Answer:
1000 MM individuals; 2000 MN individuals; 1000 NN individuals
Explanation:
The Hardy-Weinberg principle states that, under certain conditions, the frequency of alleles and genotypes in a sexually reproducing population will remain constant over generations. The Hardy-Weinberg assumptions include random mating, sexual reproduction, and the absence of evolutionary forces such as mutation, natural selection, and genetic drift (as in the example above). Under these conditions, the frequency of alleles and genotypes in a population will not change and tend to the equilibrium. In this case, under Hardy-Weinberg equilibrium, the frequency of the M allele or 'p' must be equal to 0.5, and the frequency of the N allele or 'q' must be equal to 0.5 (i.e., the sum of all allele frequencies in the population must be equal to 1). Moreover, the frequencies of the genotypes will be p², 2pq, and q² >> p² (MM genotype) = (0.5)² = 0.25; q² (NN genotype) = (0.5)² = 0.25; and 2 x p x q (MN genotype) = 2 x 0.5 x 0.5 = 0.50. In consequence, under Hardy-Weinberg equilibrium, in a population of 4000 diploid individuals (8000 alleles), we have
- 4000 M alleles (M = 0.5) and 4000 N alleles (N = 0.5);
- 1000 MM individuals [p² >> (0.5)² = 0.25]; 2000 MN individuals (2pq >> 2 x 0.5 x 0.5 = 0.5) and 1000 NN individuals [q² >> (0.5)² = 0.25].
Biology Question
If the frequency of an autosomal dominant allele is 0.6 . Calculate the frequency of recessive phenotype in a population of 6000
Answer:
First thing is that word “autonomous” given in the question is actually “ autosomal”.
Now let's start solution:
Frequency of dominant allele “A” is denoted by p.
Here p = 0.6
Frequency of recessive allele “a” is denoted by q.
According to Hardy - Weinberg equilibrium
p + q = 1
Hence q = 1 - p = 1 - 0.6 = 0.4
Now in question what is asked is about frequency of recessive phenotype (aa).
According to Hardy - Wienberg equilibrium frequency of recessive phenotype is q2 (q square).
Frequency of recessive phenotype is = q2 = 0.4 x 0.4 = 0.16
Now Number of recessive phenotype in a population of 10000 is
= frequency of recessive phenotype x total population
= q2 x 10000
= 0.16 x 10000
= 1600
I answer you other question.
I apologize. it's mistakenly
The most basic organization level of life is a ____________. A. membrane B. tissue C. cell D. organ
[tex]Hello[/tex] [tex]There![/tex]
The answer is...
C. Cell.
Hopefully, this helps you!!
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Our body needs both vitamin and mineral in a small quantity ,still they are important why?
Answer:
Vitamins and minerals are considered essential nutrients—because acting in concert, they perform hundreds of roles in the body. They help shore up bones, heal wounds, and bolster your immune system. They also convert food into energy, and repair cellular damage.
What test is used to ensure that the results you expect are the same as the results you are observing?
A. Epistatic test
B. Chi-squared test
C. Observational confirmation test
D. Punnett square test
Reset Selection
Answer:
b. Chi Squared
Explanation:
The chi squared test is best to find signifigance between differences in observed and expected values.
In a hydrogen ion pump, the energy is used to join small molecules together
to
make larger ones. Which factor most likely has the greatest effect on the
number of molecules mitochondria can produce?
Answer: The number of H+ ions moving down the channel
Explanation:
Which of these sentences is true with respect to species?
One Reason Why The Temperature needs To Be Kept Constant
Answer:
the latent heat as the heat supplied to increase the temperature of the substance is used up to transform the state of matter of the substance
Explanation:
It is due to the latent heat as the heat supplied to increase the temperature of the substance is used up to transform the state of matter of the substance hence the temperature stays constant. Hence the temperature remains constant as all the heat is used up and no external heat is released or absorbed
H. pylori cannot grow in other microenvironments of the human body because the conditions are unsuitable for its growth, but other species require different conditions.
a. True
b. False
Answer:
it should be an true statement
the citric acid cycle role is
Answer:
The citric acid cycle is the final common oxidative pathway for carbohydrates, fats and amino acids. It is the most important metabolic pathway for the energy supply to the body.
Please answer the following Anthropology questions fully, thoroughly, and in depth.
Part i. What adaptations are associated with an insectivory diet? Give an example of a primate that has this diet.
Part ii. What adaptations are associated with a gummivory diet? Give an example of a primate that has this diet.
