutation causes na+ to become nonpolar, allowing it to diffuse directly through the membrane and preventing glucose from following. the transporter still moves glucose, but fever in the kidney denatures the glucose molecule itself, making it unable to be reabsorbed.

Explanation:

• Analyze Option 1: The sodium - glucose cotransporter in proximal tubule cells requires the negative region for \(Na^+\) (positive) to bind. If the negative region is removed, \(Na^+\) can't bind, and since glucose transport is coupled to \(Na^+\) transport, glucose also can't enter the cell. Both stay in the filtrate, leading to glucosuria. This matches the mutation's effect.
• Analyze Option 2: Glucose is polar and can't freely diffuse across the lipid - bilayer membrane; it needs a transporter. So this option is incorrect.
• Analyze Option 3: A mutation in the transporter doesn't change the chemical nature of \(Na^+\) (it remains charged and can't diffuse through the non - polar membrane directly). So this option is wrong.
• Analyze Option 4: The mutation affects the transporter's ability to bind \(Na^+\) and glucose, so the transporter can't move glucose properly. Also, there's no indication of fever denaturing glucose. This option is incorrect.

Answer:

The first option (Without the transporter's negatively charged region, the positively charged \(Na^+\) cannot interact with the protein, so neither \(Na^+\) nor polar glucose can enter the cell; both remain in the filtrate, and glucose is lost in the urine.)