Answer:
D) elastic cartilage
Explanation:
Cartilage is a connective tissue which is fibrous in composition which provides its physical properties like the softness, elasticity and strength.
There are three types of cartilages- the elastic, hyaline and the fibrocartilage. Among these three the cartilage which forms the external ear is the elastic cartilage. The elastic cartilage is composed of the elastin protein, type II collagen and the elastic fibres. The presence of the elastin proteins and fibres allows the ear to be an elastic organ.
Thus, Option-D is the correct answer.
Answer:
You need to compare the location of bands for X and X's child. X is the mother, so the child will have half of all its bands from its mother, and the other half must be from the father. Some bands that X and X's child have in common are at around 185bp and 130bp (it is difficult for me to determine the exact position). Then look at the bands which X and X's child do not have in common. Those bands must have come from the father. So now you compare the remaining bands to all the bands of Megabucks and see if they match up. You can see a band at around 60bp that the child has. The mother did not have this band, so it must have come from the father. Megabucks does not have this band, so he is not X's father.
Explanation:
I hope this has helped you a little. The main thing to know is that a child's bands come from their mother and father, so if half the bands match up to the mum, the other half have to match up to some of the father's bands. But a child will never have the same set of bands as one of their parents - it will be a mix of both parents' bands.
Answer:
a. preventing the cross-linkage of NAM subunits
Explanation:
The bacterial cell wall is made of peptidoglycan which is a polymer of alternating N-acetylglucosamine (NAG) and N-acetylmuramic acid (NAM) residues. The NAM residues are crosslinked to impart additional strength to the cell wall.
Most of the antibiotics that target the cell wall synthesis in bacteria inhibit the enzyme of transpeptidation which in turn cross-links the NAM residues of the polysaccharide chains of the bacterial cell wall. Examples of these antibiotics include penicillins and cephalosporins.
A. X cells I animal cells and Y cells are plant cells
The width of the cloud is : 384 m
<u>Given data :</u>
Cloud altitude = 2200 m
Temperature = 10°
<h3>
</h3><h3>Determine how wide the cloud is </h3>
applying the equation below
I = d
where : I = linear size , d = 2200 m , 
therefore :
I = 2200 ( 10 / 57.3 )
= 383.94 m
≈ 384 m
Hence we can conclude that the width of the cloud is : 384 m
Learn more about linear size : brainly.com/question/1048150
