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2 years ago

HELPPP I GIVE BRAINLIEST!!!

Biology

1 answer:

alexdok [17]2 years ago

4 0

Answer:

B. Rub a soft rock on a rough surface

Explanation: because the force will cause the other rock to break plz mark brainlyes :P

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Cellular respiration is balanced by photosynthesis; the two processes work

Finger [1]

Answer:

A is the answer

Explanation:

Carbon dioxide and water are the reactants of both processes

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3 years ago

7. Red-cockaded woodpeckers live in mature open pine forest with little understory. One aspect of managing this endangered speci

Anni [7]

<h2>The Red-Cockaded Woodpeckers</h2>

Explanation:

Red-cockaded woodpeckers are sharp, socially adroit, and adorable.
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Snakes are a portion of their most risky home predators, so the winged animals peck the pine trees' bark to make sap overflow down the tree trunk. The smooth of sap stops winds and protects the nest cavities above.
Each pine tree in a backwoods into timber or turpentine, it sets aside a long effort for trees to become large enough to help respectable size home depressions.
The tree timberlands these flying creatures advanced in was intermittently gotten out by fire, which is currently typically stifled, prompting woods that are unreasonably thick ridiculous.

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2 years ago

Most adult amphibians breathe how?

nikitadnepr [17]

Answer:

Most adult amphibians can breathe both through cutaneous respiration and buccal pumping though some also retain gills as adults.

Explanation:

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2 years ago

What is the region of the sarcolemma across from the axon terminal at the neuromuscular junction?

polet [3.4K]

Answer:

The answer is motor end plate.

Explanation:

The region of the sarcolemma across from the axon terminal at the neuromuscular junction is the motor end plate.

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2 years ago

Cystic fibrosis (CF) is one of most common recessive disorders among Caucasians it affects 1 in 1,700 newborns. What is the expe

Phantasy [73]

Answer: The expected frequency of carriers is P(Aa)=0.046.

The proportion of childs with CF is P(aa)=0.024.

25% of having a child with CF (aa).

Explanation:

Hardy-Weinberg's principle states that in a large enough population, in which mating occurs randomly and which is not subject to mutation, selection or migration, gene and genotype frequencies remain constant from one generation to the next one, once a state of equilibrium has been reached which in autosomal loci is reached after one generation. So, a population is said to be in balance when the alleles in polymorphic systems maintain their frequency in the population over generations.

Given the gene allele frequencies in the gene pool of a population, it is possible to calculate the expected frequencies of the progeny's genotypes and phenotypes. If P = percentage of the allele A (dominant) and q = percentage of the allele a (recessive), the checkerboard method can be used to produce all possible random combinations of these gametes.

Note that p + q = 1, that is, the percentages of gametes A and a must equal 100% to include all gametes in the gene pool.

The genotypic frequencies added together should also equal 1 or 100%, and all the equations can be summarized as follows:

$p+q=1\\(p+q)^{2} = p^{2} +2pq+q^{2} = 1\\P(AA)=p^{2} \\P(aa)=q^{2} \\P(Aa)=2pq1$

So, there are 1700 individuals and only one is affected. Since it is a recessive disorder, the genotype of that individual must be aa. So the genotypic frequency of aa is 1/1700=0.000588.

Then, $P(aa)=q^{2}=0.000588$ . And with that we can calculate the value of q,

$P(a)=q=\sqrt{0.000588}=0.024$

And since we know that p+q=1, we can find out the value of p.

$p+0.024=1\\1-0.024=p\\p=0.976$

Next, we find out the genotypic frequency of the genotype AA:

$P(A)=p=0.976\\P(AA)=p^{2} = 0.976^{2}=0.95$

Now, we can find out the genotypic frequency of the genotype Aa:

$P(Aa)=2pq=2 x 0.976 x 0.024 = 0.046$

Notice than:

$p^{2} + 2pq + q^{2} = 1\\x^{2} 0.976^{2} + 2 x 0.976 x 0.024 + 0.024^{2} = 1$

Then, the expected frequency of carriers is P(Aa)=0.046

The proportion of childs with CF is P(aa)=0.024

If two parents are carriers, then their genotypes are Aa.

Gametes produced by them can only have one allele of the gene. So they can either produce A gametes, or a gametes.

In the punnett square, we can see that there genotypic ratio is 2:1:1 and the phenotypic ratio is 3:1. So, there is a probability of 25% of having an unaffected child, with both normal alleles (AA); 50% of having a carrier child (Aa) and 25% (0.25) of having a child with CF (aa).

5 0

3 years ago