Answer:
they don't have back bone.
A stem cell is
A. A specialized cell that can divide limitlessly but cant differentiate
B. A specialized cell that can divide limitlessly.
C. An unspecialized cell that cant divide limitlessly nor differentiate into specialized cell.
Answer:
B
Explanation:
They divide to form more cells called daughter cells under proper conditions
all dicots are annuals true or false
Answer: The answer is false
A boatload of Swedish tourists, all of whom bear the MM blood group, is marooned on Haldane Island, where they are met by an equally sized population of Islanders, all bearing blood group NN. In time, the castaways become integrated into Island society. Assuming random mating, no mutation, no selection (based on blood group), and no genetic drift, what would you expect the blood group distribution to be among 4000 progeny of the new Haldane Island population
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].
what is the difference of biology and human and social biology?????
Answer:
Man's health (human biology) affects and effects change on society (social biology). Human and social biology scrutinizes the human body, disease, health, nature and the environment's influence on biology.
Answer:
Man's health (human biology) affects and effects change on society (social biology). Human and social biology scrutinizes the human body, disease, health, nature and the environment's influence on biology.
Hope this helped! Good luck! :)
Also do you think you could pls give me the brainly crown? Its totally fine if not! :)
Starch and protein digestion in a single stomach?
Answer:
Explanation: Protein digestion occurs in the stomach and the duodenum through the action of three main enzymes: pepsin, secreted by the stomach, and trypsin and chymotrypsin, secreted by the pancreas. During carbohydrate digestion the bonds between glucose molecules are broken by salivary and pancreatic amylase.
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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What are three ways in which bacteria obtain food
Answer:
The three ways by which bacteria obtain food are photosynthesis, chemosynthesis, and symbiosis.
Explanation:
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.
What is the biological impact of minimum catch sizes on a population of fish?
a. The population comes to be dominated by smaller, slower-growing individuals.
b. Older, less productive adults are removed, improving the population's health.
It applies a selective pressure for larger, faster growing fish.
d. The fish in the population produce more and healthier eggs to compensate.
C.
Please select the best answer from the choices provided
Ο Α
OB
Ос
OD
Answer:
answer is A.) The population comes to be dominated by smaller, slower- growing individuals
Which of the following shows the correct energy flow during aerobic respiration?
A. Lipids, Glucose, ATP, cells
B. ATP, Glucose, cells
C. Carbohydrates, Glucose, ATP, cells
D. Carbohydrates, Glucose, Lactic acid, ATP
The correct energy flow during aerobic respiration is ATP, Glucose, and cells. Thus, the correct option is B.
What is Aerobic respiration?Aerobic respiration may be defined as the process of conversion of glucose into carbon dioxide, water, and energy via the intermediate product of pyruvate in presence of air.
The release of energy in aerobic respiration is found to be greater as compared to anaerobic respiration.
This released energy is immediately used to synthesize a molecule called ATP via glucose that fuels the cells for metabolic activities.
Therefore, the correct option for this question is B.
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what is the source of food for germinating seed ?
You are in charge of genetically engineering a new organism that will derive all of its ATP from sunlight by photosynthesis. Will you put the enzymes of the citric acid cycle in this organism
Answer:
Yes.
Explanation:
I will put the enzymes of the citric acid cycle in this organism so that in the absence of sunlight, it also produces energy to grow quickly. Citric acid cycle is a series of chemical reactions which release stored energy through the oxidation of acetyl-CoA derived from carbohydrates, fats, and proteins. This cycle is also called Krebs cycle so incorporation of this cycle is good for the organism.
Which of the following is not a natural disaster ?
1 . Tsunami
2. Earthquake
3. Terrorism
4. Hurricane
Answer:
terrorism
Explanation:
everyhthing else is natural because hurricanres come from wind and tide overflow tsunamis are caused by earth quakes and earthquakes are natural
The one that follows that is not a natural disaster is the terrorism that is present in Option 3, as the terrorism is done by man while the natural disasters are caused by nature, such as the hurricane and earthquake.
