the life is very beautiful thing to gift by God. a some are happy. b. some are sad c. some are smile. choose any one

The benefit of measuring the initial rate of a reaction, V0, is that early in the course of the reaction, A) Km is the substrate

aleksandr82 [10.1K]

Answer:

Rate of product formation is linear and [S] has not been lowered significantly.

Explanation:

The rate of enzyme-catalyzed reactions is affected by several factors, the contraction of substrates [S] is one of them. The substrate concentration keeps on changing as the reaction proceeds. This is why the reaction rate is measured at the initial stages of reactions when the substrate concentration [S] is much greater than the concentration of the enzyme. It is called the initial rate or initial velocity.

Under the conditions of higher substrate concentration and relatively much lower enzyme concentrations, only a few molecules of substrates are being converted into product. At a relatively higher substrate concentration, the rate of product formation increases linearly.

4 0

3 years ago

20 POINTS!!

Naily [24]

Answer:

Because Intercalary meristems are found in the the stem part of the sugarcane in whole length.

Explanation:

As we have studied about meristem regions, that these are the regions found at the nodal areas which are able of cell division and growth. Intercalary meristem are present in the nodes of sugarcane stem which lead to the rapid increase in elongation of stem.
Intercalary meristem are located at the base of leaves and nodes that leads to the increase in length.
On the tip of sugarcane, apical or tip region meristem are found. When the tip region is removed, the presence of intercalary meristems play the role by facilitating the elongation of stems in sugarcane.

Please see attached image for better understanding.

Hope it helps!

4 0

3 years ago

How are bryophytes and seedless vascular plants alike?

Llana [10]

Answer:

both undergo vegetative propagation.

Explanation:

7 0

3 years ago

The Offspring produced by a cross between two given types of plants can be any of the three genotypes denoted by A, B, and C. A

ratelena [41]

Complete question:

The offspring produced between two given types of plants can be any of the three genotypes, denoted by A, B and C. A theoretical model of gene inheritance suggests that the offspring of types A, B and C should be in a 1:2:1 ratio (meaning 25% A, 50% B, and 25% C). For experimental verification, 100 plants are bred by crossing the two given types. Their genetic classifications are recorded in the table below.

Genotype Observed frequency

A → 18 individuals

B → 55 individuals

C → 27 individuals

Do these contradict the genetic model?

Use a 0.05 level of significance.

Determine the chi-square test statistic.

Answer:

Do these contradict the genetic model? No, according to the chi-square test, there is not enough evidence to reject the null hypothesis of the population being in equilibrium.

Explanation:

Available data:

Crossed genotypes: two
Genotypes among the offspring; Three → A, B, and C
Expected phenotypic ratio → 1:2:1
Total number of individuals, N = 100
A = 18 individuals
B = 55 individuals
C = 27 individuals

So, let us first state the hypothesis:

H₀= the population is equilibrium for this locus → F(A) = 25%, F(B) = 50%, F(C) = 25%
H₁ = the population is not in equilibrium

Now, let us calculate the number of expected individuals, according to their expected ratio.

4 -------------- 100% -------------100 individuals

1 --------------- 25% -------------X = 25 individuals A

2 -------------- 50% -------------X = 50 individuals B

1----------------- 25%--------------X = 25 individuals C

A B C

Observed 18 55 27
Expected 25 50 25
(Obs-Exp)²/Exp 1.96 0.5 0.16

(Obs-Exp)²/Exp

A) (18 - 25)²/25 = 49/25 = 1.96

B) (55 - 50)² / 50 = 25/50 = 0.5

C) (27 - 25)²/25 = 4/25 = 0.16

Chi square = X² = Σ(Obs-Exp)²/Exp

∑ is the sum of the terms
O are the Observed individuals: 2 in chamber B, and 18 in chamber A.
E are the Expected individuals: 10 in each chamber

X² = ∑ ((O-E)²/E) = 1.96 + 0.5 + 0.16 = 2.62

Freedom degrees = 2

Significance level, 5% = 0.05

Table value/Critical value = 5.99

X² < Critical value

2.62 < 5.99

These results suggest that there is not enough evidence to reject the null hypothesis. We can assume that the locus under study in this population is in equilibrium H-W.

6 0

3 years ago

All of the following drugs will lead to physical dependence except;

Xelga [282]

Answer:

B

Explanation:

5 0

2 years ago

the life is very beautiful thing to gift by God. a some are happy. b. some are sad c. some are smile. choose any one​

the life is very beautiful thing to gift by God. a some are happy. b. some are sad c. some are smile. choose any one