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3 years ago

PLEASE HELP!!!!!!!!!!!!

Biology

2 answers:

Marat540 [252]3 years ago

4 0

it's the structure of the nucleotides

the basic building unit of nucleic acid

dezoksy [38]3 years ago

3 0

Answer:

This is a nucleic acid structure. {Nucleotides} DNA/RNA

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svetlana [45]

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A botanist who was studying a new plant ground up some of the leaf cells, then centrifuged the mixture. The organelles became fr

Harrizon [31]

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2 years ago

In a particular woodland community, two species of rodents coexist: mice and chipmunks. Both species compete for the same kinds

scZoUnD [109]

Answer:

First, we take the owl out, the total number of mouse increases at a higher rate and the chipmunks are also decreasing in numbers. In an ideal ecosystem, both mice (lets call it "a") and chipmunks (lets call it "b") should increase since the restraining factor has been removed. But that is the opposite of what we should see. In that case, we will take the owl to be the "limiting factor" for the rodents.

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3 years ago

Barnacles and mussels are filter feeders that attach themselves to rocks in the intertidal zone. A scientist observed a dense co

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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