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Troyanec [42]
3 years ago
6

Polysaccharide structure can be varied by differences in ________.

Biology
1 answer:
Scilla [17]3 years ago
7 0

Answer:

monomers of MONOSACCHARIDES

Explanation:

Polysaccharides are large molecules formed from chains of POLYMERS linked together by glyosidic bonds. <u>MONOMERS are small sub units that formed polymers, they are therefore the building block of a polysaccharides.  The monomers of polysaccharides are called monosaccharid</u>es (1 sugar molecule.) when two of these are joined together they formed disaccharides (two sugars.)

Polysaccharides are fromed by joining  together condensation, (loss of water molecules,)  of mutiple monosaccharides units and the reversal of this to add water molecules to sepate them to monosaccharies is  sugar Hydrolysis.

Example of polysaccharides are starch, glycogen cellulose

Example of monosaccharides are glucose, galactose.

Disaccharides are common table sugar, sucrose, maltose, lactose  

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Nucleotide!

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Suppose a species of bird called the red-crested warbler has a plumage length that is controlled by a single gene. The Plm allel
AleksAgata [21]

We are going to have different allele frequencies for the two populations

North American:

72 birds total => 144 alleles

72 - 55 birds = 17 short plume birds.

q^2 = (2*17) / 114 = 0.236

Freq(short plume allele) = q = 0.486

Freq(long plume allele) = p = 1 - q = 0.514

From this North American population we get

0.486 * 114 = 55 long plume alleles

0.514 * 114 = 59 short plume alleles

South American:

252 birds total => 504 alleles

252 - 75 birds = 177 short plume birds.

q^2 = (2*177) / 504 = 0.702

Freq(short plume allele) = q = 0.838

Freq(long plume allele) = p = 1 - q = 0.162

From this South American population we get

0.162 * 504 = 82 long plume alleles

0.838 * 504 = 422 short plume alleles

Blended population has

55 + 82 = 137 long plume alleles

59 + 422 = 481 short plume alleles

137 + 481 = 618 total alleles

p = 137/618 = 0.222

q = 481/618 = 0.778

The new population has the p and q from the blended population. The new population has 1000 individuals. The portion of long plumed will be homozygote long plumage + heterozygote, which is p^2 + 2pq, and you multiply that by the population size 1000 to get the final answer.

population size * (p^2 + 2pq) = 1000 * (0.222^2 + 2*0.222*0.778) = 395 birds (answer)

6 0
3 years ago
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