Answer: 48.93 mL of sprite
Volume of blood in 7 Kg human = 5 L
Percentage of plasma in blood = 55%
Volume of plasma in 5 L blood = (1 L = 10 dL)
Concentration of glucose in plasma = 80 mg/dL
Amount of glucose present in 27.5 dL : 80 mg /dL × 27.5 dL= 2200 mg
Let the volume of sprite with 2200 mg glucose be x
Concentration of glucose in sprite = 44.96 mg/mL
Explanation:
B) They should test their hypothesis before coming to that conclusion.
Answer: The following is a <u>postzygotic isolating barrier </u>:
- The hybrid offspring of two species of crocodiles can produce normal gametes but cannot obtain a mate.
Explanation:
<em><u>Reproductive isolation </u></em>occurs when <em>barriers prevent</em> two populations from interbreeding , <u>keeping their gene pools separate.</u>
We can find the Prezygotic isolation that occurs<u> before fertilization </u>can happen.
And the Postzygotic insolation barriers are the reproductive insulation processes that<u> act after the mating</u>. They are all that concern the viability of the individuals produced.
Prezygotic isolation barriers can be temporal, behavioural, geographic , ecological or mechanical; whereas postzygotic isolation barriers include the inviability, infertility or breakdown of hybrid organisms
