All categories

3 years ago

What is the farming method called monoculture?

Biology

2 answers:

Sholpan [36]3 years ago

5 0

Large fields to be cleared, plowed, and planted with a single crop year after year.

I hoped this helped ;)

Effectus [21]3 years ago

4 0

Hello :)

Monoculture is the agricultural practice of producing a single crop, plant, or livestock species, variety, or breed in a field or farming system at a time.

Hope this helps some, good luck!! ;)

You might be interested in

A child is brought to the emergency department experiencing an anaphylactic reaction to a bee sting. place in correct sequence w

GaryK [48]

The nurse should perform the ABC's, which starts out with A- Airway B- Breathing, and C- Circulation, if any of these are not within normal limits then the child will be given a dose of epinephrine (1/1,000) .3mg IM, Benadryl 25/50 mg followed by a dose of solumedrol

5 0

3 years ago

erica [24]

Alright my friend the answer to your question will be true let's put an example in to inform why it is alright so say you and your cousin look a like that's because the genetics are similar but not exactly the same

6 0

3 years ago

Which type of direct increases glycogen stored in muscle and endurance time at marathon speed most? high fat mixed high carbohyd

tino4ka555 [31]

The type of direct that increases glycogen stored in muscle and endurance time at marathon speed most is high carbohydrate. Option C.

<h3>What is glycogen?</h3>

The term glycogen refers to the form in which carbohydrate is stored in the body. The energy that is stored as glycogen in the body can be slowly released when required by hydrolysis of the glycogen molecule.

The type of direct that increases glycogen stored in muscle and endurance time at marathon speed most is high carbohydrate. Option C.

Learn more about glycogen:brainly.com/question/14466525

#SPJ4

Missing parts;

Which type of direct increases glycogen stored in muscle and endurance time at marathon speed most? A. high fat B. mixed C. high carbohydrate D. fasting

4 0

2 years ago

What must happen for the lungs to expand allowing you to inhale

malfutka [58]

Ur intercostle muscle between ur ribs

3 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