QUESTION 11 1. Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 85 individuals and H5/H5=24 individuals. Using the allele's frequencies experimentally derived, calculate the frequency of the H4/H5 genotype that would be expected if the class were a population in Hardy-Weinberg equilibrium. 1. 0.28 2. 0.51 3. 0.19 4. 0.72 5. 0.14 6. 0.24 7. 0.41 QUESTION 12 1. Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 85 individuals and H5/H5=24 individuals. Using the genotype frequencies derived assuming that the class were a population in Hardy-Weinberg equilibrium, calculate the number of H4/H4 individuals that would be expected in the class (rounded numbers). 1. 19 2. 57 3. 72 4. 147 5. 171 6. 120 7. 96 QUESTION 13 1. Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 85 individuals and H5/H5=24 individuals. Considering the Hardy Weinberg equilibrium and comparing the observed and the expected number of individuals for the three genotypes, calculate the value of the Chi-square statistic 1. 2.69 2. 0.05 3. 28.67 4. 14.59 5. 0.50 6. 22.31 7. 3.84 QUESTION 14 1. Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 85 individuals and H5/H5=24 individuals. Considering the Hardy Weinberg equilibrium, which is the (correct) null hypothesis tested by Chi-square? 1. The whole class represents a population that is in Hardy-Weinberg equilibrium 2. The whole class represents a population that may not be in Hardy-Weinberg equilibrium 3. The whole class represents a population that is not in Hardy-Weinberg equilibrium 4. The whole class represents a population that may be in Hardy-Weinberg equilibrium QUESTION 15 1. Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 85 individuals and H5/H5=24 individuals. Considering the Hardy Weinberg equilibrium and calculating the Chi-square statistic, do you reject or fail to reject the null-hypothesis? 1. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P>0.05. Hence, I reject the null hypothesis. 2. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom, I conclude that P>0.05. Hence, I fail to reject the null hypothesis. 3. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P<0.05. Hence, I reject the null hypothesis. 4. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P<0.05. Hence, I fail to reject the null hypothesis.

Answers

Answer 1

According to Hardy-Weinberg, when a population is in equilibrium, it will have the same allelic frequencies generation after generation, meaning that they are stable, they are not evolving.

When any evolutive force is acting on the population, this equilibrium breacks, and allelic and genotypic frequencies change through generations, differing from the expected ones.  

A) Option 7 is the correct answer ⇒ 0.41

B) Option 6 is the correct answer ⇒ 120

C) Option 7 is the correct answer ⇒ 3.84

D) Option 1 is the correct answer ⇒ The class represents a population that is in H-W equilibrium

E) Option 1 is correct. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for  the appropriate degrees of freedom I conclude that P>0.05. Hence, I reject the null hypothesis.

-------------------------------------------

Allelic frequencies in a locus are represented as p and q, referring to the

allelic dominant or recessive forms. The genotypic frequencies after one generation are p² (H0m0zyg0us  dominant), 2pq (H3ter0zygous), (H0m0zyg0us recessive). Populations in H-W equilibrium will get the same

allelic frequencies generation after generation.

The sum of the allelic frequencies equals 1, this is p + q = 1.

In the same way, the sum of genotypic frequencies equals 1, this is

p² + 2pq + q² = 1

Being

 p the dominant allelic frequency,

 q the recessive allelic frequency,

 p² the h0m0zyg0us dominant genotypic frequency

 q² the h0m0zyg0us recessive genotypic frequency

 2pq the h3ter0zyg0us genotypic frequency

 

Situation: Through PCR, we have determined the PER3 genotypes for a class of students as follows:

 H4/H4 = 125 individuals;

 H4/H5 = 85 individuals;

 H5/H5=24 individuals.

⇒ Total number of individuals= 125 + 85 + 24 = 234

⇒ Genotypic frequencies, F(xx):

 F(H4/H4) = 125/234 =0.534

 F(H4/H5) = 85/234 = 0.363

 F(H5/H5) = 24/234 = 0.102

⇒ Allelic frequencies, f(x):

 f(H4) = p = F(H4/H4) + 1/2 F(H4/H5) = 0.534 + 0.363/2 = 0.534 + 0.182 = 0.716

 f(H5) = q = F(H5/H5) + 1/2 F(H4/H5) = 0.102 + 0.363/2 = 0.102 + 0.182 = 0.284

Questions:

A)  According to the theoreticall frame, we know that 2pq is the h3ter0zygous genotypic frequency. So,  

F(H4/H5) = 2pq = 2 x 0.716 x 0.284 = 0.408 ≅ 0.41 ⇒ Option 7 is the correct answer.

-------------------------------------------------------------------------------------------------------------

B)  According to the theoreticall frame, we know that p² is the h0m0zyg0us genotypic frequency. So,

p = 0.716

p² = 0.5126 ≅ 0.513 ⇒ This is the genotypic frequency.

To calculate the number of individuals carrying this genotype, we need to multiply it by the total number of

individuals.

