The Krebs Cycle can best use acetyl coA to make ATP and pyruvate breaks down into acetyl CoA and CO2 this conversion do to the pyruvate molecules.
The required details for pyruvate molecules in given paragraph
Pyruvate is a flexible organic molecule that includes 3 carbon atoms and useful groups - a carboxylate and a ketone group. Pyruvate is concerned in some of key biochemical processes, together with gluconeogenesis, that's the synthesis of glucose, in addition to the synthesis of different key biochemicals. Two molecules of pyruvate are transformed into molecules of acetyl CoA begin text, C, o, A, quit text.
Two carbons are launched as carbon dioxide—out of the six firstly found in glucose. The number one characteristic of pyruvate is to function the transporter of carbon atoms into the mitochondrion for entire oxidation into carbon dioxide.
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The answer is <span>b. wolf.
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Organisms from the higher trophic levels consume organisms from the lower trophic level. In this process, energy is lost as metabolic heat. Thus, primary producers, such as sedge, contain the greatest amount of energy originally from the sunlight. The next trophic level belongs to primary consumers, such as arctic hare, that consume primary producers resulting in less energy. Similarly, arctic fox eats arctic hare, and energy is lost again. The highest trophic level is tertiary consumers such as wolf, therefore, the wolf contains the least amount of energy.
Calculate the average metabolic rate of a 65-kg person who sleeps 8.0h, sits at a desk 6.0h, engages in light activity 6.0h, watches TV 2.0h, plays tennis 1.5 h, and runs 0.50h daily, If the metabolic rate for sleeping is 70 J/s, for sitting is 115 J/s, for engage in light activity is 230 J/s, for watching TV is 115 J/s, for playing tennis is 460J/s and for running is 1150 J/s.
Answer:
171.8 W
Explanation:
The Average metabolic rate can be deduced by multiplying each entity performed at a some particular hour of time with their corresponding energy, all divided by 24 hours.
Average metabolic rate = 
= 171.8 W
1) B- brown; b- red hair
Having brown hair is a dominant trait while red hair is recessive.
If the parents had a red-haired child it means they were carriers of the red hair trait, so their genotype is
Bb
2)Getting their genotypes together would result in the possibilities that you see at the image below.
1/4 = 25% BB - brown
2/4 = 50% Bb brown but carrier
1/4 = 25% bb red hair
Answer: 25%
3) The genotype of the child, as i said before would be
bb. By having both alleles being recessive the child will show red-hair. If it was any other genotype( BB or Bb), the child would have brown hair, like their parents.