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IRINA_888 [86]
3 years ago
9

QUESTION 15 List the 5 characteristics shared by all living things.

Biology
1 answer:
Xelga [282]3 years ago
8 0

Answer:

movement

sensitivity

excretion

respiration

reproduce

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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. <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
What can you infer about copper and silver based on their position relative to each other
IRINA_888 [86]
They are both metals

5 0
3 years ago
Which of these actions is the best way for an individual to conserve water?
BigorU [14]

Answer: B. Take short showers instead of baths.

Explanation:

The water is the most important non-living environmental factor which is the main requirement for survival of every living being. Even though water is a renewable resource which can be replenish by the mode of water cycling in nature but with the increase in the global population the demand of freshwater has increased. Also with the increase in the causes and sources of water pollution the levels of freshwater sources are decreasing. Therefore, the issues of water conservation should be taken into consideration.

According to the given situation, if an individual needs to conserve water, take short showers instead of baths is the correct option because of the fact that short showers will prevents the wastage of large amount of water which could be wasted while taking baths.  

4 0
3 years ago
In mice brown fur (B) is dominant to white fur (b). Two brown mice mate. Their genotype in unknown. You observe the offspring an
Brilliant_brown [7]

Answer:

bb bc there more alike

but i am not so sure

7 0
3 years ago
The table shows the characteristics of three different classes of organisms. Based on the given characteristics, name each class
Yanka [14]
<span> 1 = Birds, 2 = Mammals, 3 = Amphibians.

I took the test 

</span>
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3 years ago
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