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3 years ago

What creates global wind patterns on earth

Biology

1 answer:

vovangra [49]3 years ago

6 0

Answer:

Global winds

Explanation:

Earth's orbit around the sun and its rotation on a tilted axis causes some parts of Earth to receive more solar radiation than others. This uneven heating produces global circulation patterns. ... Six of these large convection currents cover the Earth from pole to pole.

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The area of biology devoted to the study of fungi is known as?

MakcuM [25]

Answer: mycologist []

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2 years ago

A leaf in which small veins extend outward from a single main veins has what type of ventaion?

e-lub [12.9K]

Well...This might helps Parallel venation is characteristic of monocots, the veins are usually parallel to each other along the length of the leaf. 
Pinnate venation is the veins are in a branching pattern, characterized by one major vein(called midrib) with smaller veins extending outward from it.
Palmate venation is characterized by two or more major veins extending outward from one point like the fingers extending form the palm of a hand

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3 years ago

Functional analysis of the genome of a prokaryote shows that 20% of the genome is for metabolism, 1% for biosynthesis of amino a

Zarrin [17]

Answer:

This particular organism lives in an environment that is rich in organic material, particularly amino acids and proteins. Also the organism has low amount of amino acid biosynthesizing genes and relatively higher amount of peptide transporters.

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2 years ago

Read 2 more answers

Why did Mendel want to control the pollination of his pea plants?

Viefleur [7K]

Mendel wants to control the pollination of his pea plants because traits were not blended but they remain separate in the subsequent generations and also it is contrary to scientific opinion during his time

Explanation:

Mainly, he wants to control because they were against the scientific opinion during his time.
Mendel takes a pea plant for his experiment because it can be easily observable.
Pea plant has seven traits and it can be pollinated by both the ways. One is self-pollination and the other is cross-pollination.
Mendel did not know about the genes but he speculates the factors of formation in the traits.
Mendel produces three laws, the law of dominance, the law of segregation, the law of independent assortment.

4 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