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3 years ago

True or false the cardiovascular system does not interact with any other body system

Biology

2 answers:

Luda [366]3 years ago

6 0

False it interacts with the respiratory system

balandron [24]3 years ago

5 0

False, the cardiovascular System interacts with the respiratory system

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when two individuals who are heterozygous for a gene are mated, we would predict a dominant to recessive phenotypic ratio of____

Scilla [17]

Answer:

Explanation:

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3 years ago

Being a visual learner have you got any tips on how to study science?

sukhopar [10]

Even being a visual learner the trick is to take notes, even if you never look at the notes ever again its good to write them down, you are more likely to remember something you have written down rather than to just look at it. Science is something that could be easy, you just have to take it at your own pace, don't try to learn science by giving it a quick read or glance at it.

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3 years ago

2. Chimpanzees and gorillas more closely resemble each other anatomically than either resembles humans. For example, the hands o

Ganezh [65]

Answer:

Human evolution in the hominid family occurred faster than between Chimpanzees and gorillas. Neutral mutations may result in sequence divergence between organisms that are structurally similar (in this case, chimpanzees and gorillas)

Explanation:

Human and chimpanzee share higher sequence identity than predicted in the hominid family since observed mutations are necessarily associated with functional divergence

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2 years ago

Read 2 more answers

what is crossing a hybrid with a homozygous dominant or a homozygous recessive organism resulting in a 1:1 ratio called?

Ghella [55]

That would be a test cross.

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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