The diagram shows the inside of a eukaryotic cell

Why is it important that the heart and lungs kept healthy

Genrish500 [490]

Essentially, when you breathe in the lung take in oxygen and pass it along to the bloodstream, which brings it to your bodies cells. In return it picks up carbon dioxide, which the lungs dispose of. Every time you breathe this process is happening. That is why it's crucial to keep your lungs healthy.

4 0

2 years ago

What is the vent in earths surface through which magma and gases are expelled

lutik1710 [3]

Answer:

A volcano

Explanation:

Is a vent or fissure in the Earth's surface through which molten rock and gases are expelled. Molten rock is called magma.

5 0

2 years ago

Blue jays and crows eat caterpillars and mice. if the number of blue jays increases, what are the two likely effects on other po

suter [353]

If the number of blue jays increases, the two likely effects on other populations are both the number of mice and the number of caterpillars will decrease.

<h3>What are two likely effects on other populations due to the increment in Blue jays population?</h3>

An organism that preys on other species in order to consume them as food is referred to as a predator.
Prey is a term used to describe an organism that a predator kills.
An ecosystem's ability to balance an expanding prey population depends on the interaction between predators and their prey.
Blue jays and crows both rely on mice and caterpillars as food sources, as indicated by the current circumstance.
Mice, caterpillars, and blue jays are the prey, while crows and crows are the predators. The two prey populations will decline if the number of blue jays rises.

Hence, both the number of mice and the number of caterpillars will likely drop as the number of blue jays rises, having two possible implications on other populations.

To learn more about Predator refer to:

brainly.com/question/12655080

#SPJ4

5 0

1 year ago

The vascular tissue that transports water and dissolved nutrients from the roots up through the plant is called

lisov135 [29]

It is Xylem of the plants that help transport the water and other nutrients from the roots up through the plant. The phloem basically transport food synthesised by the plant but not the water.

8 0

2 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