23 points!!! Answer 1,2,3,4,5

PLEASE HELP : Do you think intelligence is linked more to someone's DNA or how someone is raised/type of education? Why? At leas

Aleksandr-060686 [28]

Answer:

In physical appearance can be changed by different DNA but for mental not very sure.But mental state can be altered when the DNA has anomaly or mutation. Mostly, people whom has high education always study hard for example students even they have physical problem but they still have good mental

4 0

3 years ago

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How is the earth oriented to the sun

Anon25 [30]

Answer:

the orientation of Earth's tilt with respect to the sun – our source of light and warmth – does change as we orbit the sun. In other words, the Northern Hemisphere is oriented toward the sun for half of the year and away from the sun for the other half.

Explanation:

7 0

3 years ago

Why is this study conducted in a vacuum zone.

svet-max [94.6K]

Your answer would be A

5 0

2 years ago

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Use an example from nature to explain how natural selection causes populations to change as their environment change include the

ivanzaharov [21]

Answer:

Galapagos finches have various beak sizes that make foraging for food more successful.

Explanation:

Organisms evolve over time due to changes in their genome. These are pontaneous, and occur in DNA at random. These changes are called mutations and form alleles or different forms of a gene.

Over time within a population, the number alleles increase the variation of the population. These variants may confer specific traits within an individual, that may confer a biological advantage.

Thus, the trait may make the organisms more capable of obtaining food, shelter a mate etc. or ensure survival, i.e. they are able to pass on their genes to the next generation.

5 0

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