H4/H4 individuals = p² x total number of individuals = 0.513 x 234 = 120

Option 6 is the correct answer.

-----------------------------------------------------------------------------------------------------------

C)  Up to here we know that 2pq = 0.41 and p² = 0.513

Now we need to calculate q ²

q = 0.284, then q² = 0.284² = 0.08

These are the expected frequencies if the population was in H-W equilibrium.

The expected number of individuals with each genotype are:

 H4/H4 = 0.513 x 234 = 120 individuals

 H4/H5 = 0.41 x 234 = 96 individuals

 H5/H5= 0.08 x 234 = 18 individuals

The observed number of individuals with each genotype are:

 H4/H4 = 125 individuals

 H4/H5 = 85 individuals

 H5/H5=24 individuals

X² = ∑ (Observed - Expected)²/Expected)

X² = ((125-120)²/120) + ((85 - 96)²/96) + ((24-18)²/18)

X² = 0.21 + 1.26 + 2 =

X² = 3.47

The clossest option is option 7 = 3.84. The difference might be related to decimals and rounding.

-------------------------------------------------------------------------------------------------------------

D)  The correct answer is  1 ⇒ The whole class represents a population that is in Hardy-Weinberg equilibrium

The null hypothesis always predict that populations are in H-W equilibrium.

-----------------------------------------------------------------------------------------------------------

E)  

 X² = 3.47

 Freedom degrees = n - 1 = 3 - 1 = 2

 Table p value: 7.82

 Significance level, 5% = 0.05

 Table value/Critical value = 5.991

5.991 > 0.347

Meaning that the difference between the observed individuals and the expected individuals is statistically  significant. Not probably to have differe by random chances. There is enough evidence to reject the null

hypothesis.

Option 1 is correct. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for  the appropriate degrees of freedom I conclude that P>0.05. Hence, I reject the null hypothesis.

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Related Questions

help me please like rn :(
Which option distinguishes what a commodities trader would do with the information in the following scenario?
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cancel all options on future sales of wheat

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Answers

The commodities trader would immediately buy futures contracts at their current prices. The correct option is C.

Thus, a commodities trader would probably predict a decline in wheat supply if they learned about an anticipated heat wave and drought that may affect wheat plants. They would then buy futures contracts at their present levels in response.

By taking this move, the trader is able to protect himself against prospective price rises brought on by the decreased supply of wheat by securing the right to purchase wheat at a certain price in the future. The trader may be able to take advantage of the predicted market conditions before the impact of the heat wave is fully felt by purchasing futures contracts.

Thus, the ideal selection is option C.

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Answers

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I believe the one with 2 cups of water will be saltier

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When you change the fractions to a common denominator, which is 12, you see the recipe with 1 cup should be half of the recipe with 2 cups. But the one with 2 cups is more than double the recipe with 1 cup.

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Question 5 of 25
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Answers

Answer:

A.

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Answers

In order to serve as evidence that the hypothesis put forward by the Biochemists is true, they will have to observe that the onion cells stop multiplying and the roots do not grow.

The primary function of microtubules and centrioles in a cell (or at least one that they have in common) is their participation in cellular division, that is, mitosis and the formation of a cytoskeleton.

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Answers

When a cross between 2 contrasting parents produces a new phenotype, it is known as incomplete dominance. In diploid (2n) organisms, individuals receive one allele or gene variant from each parent.

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Incomplete dominance occurs when both alleles are expressed in heterozygous organisms.

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Answers

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Answers

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Answers

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Answers

Answer:

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Explanation:

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Answers

Answer:

An example of a scientific investigation design that uses independent variable and dependent variable in its design is a research that seek to find out the effect of two different fertilizers on the growth of a tuber crop (yam). The independent variable in this case is the type of fertilizer used while the dependent variable is the corresponding growth of the tuber crop. In a research set up, the independent variable is the variable that cause a change in the dependent variable.

Explanation:

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Explanation: (if needed)

The studies of Earth’s physical structure and substance, its history and the processes that act on it. People who work in this field are known as geologists. Planetary geology is the science of studying the same things about other planets. It includes the study of organisms that have inhabited our planet. An important part of geology is the study of how Earth's materials, structures, processes, and organisms have changed over time. Geology involves methods and knowledge from biology, chemistry, physics, and mathematics.

Answer:

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Explanation:

Geology is a branch of Earth science concerned with both the liquid and solid Earth, the rocks of which it is composed, and the processes by which they change over time. Geology can also include the study of the solid features of any terrestrial planet or natural satellite such as Mars or the Moon.

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We can arrive at this answer because:

The management of a company is essential for its proper functioning and the success of the production.One of the first problems that lack of management causes is the difficulty in keeping up with market changes. This makes the company outdated and inefficient.Talking about the production and distribution of the product harms sales and reduces the number of customers.Sectoral problems hinder organized production, which will also affect sales.Problems in managing employees will also be disadvantageous, as it will affect all of the company's compositional factors.

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Answers

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Comparing the location of the prepuce - The prepuce is in the same location in fetal pigs as it is in humans.

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