I can't figure out the controlled variables please help

The small intestine is about 20 feet (6 meters) long and folds many times to fit in the abdomen. Although it is longer than the large intestine, it is called the small intestine because it is smaller in width.

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5 0

3 years ago

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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

4 years ago

Explain how osmosis will affect animal cells when the cells are placed into and environment with a low water potential compared

matrenka [14]

Answer:

Explanation:

The cytoplasm in animal cells is surrounded by a membrane, which delimits the cell and its surroundings. As opposite from the plant cells, animal cells do not have a cell wall.

Osmosis is the process in which a solvent diffuses across a semipermeable membrane (like the cell membrane) from the zone with the lower concentration of solutes to the zone with the higher concentration. That means that water, in this case, will diffuse spontaneously across the cell membrane from the place with more water molecules (usually, the exterior of the cell) to the place which has less water molecules, or is more concentrated (in this case, the cytoplasm). This diffusion will continue until an equilibrium is reached.

If we place an animal cell, which does not have a cell wall, into an environment with a low water potential, the water will flow from the exterior to the interior. If the water potential is very low compared to the one of the cytoplasm, the cell can receive too much water and explode. This process is called cytolysis.

8 0

3 years ago