Answer:
HPRT
Explanation:
HPRT catalyzes the salvage reactions of hypoxanthine and guanine with PRPP to form IMP and GMP
The formation of GMP from IMP requires oxidation at C-2 of the purine ring, followed by a glutamine-dependent amidotransferase reaction that replaces the oxygen on C-2 with an amino group to yield 2-amino,6-oxy purine nucleoside monophosphate, or as this compound is commonly known, guanosine monophosphate.
There would be no more because everything had decrease to 0
It would end up being 14.2
Answer:
108 kPa
Step-by-step explanation:
To solve this problem, we can use the <em>Combined Gas Laws</em>:
p₁V₁/T₁ = p₂V₂/T₂ Multiply each side by T₁
p₁V₁ = p₂V₂ × T₁/T₂ Divide each side by V₁
p₁ = p₂ × V₂/V₁ × T₁/T₂
Data:
p₁ = ?; V₁ = 34.3 L; T₁ = 31.5 °C
p₂ = 122.2 kPa; V₂ = 29.2 L; T₂ = 21.0 °C
Calculations:
(a) Convert temperatures to <em>kelvins
</em>
T₁ = (31.5 + 273.15) K = 304.65 K
T₂ = (21.0 + 273.15) K = 294.15 K
(b) Calculate the <em>pressure
</em>
p₁ = 122.2 kPa × (29.2/34.3) × (304.65/294.15)
= 122.2 kPa × 0.8542 × 1.0357
= 108 kPa
Answer:
<u>1</u><u>.</u><u> </u><u>0</u><u>0</u><u>0</u><u>.</u><u>2</u>
Explanation:
<u> </u><u> </u><u>a</u><u>l</u><u>u</u><u>m</u><u>i</u><u>n</u><u>u</u><u>m</u><u> </u><u>o</u><u>x</u><u>i</u><u>d</u><u>e</u><u> </u><u>m</u><u>a</u><u>d</u><u>e</u><u> </u><u>o</u><u>f</u><u> </u><u>m</u><u>o</u><u>l</u><u>e</u><u>s</u><u> </u><u>o</u><u>r</u><u> </u><u>a</u><u>t</u><u>o</u><u>m</u><u>s</u><u> </u><u>l</u><u>e</u><u>s</u><u>s</u><u> </u><u>t</u><u>h</u><u>e</u><u>n</u><u> </u><u>o</u><u>t</u><u>h</u><u>e</u><u>r</u><u> </u><u>e</u><u>l</u><u>e</u><u>m</u><u>e</u><u>n</u><u>t</u><u>s</u><u> </u><u>n</u><u>e</u><u>e</u><u>d</u><u> </u><u>t</u><u>o</u><u> </u><u>d</u><u>e</u><u>c</u><u>r</u><u>e</u><u>a</u><u>s</u><u>e</u>