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Oduvanchick [21]
3 years ago
5

The table describes two methods of heat transfer. Methods of Heat Transfer Method A Method B Transfer of heat is through vibrati

ons of the molecules of the medium Transfer of heat is through electromagnetic waves Heat is transferred in solids by this method Medium particles do not play any role in heat transfer Which statement is correct? Method A is convection and Method B is radiation. Method A is conduction and Method B is radiation. Method A is convection and Method B is conduction. Method A is conduction and Method B is convection.
Biology
2 answers:
Virty [35]3 years ago
3 0

Answer:

Method A is conduction and Method B is radiation.

Explanation:

Driving the situation in which the heat spreads through a "conductor". That is, although it is not in direct contact with the heat source, a body can modify its thermal energy if there is heat conduction by another body, or by another part of the same body. For example, while cooking something, if we leave a spoon against the pan, which is on the fire, after a while it will heat up too.

Heat is energy in transit. This energy is transported by means of electromagnetic waves in the infrared frequency. In the transmission of heat by radiation, this fact is more evident. All bodies emit radiation, just having a temperature. The only difference between light and heat is the frequency of radiation. Thermoses are a good example of thermal radiation. The walls of the thermos are double and silver. The double walls are separated by practically a vacuum, in such a way that the heat exchanges by conduction or convection are minimized, as they need a material medium to occur. The silver walls, on the other hand, minimize heat exchanges by radiation, thus making sure that the liquid inside the bottle does not lose or receive heat.

alukav5142 [94]3 years ago
3 0

Answer:                                                

           

Explanation:

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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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