All categories

4 years ago

True or false molecules in a mixture are all alike.

Biology

2 answers:

Over [174]4 years ago

6 0

Answer:

False

Explanation:

notsponge [240]4 years ago

4 0

Answer: True

Explanation:

You might be interested in

Tropical rainforests generally have a higher density of organisms compared with other biomes. Based on this information, which o

tester [92]

The answer is I think d

5 0

2 years ago

In a condition known as sunstroke, the victim appears flushed, the skin is warm and dry, and the body temperature rises dramatic

katrin [286]

Answer:

In case of sunstroke the body is unable to sweat because the body is dehydrated.

The temperature of the body won't increase further because there can be no more evaporation in body as it cannot sweat more from the skin surface.

The skin dissipates the heat and cools the body preventing the overheating of the body.

The skin appears to be red in color because the blood is dilated and dissipiated into the whole body for cooling.

8 0

3 years ago

Suppose a species of bird called the red‑crested warbler has a plumage length that is controlled by a single gene. The P????mPlm

andreev551 [17]

Answer:

Check the explanation

Explanation:

North American:

112 birds total => 224 alleles

112 - 55 birds = 57 short plume birds.

$q^2$ = (2*57) / 224 = 0.5089

Freq(short plume allele) = q = 0.7134

Freq(long plume allele) = p = 1 - q = 0.2866

From this North American population we get

0.2866 * 224 = 64 long plume alleles

0.7134 * 224 = 160 short plume alleles

South American:

452 birds total => 904 alleles

452 - 75 birds = 377 short plume birds.

$q^2$ = (2*377) / 904 = 0.8341

Freq(short plume allele) = q = 0.9133

Freq(long plume allele) = p = 1 - q = 0.0867

From this South American population we get

0.0867 * 904 = 78 long plume alleles

0.9133 * 904 = 826 short plume alleles

Blended population has

64 + 78 = 142 long plume alleles

160 + 826 = 986 short plume alleles

142 + 986 = 1128 total alleles

p = 142/1128 = 0.1259

q = 986/1128 = 0.8741

All right. 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.