The stomach lining is made up of deep muscular grooves.How do you think these structures help the stomach to breack down food.

MARKING BRAINLIEST !!!

photoshop1234 [79]

Answer: D. A river separates members of a squirrel population that used to occupy the same geographical area.

Explanation:

Allopathic Speciation occurs when a geographic barrier (river, mountain, or canyon) separates members of a population

Hope this helps!

3 0

3 years ago

In horses, black color (B) dominates chestnut color (b). The trotting gait (T) dominates the pacing gait (t). A cross is made be

Gre4nikov [31]

Black B is dominant over chestnut color b and trotting gait T is dominant over pacing gait t. The answer is 100%. Since all offspring have at least one dominant black hair allele that overwhelms the brown hair allele, all offspring are expected to have black hair.

Phenotype refers to the physical color of a horse's coat or the way genes are expressed. Phenotype II FMF in Familial Mediterranean Fever is the onset of amyloidosis prior to the onset of his FMF with typical attacks or is an isolated finding in members of the FMF family.

White horses have pink skin. White fur mane tail, and hooves. and dark brown eyes. White is the dominant trait W at the white locus. White horse genotype must be heterozygous dominant We. Homozygous dominant because WW is more likely to cause early embryonic loss. Scientists conclude that horses were first domesticated 6,000 years ago in the western part of the Eurasian steppe, in what is now western Ukraine and Kazakhstan.

Learn more about Horse phenotype here:-brainly.com/question/26406162

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7 0

1 year ago

If the producer level of the energy pyramid start with 5000 calories how many calories will transfer to the secondary consumers

Strike441 [17]

Explanation:

Only 10% of calories will be transfered to the next higher level.

5000 * 0.1 = 500.

500 calories will transfer to the secondary consumers.

8 0

2 years ago

Identify the category each organism belongs to, based on the food it consumes.

Nikolay [14]

Cheetah is Carnivore

Tree is producer

The mushroom thingy is decomposer

Gazille is herbivore

4 0

3 years ago

Read 2 more answers

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