Please, I need HELP!!!!!

Why does the lipid bilayer self assemble?

Oduvanchick [21]

Hkbk hkcb Jackie jkhvm hlcfb give.

8 0

3 years ago

How do enzymes act as catalysts to help maintain homeostasis for the cell.

Alenkasestr [34]

Enzymes are the things in our bodies that facilitates reactions. They break down bonds in carbohydrates, lipids and proteins. When the food is being broken down, it is easier for the body to use. When the body is conducting cellular respiration, it uses monosaccharides (singular carbohydrate), to form ATP which is useable energy for the body. The body uses ATP everywhere for all bodily functions, which maintains homeostasis. <span />

3 0

4 years ago

The monomers of Proteins are

Maslowich

Answer:

The answer of the question is c

7 0

3 years ago

Read 2 more answers

Imagine that you attempted to recreate Mendel's work with garden peas. You began by crossing true breeding violet-flowered, tall

salantis [7]

Answer:

(a) 90

(b) 30

(c) 30

(d) 10

F2 data for stem length (tall:dwarf) is consistent with Mendel's law of segregation.

F2 data for stem length (tall:dwarf) and flower color (violet:white) is consistent with Mendel's law of independent assortment

Explanation:

The F2 phenotypes are supposed to be in 9:3:3:1 according to Mendelian law. The total number of F2 progeny is: 80 + 36 + 39 + 5 = 160

(a) Hence, the expected number of tall, violet plants will be:

9/16 x 160 = 90

(b) Expected number of tall, white plants will be:

3/16 x 160 = 30

(c) Expected number of dwarf, violet plants will be:

3/16 x 160 = 30

(d) Expected number of dwarf, white plants will be:

1/16 x 160 = 10.

Total number of tall F2 plants = 80 + 36 = 116

Total number of dwarf F2 plants = 39 + 5 = 44

Expected ratio of tall:dwarf plants according to Mendel = 3:1

Expected number of tall plants = 3/4 x 160 = 120

Expected number of dwarf plants = 1/4 x 160 = 40

Chi square $X^2$

= $\frac{(observed frequency - expected frequency)^2}{expected frequency}$

$X^2$ for tall = $\frac{(116 - 120)^2}{120}$

= 0.1333

$X^2$ for dwarf = $\frac{(44 - 40)^2}{40}$

= 0.40

Total $X^2$ = 0.1333 + 0.4

0.5333

Degree of freedom = n - 1

2 - 1 = 1

Tabulated $X^2$ (α = 0.05) = 3.841

Since the tabulated $X^2$ is more than the calculated $X^2$ , the hypothesis that the F2 data for stem length (tall:dwarf) is consistent with Mendel's law of segregation is accepted.

Observed Expected $X^2$

tall, violet flowers 80 90 $\frac{(80-90)^2}{90} = 1.11$

tall, white flowers 36 30 $\frac{(36-30)^2}{30} = 1.2$

dwarf, violet flowers 39 30 $\frac{(39-30)^2}{30} = 2.7$

dwarf, white flowers 5 10 $\frac{(5-10)^2}{10} = 2.5$

Total 160 160 7.51

Degree of freedom = 4 - 1 = 3

Tabulated $X^2$ (α = 0.05) = 7.815

The tabulated $X^2$ value is more than the calculated $X^2$ value. Hence, F2 data for stem length (tall:dwarf) and flower color (violet:white) is consistent with Mendel's law of independent assortment.

3 0

3 years ago

75 soruluk bi biyoloji testi lütfen yardım et

podryga [215]

Answer:

Wut-......................

8 0

3 years ago