What role does a natural disaster play in the ecosystem?Natural disasters are phenomena caused by nature, such as cyclones, earthquakes, and tsunamis, and they cause significant damage, which may be indirectly caused by man as man causes global warming, climate change, and so on. The terrorism is man-made or artificial, as certain people are involved and harm other people or the country.
Hence, the one that follows that is not a natural disaster is the terrorism that is present in Option 3, as the terrorism is done by man while the natural disasters are caused by nature, such as the hurricane and earthquake.
Learn more about the natural disaster here.
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Match the items of column 'A' with those of column 'B':
Turbidity
Reduces the light in the water column
Natural pollution
Choose
Nitrates and phosphates
Air pollution
Biochemical demand for oxygen
Contaminants into a natural environment
Chemical fertilizers
Pollution
Floating materials
Industrial wastes
Emissions pollution
Reduces the light in the water column
Concentration of hydrogen pH
Eutrophication
CFCS
New species invasion
Temperature
Harmful algal blooms
Oils fats, and foam
Please answer all parts of th Hypertrophication
Toxicity
Volcanic
In crude oil affect eggs and larvae of fish a
Atomic number
a. Hydrocarbon
Answer:
Match the items of column 'A' with those of column 'B':
Turbidity
Reduces the light in the water column
Natural pollution
Choose
Nitrates and phosphates
Air pollution
Biochemical demand for oxygen
Contaminants into a natural environment
Chemical fertilizers
Pollution
Floating materials
Industrial wastes
Emissions pollution
Reduces the light in the water column
Concentration of hydrogen pH
Eutrophication
CFCS
New species invasion
Temperature
Harmful algal blooms
Oils fats, and foam
Please answer all parts of th Hypertrophication
Toxicity
Volcanic
In crude oil affect eggs and larvae of fish a
Atomic number
a. Hydrocarbon
Which of the following processes releases heat?
Evaporation
Freezing
Melting
Sublimation
Answer:
Freezing
Explanation:
I got 123% on the quiz (Jokes aside yes it is freezing)
I need help with the work
Answer:
Gw
Explanation:
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:
5x_=1 jdjdjcjdjdjjdkd
Answer:
what...
Explanation:
give a reason why lack of roughage in diet often leads to constipation
Answer:
The main cause of constipation is intake of a low fiber diet.the bulk and soft texture of fibre helps prevent hard dry stools that are difficult to pass thereby reducing constipation.
I hope this helps
Which of these is not a component of a molecule of adenosine triphosphate (ATP)?
from brainly
cygilberts
Ambitious
34 answers
1.8K people helped
The right answer is Deoxyribose sugar
Explanation:
ATP consists of adenosine (this is an adenine ring and a ribose sugar) – and three phosphate groups (triphosphate).
Hope this helps
According to the phenotypic characters of pneumococcus considered in Griffith's
experiment of transformation, which of the following statements are correct? Choose the
correct option (i) Presence of slime layer (ii) Presence of capsule (iii) Absence of capsule
Answer:
发发发 发斯蒂芬
Explanation:
Presence of capsule. Therefore, option (B) is correct.
What was Griffith's experiment?Griffith's experiment of transformation is a landmark experiment in microbiology conducted in 1928 by British bacteriologist Frederick Griffith. The experiment aimed to determine the nature of the "transforming principle" responsible for transferring genetic material between bacteria.
Griffith used two strains of Streptococcus pneumoniae, one that was virulent and had a polysaccharide capsule (S strain) and another that was non-virulent and lacked the capsule (R strain). He found that when he injected mice with the heat-killed S strain and live R strain, the mice died, and live S strain was found in their blood. This suggested that the R strain had been transformed into the S strain, and the genetic material was responsible for the transformation.
Griffith's experiment provided the first evidence of bacterial transformation and paved the way for future research on the molecular basis of genetics.
