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3 years ago

A ___________ is the smallest unit of life?

Biology

2 answers:

ohaa [14]3 years ago

6 0

Answer:

Cell

Explanation:

A cell is the smallest unit of a living thing.

Alisiya [41]3 years ago

6 0

Answer:

A cell is the smallest unit of life.

Explanation:

Cell is the smallest and basic unit of all living organisms. Some organisms are unicellular i. e. made of one cell and some organisms are multicellular i. e. made of more than one cell. In Unicellular organisms, single cell perform all functions and activities of the cell such as excretion, reproduction and digestion etc, while in multicellular organisms, there are specific cells for each activity.

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In the Cori cycle, when glucose is degraded by glycolysis to lactate in muscle, the lactate is excreted into the blood and retur

Kay [80]

Answer:

Explanation:

Glycolysis is the breakdown of glucose while gluconeogenesis is the formation of glucose. Both pathways are almost the same except that the non-reversible reactions in glycolysis requires another enzyme during gluconeogenesis.

Glycolysis is a 10-step multi-enzyme pathway, in which three of these 10 steps are non-reversible. It is these three steps that basically differentiates the glycolytic pathway from gluconeogenesis. These three steps are bypassed by different enzymes during gluconeogenesis

1. The correct option here is c

Glyceraldehyde 3-phosphate dehydrogenase converts glyceraldehyde 3-phosphate to 1,3-bisphosphoglycerate in glycolysis. This reaction is however a reversible reaction, hence this enzyme also converts 1,3-bisphosphoglycerate to glyceraldehyde 3-phosphate during gluconeogenesis.

2. The correct option here is b

Glucose 6-phosphatase converts glucose 6-phosphate to glucose during gluconeogenesis

3. The correct option here is d

Alcohol dehydrogenase converts ethanol into acetaldehyde. This reaction is not present in both glycolysis and gluconeogenesis and hence the enzyme is not involved/present in these pathways.

4. The correct option here is b

Phosphoenol pyruvate carboxykinase converts oxaloacetate to phosphoenol pyruvate during gluconeogenesis

5. The correct option here is a

Phosphofructokinase-1 converts fructose 6-phosphate to fructose 1,6-bisphosphate during glycolysis

6. The correct here is c

Phosphoglycerate mutase converts 3-phosphoglycerate to 2-phosphoglycerate during glycolysis. This reaction is however a reversible reaction and hence this same enzyme converts 2-phosphoglycerate to 3-phosphoglycerate during gluconeogenesis.

7. The correct option here is a

Hexokinase converts glucose to glucose 6-phosphate during glycolysis.

8. The correct option here is d

Pyruvate dehydrogenase converts pyruvate to acetyl-coA. This reaction is one of the fates of pyruvate. This reaction is however not present in both glycolysis and gluconeogenesis and hence the enzyme is not involved/present in both also.

7 0

3 years ago

What is found at the center of a syncline?

Marianna [84]

(A) the softest layer of rock?

8 0

3 years ago

The LMand LNalleles at the MN blood-group locus exhibit codominance. Give the expected genotypes and phenotypes and their ratios

Dmitry [639]

Answer:

Please find the expected genotypes and phenotypes of the progenies of each cross below.

Explanation:

In the MN blood-group locus of the gene, alleles LM and LN exhibit codominance i.e one allele is not dominant or recessive to the other, hence, they are both expressed when they occur in a heterozygous state (MN).

Considering the following crosses (find the punnet square attached);

a)LMLM x LMLN - The progeny are LMLM and LMLN in the genotypic ratio 1:1. Phenotypic ratio is Blood type M (1) : blood type MN (1)

b) LNLN x LNLN - The progeny are all LNLN offsprings with a phenotypic and genotypic ratio 4:0. All offsprings will have a blood type N (4)

c) LMLN x LMLN - The progenies are LMLM, LMLN and LNLN in the genotypic ratio 1:2:1 respectively. The phenotypic ratio is Blood type M (1) : L

Blood type MN (2) : Blood type N (1)

d) LMLN x LNLN - The progeny are LMLN and LNLN with genotypic ratio 1:1 and phenotypic ratio blood type MN (1) : blood type N (1)

e) LMLM x LNLN - The progeny are all LMLN offsprings with penotypic ratio blood type MN (4)

6 0

3 years ago

Echinoderms _____. _____ Select one: a. have an exoskeleton of hard calcareous plates b. often use tube feet to move around in t

atroni [7]

Answer:

B. often use tube feet to move around in their environment

Explanation:

Tube feet are tiny tubular projections of echinoderms on the underside (oral side). They are a member of the echinoderm water vascular system.

Tubular feet are used for feeding, breathing and shifting. They are arranged around the sides, in grooves. They work by hydraulic pressure. They are used to transfer food in the centre to the oral mouth, and may stick to surfaces. Tube feet allow certain animals to stick and travel slowly to the ocean floor. for example starfish uses tube feet for the above functions.

Hence, the correct option is B.

4 0

3 years ago

Helpppp plssss !!! if u can’t see the pic don’t bother answering !!! hurryyy

FinnZ [79.3K]

Answer:b

Explanation:

7 0

3 years ago

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