He finds that those who took vitamins are, on average, 2 cm taller. What is a major limitation of his experiment

What provides instructions to a eukaryotic cell about living and growing?

Damm [24]

The answer to this is DNA.

3 0

3 years ago

Distinguish between directional, stabilizing, and disruptive selection. Give examples.

Pepsi [2]

Distribution of phenotypes within a population may be impacted by different types of natural selection. These types of selection include;
Directional selection involves a change in the environment shifts of phenotypes observed. Example is the giraffe neck-length, where the length got longer and longer over time to reach the leaves at the top of trees.
Stabilizing selection , this is a selection where an average phenotype is favored. example is number of eggs laid by birds such that too many eggs will be a waste of energy because some chicks would starve.
Disruptive selection is a selection where individuals with an average value of a trait are affected most while those with values below average are affected least. Example small individuals are capable of hiding from predators while big individuals can not fit in the hiding place but instead will fight with predators.

8 0

3 years ago

One character in peas that Mendel studied was yellow versus green seeds.

weqwewe [10]

Answer:

Mendel was an Austrian monk whose researches laid the foundation of genetics. The experiments conducted by Mendel led him to the foundation of two laws which are named as the law of segregation and law of independent assortment.

According to the law of segregation, the two alleles of a gene segregate during the time of gamete formation and there are 50-50% chances of each of the alleles to be received by the gametes. Hence, there are 50% chances for Y gametes to be produced and 50% chances for y gametes to be produced.

3 0

3 years ago

What are the the similarities and differences of ovary and ovulation?

Wewaii [24]

Answer:

Ovary is an organ of the female reproductive system which is involved in the production and releasing of an ovum or egg. From the ovaries, the egg is released into the Fallopian tube in the uterus.

Ovulation can be described as the process in which an egg is released from the ovary. It may or may not be fertilized after being released.

Similarities of the ovary and ovulation are that both refer to releasing of the egg.

4 0

3 years ago

Most black bears (Ursus americanus) are black or brown in color. However, occasional white bears of this species appear in some

Sphinxa [80]

Answer and Explanation:

The number of observed individuals:

AA 42

AG 24

GG 21

Total number of individuals, N= 87 = 42 + 24 + 21

Allelic frequencies:

f(p) = (2 x AA + AG)/ 2 x N

f (p)= (2 x 42 + 24) /2 x 87

f (p) = (84 + 24) / 174

f (p)= 108 / 174

f (p) = 0.62

f (q) = (2 x GG + AG)/2 x N

f (q) = (2 x 21 + 24 )/2 x 87

f (q) = (42 + 24)/ 174

f (q) = 66/174

f (q) = 0.38

p + q = 1

0.62 + 0.38 = 1

The expected genotypic frequency:

F (AA)= 0.62 ² = 0.3844

F (AG) = 2 x A x G = 2 x 0.62 x 0.38 = 0.4712

F (GG) = 0.38 ² = 0.1444

AA + AG + GG = 0.3844 + 0.4712 + 0.1444 = 1

The number of expected individuals:

AA= (0.62)² x 87 = 0.3844 x 87 = 33.44

AG= (0.4712) x 87 = 40.99

GG= (0.38)² x 87 = 12.563

Total number of expected individuals = 33.44 + 40.99 + 12.563 = 87

Chi square= sum (O-E)²/E

AA= (O-E)² /E

AA=(42 - 33.44) ² / 33.44

AA= 2.2

AB= (O-E)² /E

AB= (24 - 40.99)²/ 40.99

AB=7.04

BB=(O-E)² /E

BB= (21-12.563)²/12.563

BB= 5.66

Chi square= sum ((O-E)²/E) = 2.2 + 7.04 + 5.66 = 14.9

Degrees of freedom = genotypes - alleles = 3 - 1 = 2

p value less than 0.05

There is enough evidence to reject the nule hypothesis. The genotype frequencies are not in equilibrium.

5 0

3 years ago