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ddd [48]
3 years ago
12

__________ is indigestible by animals because they lack __________, which is a __________.

Biology
1 answer:
alekssr [168]3 years ago
3 0

Cellulose is indigestible by animals because they lack the enzyme Cellulase.

Cellulose is the polysaccharide which is the main constituent of plant cell wall. It consists of chains of glucose monomers. Animals like human beings lack the enzyme Cellulase which is necessary for the digestion of cellulose.  

Hay and grasses are rich in cellulose and therefore, both are indigestible by human beings. Animals like cows and horses also do not possess the enzyme cellulase, but they harbor microbes which help in the digestion of cellulose.

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What is the difference between a virus and bacteria in causing infection?
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As you might think, bacterial infections are caused by bacteria, and viral infections are caused by viruses. Perhaps the most important distinction between bacteria and viruses is that antibiotic drugs usually kill bacteria, but they aren't effective against viruses.

Explanation:

As you might think, bacterial infections are caused by bacteria, and viral infections are caused by viruses. Perhaps the most important distinction between bacteria and viruses is that antibiotic drugs usually kill bacteria, but they aren't effective against viruses.

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A person driving a car applies the brakes. This produces friction, which stops the car. Into which type of energy is the mechani
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What are the correct labels for this diagram of glucose catabolism
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Ted runs from the biology laboratory straight to his therapist's office. He is sweating and fear is etched on his face. He asks
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5 0
3 years ago
The following F2 phenotypic data was obtained from a Drosophila testcross using an F1 offspring. Assume red eye and brown body a
dmitriy555 [2]

Answer:

The genotype of the F1 was wy+/w+y.

Explanation:

One of the given options has a typo: the red eye-brown body offspring count should be 56 instead of 561.

<u>We have two genes with two alleles each:</u>

Red eyes (w+) is dominant over white eyes (w).

Brown body (y+) is dominant over yellow body (y).

The phenotypes of the F2 tesulting from a test cross (F1 x wy/wy) are:

  • wy+/ey (white-eye, brown body): 670
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  • wy/wy (white-eye, yellow body): 38
  • w+y+/wy (red-eye, brown body 56

If the genes w and y are linked, two phenotypes in the F2 will be much more abundant than the other two. Recombination during meiosis is a rare event, so the most abundant phenotypes are the parentals (the ones present in the F1 parent).

Every individual in the offpsring has a <em>wy</em> chromosome, as this was the gamete inherited from the test cross individual.

In this case, the most abundant gametes are wy+ and w+y, so the genotype of the F1 was wy+/w+y.

Notice how when recombination occurs in the F1 parent, the recombinant gametes appear: wy and w+y+, which are the less abundant in the F2 progeny.

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