Answer:
There are seven main classes of nutrients that the body needs. These are carbohydrates, proteins, fats, vitamins, minerals, fibre and water. It is important that everyone consumes these seven nutrients on a daily basis to help them build their bodies and maintain their health.
these are the 7 main classes of Nutrition that the body needs hope it will help you
Occurs in mitochondria.
Select one:
a. Both Photosynthesis and Cellular Respiration
b. Neither Photosynthesis nor Cellular Respiration
c. Photosynthesis
d. Cellular Respiration
Answer:
d. cellular respiration
Explanation:
mitochondria is where cellular respiration occurs, and photosynthesis occurs in the chloroplast, so d would be the most reasonable answer.
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.
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), q² (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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Select the statement about human impact on the environment that is FALSE. The number of humans engaging in a specific activity impacts the severity of the issue. Damaging the environment threatens human and ecosystem health. Pesticide and fertilizer use by humans is particularly damaging to the environment. Urban growth leads to increased waste, which leads to climate change.
Answer:
the answer is should be True
The mass of the Sun is 1.99 × 1030 kg. Jupiter is 7.79 × 108 km away from the Sun and has a mass of 1.90 × 1027 kg. The gravitational force between the Sun and Jupiter to three significant figures is × 1023 N.
Answer:
4.16
Explanation:
edge2021
Answer:
4.16
Explanation:
What variable should Anurag change in his experiment
Answer:
For geological carbon sequestration, the reaction of aqueous CO2 with silicate rock permits carbonate formation, achieving permanent carbon sequestration.A
Explanation:
A man bought a goldfish in a pet shop. Upon returning home, he put the goldfish in a bowl of recently boiled water that had been cooled quickly. A few minutes later the fish was found dead. Explain what happened to the fish
Answer:
lack of oxygen in the water
Explanation:
The fish most likely died from lack of oxygen in the water. This is because fishes actually use their gills to extract and breathe in the oxygen from the water while also expelling carbon dioxide from their lungs. Similar to how humans breathe. When the water was boiled it caused the dissolved gases to be expelled, which includes oxygen. Therefore, without the necessary oxygen in the water, the fish ultimately suffocated.
Boiling the water reduce its amount of dissolved oxygen which is needed by the fish to breathe, that's why the fish died after few minutes.
What is dissolved oxygen?Dissolved oxygen is the amount of oxygen present in the water.
The organisms live to consume dissolved oxygen to breathe.
The amount of dissolved oxygen is high in the current water like rivers than in the still water like pond.
If the amount of DO is high in the water, it causes bubble gas disease in the aquatic organisms.
If the amount of DO is low in the water than, fishes and other aquatic organism cant survive due to low oxygen level.
Thus, boiling the water reduce its amount of dissolved oxygen which is needed by the fish to breathe, that's why the fish died after few minutes.
Learn more about goldfish
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Is cell division a continuous process?
Illustrate with an example
Answer:
Cell division is the means of reproduction in unicellular organisms whereas it is the means of tissue growth and maintenance in multicellular organisms. ... In adults, cell divisions are involved in renewing old tissues rather than growth.,ya it is continuously processing
The meaning of ALARA in radiation?
Answer:
The guiding principle of radiation safety is “ALARA”. ALARA stands for “as low as reasonably achievable”.
Ecosystems rely on interdependence between species to keep balance. Which of the following is a threat to a stable
ecosystem?
A. Loss of biodiversity
B. High biodiversity
C. Low biodiversity
D. Increase in biodiversity
Answer:
loss of biodiversity
Explanation:
Biodiversity- refers to the variety of life on Earth at all its levels, from genes to ecosystems, and can encompass the evolutionary, ecological, and cultural processes that sustain life.
loss in biodiversity affect food chains greatly
thanks
hope it helps
.1.2 The type of control shown in the diagram is known as
Answer:
which diagram
Explanation:
I hope you understand
Importances of bacteria to humans
Answer:
Bacteria are very important to humans, both for good and for bad, due to their chemical effects and the role they play in spreading disease. In their beneficial effect, some bact
¿Qué afirmación(es) es(son) correcta(s)?:
a) El dióxido de azufre de la quema de carbón en centrales
eléctricas protege contra la lluvia ácida.
b) Los óxidos de nitrógeno de las emisiones de combustión
provocan el aumento del pH en el agua de lluvia.
c) Una solución ácida de dióxido de carbono en agua de lluvia se
llama lluvia ácida.
d) El dióxido de azufre junto con los óxidos de nitrógeno son las
principales causas de la lluvia ácida.
Explanation:
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i need help in biology questions please G10?
Answer:
ok where is it
we can help only if there is something attached
What component of Earth's atmosphere exists entirely as a result of photosynthesis?
oxygen pas
n mas
O water vapor
O nitrogen gas
O carbon dioxide gas
Answer:
Explanation:
Carbon dioxide
Why should body temperature not be allowed to fluctuate too much?
