20 points and I will mark you as brainliest

Environmental factors carn have various effects on ecosystems. In August 2005, Hurricane Katrina caused signifcant damage in sev

matrenka [14]

Assuming this is supposed to be a multiple choice question- I think the correct answer would be “early onset of spring change in marine ecosystems” because it looks like the question is asking what it wouldn’t affect, and I’m sure if the people had to leave, that animals would’ve all had to leave as well!

4 0

2 years ago

Read 2 more answers

If chromosomes fail to separate during cell division ________________ can occur. This can result in cells with the incorrect num

storchak [24]

Answer:

1. NONDISJUNCTION

2.Aneuploidy

3.Trisomy

4.Monosomy.

5.3.

6.Trisomy 21.

Explanation:

8 0

3 years ago

Describe the behavior of the frog heart that you can elucidate from the force recording (e.g., mechanical events). How can you u

Sladkaya [172]

Answer:

P wave and R wave

Atrial contraction = P wave

Ventricular contraction = R wave

Explanation:

The behavior of the frog heart from force recording can be classified into ; P wave and R wave.

utilizing the ECG ( Electrocardiogram) data the aspect of the force recording that is due to atrial contractions is the ; P wave while the force recording that is due to ventricular contractions is the ; R wave the ventricular contractions is also known as QRS

ECG is used to record the electrical activities of a heart at rest

3 0

2 years ago

The difference between tight junctions, desmosomes, and gap junctions is that

disa [49]

<span>Narrow junctions and desmosomes have different structures. Narrow junctions are made of a protein arranged inside the membrane. There are many intersecting lines that cause the cells to be sealed and together. Desmosomes contain the protein cadherin. There are cadherins in each cell and they join at this point sealing the cells together, the tight junctions have the function of preventing the materials that circulate between the cells. The desmosomes provide binding sites for the cells to be attached.</span>

6 0

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