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2 years ago

Plzzzz help i will give brainlyest Write a paragraph describing silverback gorillas

1 answer:

5 0

Silverback gorillas live high in the mountains in two protected parks in Africa. They are also referred to as mountain gorillas. Silverback gorillas continually wander through their home ranges of 10 to 15 square miles, feeding and resting throughout the day. Because gorillas are nomadic, they build new nests each day at dusk, constructing them out of bent branches in a tree or of grasses on the ground. Silverback gorillas have long black hair and their thick, shaggy coats help to keep them warm in cold climates. Adult male mountain gorillas are called Silverbacks because of the silver saddle of hair on their backs. Mountain gorillas have a stocky build, with a broad chest, long, muscular arms and wide feet and hands. Their arms are longer than their legs. Gorillas live in groups. Each gorilla family has a silverback as leader who scares away other animals by standing erect on its hind legs, tearing up and throwing plants, drumming on the chest with its hands or fists, stamping its feet, striking the ground with its palms and galloping in a mock attack on all fours!

Silverback Gorillas are around six feet tall – about the height of an adult male – while females are around five feet tall. Silverback gorillas weigh approximately 350 pounds- about twice the weight of normal adult man. Gorillas are herbivores, and eat only plants. They spend most of their day foraging for food and eating bamboo and leafy plants

Silverback gorillas were not even known to exist until 1902.Today, their habitat is being destroyed when people use the land for farming and the trees for fuel. Gorillas are also killed by poachers and sometimes get caught in poacher’s snares meant for other animals. The Silverback Gorilla is the most endangered species of Gorilla.

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lys-0071 [83]

Answer:

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Explanation:

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3 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. <u>If P = percentage of the allele A (dominant) and q = percentage of the allele a (recessive)</u>, 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).

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3 years ago

Where does antibiotic resistance come from?

inessss [21]

Answer:

Antibiotic resistance happens when germs like bacteria and fungi develop the ability to defeat the drugs designed to kill them. That means the germs are not killed and continue to grow. Infections caused by antibiotic-resistant germs are difficult, and sometimes impossible, to treat.

Explanation:

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3 years ago

***PLEASE HELP!!!!!!!!!!***

jenyasd209 [6]

The answer is the first one

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3 years ago

What is going if there is no atmospheric greenhouse

masha68 [24]

Answer:

Without naturally occurring greenhouse gases, Earth's average temperaturewould be near 0°F (or-18°C) instead of the much warmer 59°F (15°C). The concentration of greenhouse gases, especially carbon dioxide and methane, has fluctuated naturallyover geological time scales.

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2 years ago