Answer:
Because that can destroy the helpful enzymes in the body, and therefore cause a lot of problems or maybe cause death
Phân tích các quy luật hoạt động thần kinh cấp cao ở trẻ và vận dụng trong thiết lập thói quen học tập và kỉ luật ở học sinh tiểu học.
Answer:
very different than ducks do u want it is not the
Help meeee!!!!
Why is it necessary to check total aerobic microorganisms in microbiological quality control but not to test total anaerobic microorganisms?
Answer:
While many microbes are harmless to humans, others can cause serious problems. They can spoil food, introduce toxins, cause disease and lead to a host of other problems. The importance of microbiological testing is to quickly identify these contaminants and treat them before they do irreversible damage.
Explanation:
Aerobic bacteria are bacteria that can grow and live when oxygen is present.
What is the function of the mitochondria?
A. Stores the cell's DNA
B. Builds proteins
C. Produces energy for the cell by respiration
OD. Stores the cell's glucose
Reset Selection
Answer:
Produces energy for the cell by respiration
Explanation:
The glucose obtained from food is broken down to pyruvic acid in the cytoplasm. This pyruvic acid is broken down into oxygen, water and energy rich ATP molecules in the Mitochondria.
what is the definition of wildlife?
Answer:
wild animals collectively; the native fauna (and sometimes flora) of a region.
Explanation:
I hope this helped :)
List some Characteristic of fruits.
Answer:
Quality factors for fruits include the following— maturity, firmness, the uniformity of size and shape, the absence of defects, skin and flesh color. Many of the same quality factors are described for vegetables, with the addition of texture-related attributes such as turgidity, toughness, and tenderness.
Carnivore that feeds on primary consumers
Question 3 Multiple Choice Worth 3 points
(01.01 LC)
Which of the following is an example of a decomposer?
Question 5
Not yet answered
Marked out of 1.00
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Complete the following sentence: "The interior of living cells is more
than the exterior because more
ions are expelled than ions are taken in by the sodium-potassium pump."
Select one:
O a. electropositive. Nak
O b. electronegative, Na.
O C. electronegative, Na, K
O d. electropositive, Na+, K+
Question 6
Not yet answered
Marked out of 1.00
P Flag question
The high concentration of protons in the inner mitochondrial space relative to the mitochondrial matrix represent
O
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77°F
AGD (0)
10-25 PM
7/28/2021
Answer:
i really really need the brainly points
Explanation:
sry for this answer, i need the answers for myself
The biological classification of organisms is called taxonomy. Based on your own investigation, discuss the importance of taxonomy in biological research.
Answer:
mark me as brainliest
Explanation:
It helps to ascertain the number of living beings on Earth. More than one million of species of plants and animals have been discovered and classified so far. It aims to classify the living organisms. Millions of organisms are classified scientifically in categories, which helps to have a better understanding
The theory and practice of characterizing, naming, and classifying living things is known as taxonomy. Such research is critical for a basic knowledge of biodiversity and its preservation.
What is the importance of taxonomy in biological research?Taxonomy is important because:
It aids in estimating the number of living beings on the planet. nearly one million plant and animal species have been identified and classified till date.Its goal is to classify living things. Thousands of creatures are scientifically divided into categories, which aids in understanding.It aids our understanding of the characteristics seen in plants and animals.It depicts the physical development in chronological order.It provides an overview of the local fauna and flora, which aids in the identification of endemic species.learn more about taxonomy here:
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what is osmolarity of mammalian urine?
Answer:
The osmolarity of mammalian urine may vary over time. The osmolarity of mammalian urine varies little between species. Mammalian urine is always hyperosmotic to blood. The osmolarity of mammalian urine may vary over time.
Explanation:
I hope it will help you
The conversion of liquid water into gaseous water is called?
Answer:
vapourization
Explanation:
When the water is heated, it changes into water vapour which is called vapourization or sometimes we can also call it evaporation.
What can you learn about by studying DNA?
Answer:
whether these DNA tests can tell us much about an individual newborn's destiny, they are already a useful research tool that is providing new insights into how genes and environments interact, new avenues for understanding how mental illnesses (and other illnesses) develop and new pathways to explore potential ...
hope this will help you
mark me brainliest
Answer:
Explanation:
we can learn about DNA tests can tell us much about an individual newborn's destiny, they are already a useful research tool that is providing new insights into how genes and environments interact.
Human being get energy from
PLEASE HELP FAST PLEASE ASAP HURRY
Answer:
A number is a correct answer
What are the student’s observations and inferences before he starts his investigation?
Answer:
Hypothesis. A Hypothesis is an estimation of what might happen and the student's observation before moving on to investigate.
For every 100ml of deoxygenated blood delivers approximately _____ml of CO2 to the alveoli.
Answer:
For every 100ml of deoxygenated blood delivers approximately __4___ml of CO2 to the alveoli.
Answer:
4ml
Explanation:
For every 100ml of deoxygenated blood delivers approximately 4 ml of CO2 to the alveoli.
Hope it is helpful....