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valentina_108 [34]
2 years ago
10

African skulls that show both humanlike and apelike characteristics are thought to be derived from early African primates and ar

e collectively referred to as
Biology
1 answer:
Daniel [21]2 years ago
5 0

African skulls that show both humanlike and apelike characteristics are thought to be derived from early African primates and are collectively referred to as australopithecine.

<h3>What are the derived characteristics that distinguish early hominins from ancient chimpanzees?</h3>

Some features that have distinguished hominins from other primates, living and extinct, are their erect posture, bipedal locomotion, larger brains, and behavioral characteristics such as specialized tool use and, in some cases, communication through language.

Thus, they referred as australopithecine.

To learn more about australopithecine click here:

brainly.com/question/177662

#SPJ1

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metals (left side of periodic table) become positive ions so calcium would be the answer

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Which word best describes the tone of the excerpt? A. mysterious B. mocking C. frustrated D. troubled
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Why would a farmer plant a food crop in am small garden plot?
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A catabolic pathway may be which of the following? a. a set of reactions that combine monomers into larger, more energy-rich pol
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The correct answer is option c. "a set of reactions that release energy that can be used to drive cellular work".

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Catabolism is a part of metabolism at which molecules are broken down into smaller units in order to release energy that could be used in other reactions that drive cellular work. A catabolic pathway follows catabolism criteria. Catabolic pathways are the opposite of anabolic pathways, at which large molecules are synthesized with the requirement of external energy supply.

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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).

5 0
3 years ago
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