Can someone help me calculate these values? The answers are 49, 38 and 205 mM re

Question

Can someone help me calculate these values?

The answers are 49, 38 and 205 mM respectively.

You discover a new species of tropical jelly fish that lives near a volcanic vent that heats the water to 37degreeC. You run tests on a sample of the local sea water to determine its ionic composition. You then record currents from jelly fish neurons that are bathed in a solution that mimics the Na+, Cl- or K+ composition of sea water. You measure the equilibrium potential for each ion. The results of your experiments arc compiled the table below. Calculate the approximate intracellular concentrations for Na+, Cl-, and K+ (to the nearst mM).

Explanation / Answer

sodiom ion

extracellular concentration = 470mM

volts across the memberane is 60mM

formula is the E q of na = RT/Zf ln [ na] in /{na } out ,

from this equation , Na outside the cell = RT/Zf *[ na] in/ Eq   where R is the universal gas constant 8.314 , t is temperature , z = charge of ion f= faradyas constant , 96,485 by substituting the valuves we get

8.314*273/1*96485 [470/60] = 1.171 should be converted into milli volts

we will get for potassium also because possatium is also a monovalent atom , z remains the same

8.314*273/1.*96485 ln [10/80] = 2.07 ion concentration

for chloride ions ;

the valancy changes to 7 , and extracellular concentration is 540 e ion is 70

= 8.314*273/96485*7ln540/70= 8.314*273/96485*7 =0.003 ln 7.7=2.041 chloride ions

all these values should we convereted to millivolts

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.