Part iii. What adaptations are associated with a frugivory diet? Give an example of a primate that has this diet.
Part iv. What adaptations are associated with a folivory diet? Give an example of a primate that has this diet.
Part v. What adaptations are associated with vertical clinging and leaping? Give an example of a primate that practices this form of locomotion.
The examples that you need are 1-Gummivory: marmoset, tamarins; 2-Frugivores: spider monkey, ray-bellied night monkey (owl monkey); 3-Folivory: howler monkey, leaf monkey, colobine; and 4-Vertical clinging and leaping: lemurs and tarsiers
A gummivore is an animal (in this case, a primate) with a feeding strategy that depends on the sap or gum from trees. It is a type of diet that consists primarily of exudates. Some adaptations observed in these species include: 1-specialized anterior teeth in order to stab bark; 2-well-developed claws for clinging to trees
Frugivores primates have a feeding strategy mainly based on raw fruits. Some adaptations observed in these species include 1-low rounded molar cusps and 2-broad incisors.
Folivore primates (also called leaf-eating monkeys) have a feeding strategy mainly based on leaves. Some adaptations observed in these species include 1-narrow incisors, 2-broad molars (high shearing crests), 3-thin enamel molars.
Vertical clinging and leaping is an arboreal locomotor pattern practiced by lemurs and tarsiers. These animals push off from one vertical support with their hindlimbs to land on another vertical support. Some adaptations observed in these species include 1-hips, knees, feet, arms, and tails that facilitate climbing and leaping, 2-large hip extensor muscles and 3-specialized minor muscles.
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Which of the following words is generally used to describe what managers do as opposed to what leaders do b) Organize c) Inspire O d) Innovate
Organisrns that transfer diseases to hurnans are
O hosts
O pathogens
O parasites
O vectors
which enzyme breaks down lipids? which enzyme breaks down fats? which enzyme breaks down proteins?
Answer:
lipase enzymes, lipase, proteolytic enzymes
Answer:
for lipids and fat is same enzyme that is lipase enzyme. protease enzyme is for proteins.
provide the labels of the angiosperm flower shown below
Explanation:
1.peduncle
2.ovary
3.sepal
4.petal
7.stigma
12.receptacle
13.style
8.ovule with enbryo sac
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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Suggest how whooping cough spreads from person to person
Answer:
People with pertussis usually spread the disease to another person by coughing or sneezing or when spending a lot of time near one another where you share breathing space. Many babies who get pertussis are infected by older siblings, parents, or caregivers who might not even know they have the disease.
Explanation:
Answer:
Explanation:People with pertussis usually spread the disease to another person by coughing or sneezing or when spending a lot of time near one another where you share breathing space. Many babies who get pertussis are infected by older siblings, parents, or caregivers who might not even know they have the disease
Environmental scientists cite several challenges to protecting endangered species. Which of the following is NOT a difficulty faced by those working to protect an endangered species? Scientists who are focused on a single species may not recognize another species in peril. Extinction is natural in nature and its rate of occurrence has remained steady. Protection efforts may be slowed because of difficulty enforcing new regulations. Fundraising can be more difficult if the species is not perceived as valuable.
Answer:
Extinction is natural in nature and its rate of occurrence has remained steady.
Explanation:
As of right now we are going through what is known as the sixt extinction, and human acticity is one of the Main causes. (which has never happened before our time) Species are going extinct at a rapid rate, and considering how all life is connected it makes it easy to understand why the extinction rates can increase due to a biotic factor (niche) being "removed". (:
hey ,What is biological community answer it .Good night
Answer:
Community, also called biological community, in biology, an interacting group of various species in a common location. For example, a forest of trees and undergrowth plants, inhabited by animals and rooted in soil containing bacteria and fungi, constitutes a biological community.
Explanation:
hope this helps you pls mark as brainliest
Answer:
It's a group of various species in a common place.
For example: a forest of trees and undergrowth plants, lived in by animals and rooted in soil including bacteria and fungi, would make a biological community.
I hope this Helped!
Who are agree on xenotransplantation
Answer:
The people who agree on xenotransplantation are those who have severe problems such as heart problems and failure of the kidneys. Xenotransplantation, while it is still being experimented with, is a possibly life-saving option for people with these illnesses. Another reason why people agree on Xenotransplantation is that because it could potentially supply us individuals with an unlimited indefinite quantity of things such as cells, tissues, and organs for humans.