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Describe what is happening in each of the phases of cellular division with what you observe from the photomicrographs in the lab instructions (describe and compare both).
Answer:
Interphase - There are 3 subphases called - G1 , S and G2 . In these subphases, DNA material is duplicated, and essential proteins for cell division are formed.
Prophase - the nuclear envelop and nucleolus disappears and condensed DNA forms chromosomes that are short in size.
Metaphase - the form of two sister chromatids are arranged in the middle of the metaphase/equatorial line. Two centrioles are found at two poles of the cell which initiate the formation of spindle fibers.
Anaphase - spindle fibers are formed and can be observed. These fibres attach with the kinetochore present in the centromere of chromosomes and these chromatids now start separated by pulling and pushing force of spindle fibers in opposite poles
Telophase - chromatids reach at destination pole and then nuclear envelop starts creating again in both poles. Nucleolus reappears and now in a cell two nucleus are found.
Students in a science class tested different plant seeds to determine how long it took each type of seed to fall from the second- story window of the school. They tested 100 seeds of each type in order to find out which type of seed had the the parent plant. The data they collected is shown in the chart below. which of these graphs best represents the students data?
walnut/0.6
maple/4.5
redoak/0.6
ash/2.0
pine/2.4
The correct answer is C.
Explanation
According to the information provided, a bar graph (option C) is needed since this type of graph allows us to express the variation in time, distance, temperature, quantities, weight, among others. In this specific case, a bar graph allows students to analyze the time that each type of seed takes to fall from the second floor. Additionally, this is the only graph that shows the data of the chart. For example, the first bar represents the black walnut and its time is 0,6 seconds, and this value is shown by the light blue bar in the graph, which represent the same seed. According to the above, the correct answer is C.
You prepare a gel mobility assay with the following samples:
Lane 1 Radiolabeled lacO DNA + Lac Repressor protein preincubated with allolactose
Lane 2 Radiolabeled lacO DNA + Lac Repressor protein preincubated with lactose
Lane 3 Radiolabeled lacO DNA + Lac Repressor protein missing its DNA binding domain
Lane 4 Radiolabeled lacO DNA + Lac super-repressor (Is) preincubated with allolactose
Lane 5 Radiolabeled lacO DNA + lac repressor protein
On which lanes do you expect to see two bands?
a. 3
b. Lane 1 and 2
c. Lane 2, 4, and 5
d. 2 and 4
e. 1 and 5
Answer:
c. Lane 2, 4 and 5.
Explanation:
Two bands will be seen when a fraction of DNA added to gel is bound by Lac repressor protein. Radiolabeled LacO DNA will have two bands. Lac repressor is encoded with lacl gene. It has binding ability due to allolactose formation.
an organism that uses oxygen to break down food to obtain energy
Can you help me with this please?
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
How does temperature and concentration of monounsaturated phospholipids change the rate at which molecules permeate the plasma membrane? Does this change occur equally for low permeability ions the way and large molecules as it does for gasses and high permeability molecules?
Describe the way that enzymes increase the rate of chemical reactions in the cell. What other factors related to enzymes can increase or decrease this rate and why?
Answer:
я не знаю ответа извините
Explanation:
Which BEST
describes the
amount of air in
a typical soil?
A. about 90%
B. hardly any
C. about 25%
Answer:
C
Explanation:
it's about 20%-30%, so C would appropriately about that range since it the best answer :)
hope it helps
hey everyone,what is greenhouse effect
Answer:
The greenhouse effect is a natural process that warms the Earth's surface.
When the Sun's energy reaches the Earth's atmosphere, some of it is reflected back to space and the rest is absorbed and re-radiated by greenhouse gases.
The absorbed energy warms the atmosphere and the surface of the Earth causing green house effect .
Answer:
The greenhouse effect occurs when radiation from a planet's atmosphere warms the planet's surface to a higher temperature than it would be without the atmosphere. Radiatively active gases emit energy in all directions from a planet's atmosphere.
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