How does pollution affect biodiversity
Answer:
All forms of pollution pose a serious threat to biodiversity, but in particular nutrient loading, primarily of nitrogen and phosphorus, which is a major and increasing cause of biodiversity loss and ecosystem dysfunction. ... In addition, nitrogen compounds can lead to eutrophication of ecosystems.
Which of the following is an example of an enzymatic cycle?
Answer:
Catabolism
Explanation:
The process of catabolism degrades the bacterial and fungal enzymes into simple inorganic molecules.
A substance, without being a reactant, which speeds up a chemical process is referred to as a catalyst. Enzymes are known as catalysts for biological reactions in living organisms. Although ribonucleic acid (RNA) molecules behave as enzymes, they are usual proteins. Enzymes.
Enzymes carry out the essential role of reducing the activated energy of a reaction — that is, the amount of energy needed to start the process. Enzymes work by attaching and retaining reactant molecules so that the chemical bonding and bonding activities are carried out more easily.
Human being get energy from
State whether the following statements are true or False. Hormones in plants travel by the vascular bundle.
Answer:
True
Explanation:
In plants, hormones travel large throughout the body via the vascular tissue (xylem and phloem) and cell-to-cell via plasmodesmata. In contrast, many animal hormones are produced only in specific glands. Plants do not have specialized hormone-producing glands.
Ammonia produced in the intestines from the breakdown of proteins by bacterial enzymes is the primary source of plasma ammonia
Select one:
1.True
2.False
As water is cooled from 4° C to 0° C, its density
A. stays the same
B. decreases and increases
C. increases
D. decreases
Answer:
I believe that it increases (becoming more dense)
Explanation:
Well 0°C is freezing point so I think that in that state it will become a solid from a liquid and freeze into ice.
Which type of energy refers to the sum of potential and kinetic energies in the particles of a substance?
A. motion
B. stored
C. internal
D. heat
Answer:
A
Explanation:
Mechanical energy of a moving object the sum of its kinetic and potential energy
Which of these is a covalent bond in which the electrons are not evenly shared?
Answer:
polar covalent bond
Explanation:
DNA is a nucleic acid involved in heredity, or the passing down of genetic traits from one generation to the next. DNA consists of four different types of nucleotide monomers.Which part of the nucleotides' structure is responsible for the incredible variation that exists amongst all types of organisms
Nitrogenous base DNA consists of four unique nucleotides that each contain one unique nitrogenous base—adenine (A), thymine (T), cytosine (C), or guanine (G).
The specific arrangement of these four bases within the DNA of each organism gives that organism its unique traits; here are the arrangements:
-Adenine is paired with Thymine (think of A for apple and T for tree)
-Cytosine is paired with Guanine (think of C for car and G for garage)
search "DNA base pairs" and go to images for better understanding
The part of the nucleotides' structure is responsible for the incredible variation that exists among all types of organisms are Nitrogenous base DNA consists of four unique nucleotides that each contain one unique nitrogenous base—adenine (A), thymine (T), cytosine (C), or guanine (G).
Who is responsible for passing down of genetic traits from one generation to the next?DNA is a nucleic acid involved in heredity, or the passing down of genetic traits from one generation to the next. DNA consists of four different types of nucleotide monomers.
The specific arrangement of these four bases within the DNA of each organism gives that organism its unique traits and here are the arrangements are mentioned below:
Adenine is paired with Thymine (think of A for apple and T for tree)Cytosine is paired with Guanine (think of C for car and G for garage)Therefore, The part of the nucleotides' structure is responsible for the incredible variation that exists among all types of organisms are Nitrogenous base DNA consists of four unique nucleotides that each contain one unique nitrogenous base—adenine (A), thymine (T), cytosine (C), or guanine (G).
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g In order to stimulate the renin-angtiotensin-aldosterone system (RAAS), _______________ cells in the kidney detect _______________blood pressure which causes ________________ to be released from the juxtaglomerular cells.
Answer:
Macula densa cells, lower and renin.
Explanation:
Macula densa cells in the kidney detect lowers blood pressure which causes renin to be released from the juxtaglomerular cells which is an enzyme. The arterial cells observe the drop in blood pressure, and the decrease in Na concentration is transfer to them by the macula densa cells. The juxtaglomerular cells then release an enzyme called renin. Renin converts angiotensinogen which is a peptide, or amino acid derivative into angiotensin-1.
functions of insulin
Answer:
Insulin helps control blood glucose levels by signaling the liver and muscle and fat cells to take in glucose from the blood. Insulin therefore helps cells to take in glucose to be used for energy. If the body has sufficient energy, insulin signals the liver to take up glucose and store it as glycogen.
Explanation:
