Describe the experiments that revealed the structure of the genetic material. If a sample of DNA contains 8% adenine (A), then what are the percentages of thymine (T), cytosine (C), and guanine (G)

Answer:

option b. have complex grooming and courtship behaviors is correct answer.

Explanation:

According to a research paper published in American journal of primatology provided evidences that chimpanzee males, as promiscuous breeders with minimal costs to mating, should show little or no preference when choosing mating partners (e.g. should mate indiscriminately).

Reference: Proctor, D. P., Lambeth, S. P., Schapiro, S. J., & Brosnan, S. F. (2011). Male chimpanzees' grooming rates vary by female age, parity, and fertility status. American journal of primatology, 73(10), 989–996. doi:10.1002/ajp.20964

Answer: b) [tex]10^{4}M2[/tex] and [tex]-5.5 kcal/mol[/tex]

Explanation:

Equilibrium concentration of [tex]A[/tex] = [tex]10^{-2}M[/tex]

Equilibrium concentration of [tex]B[/tex] = [tex]10^{-2}M[/tex]

Equilibrium concentration of [tex]C[/tex] = [tex]10^{-2}M[/tex]

Equilibrium concentration of [tex]ABC[/tex] = [tex]10^{-2}M[/tex]

The given balanced equilibrium reaction is,

                            [tex]A+B+C\rightleftharpoons ABC[/tex]

The expression for equilibrium constant for this reaction will be,

[tex]K_c=\frac{[ABC]}{[A][B][C]}[/tex]

Now put all the given values in this expression, we get :

[tex]K_c=\frac{10^{-2}}{(10^{-2})^3}[/tex]

[tex]K_c=10^4M^2[/tex]

[tex]\Delta G^o=-2.303\times RT\times \log K_c[/tex]

where,

R = universal gas constant = 2 cal/K/mole

T = temperature = 300 K

[tex]K_c[/tex] = equilibrium constant = [tex]10^4[/tex]

[tex]\Delta G^o=-2.303\times 2\times 300\times \log (10^4)[/tex]

[tex]\Delta G^o=-5527.2cal/mol=-5.5kcal/mol[/tex]

Thus the equilibrium constant and the standard free energy of this association reaction at T=300 K are [tex]10^4M^2[/tex]  and [tex]-5.5kcal/mol[/tex]

Answer:

E.All chordates have ventral l, hollow nerve cord

D.All chordates are bilaterally symmetrical animals

Answer: Option C and E are correct.

Explanation:

Phylum molluscs are invertebrates species. They are the second largest phylum in the animal kingdom .Most mollusc live in marine and others live in freshwater and terrestrial habitat. Molluscs are ceoomates and they have three universal structures which are the mantle (breathing and excretion),radula and nervous system. The molluscs include snail, oyster,clams, periwinkles,octopus, squid e.t.c.

The larvae of molluscs are bilateral symmetrical while the adults are asymmetrical. Molluscs can produce shell, the presence of shell is an inclusion in the phylum.

Molluscs have three body part, the head, visceral mass and muscular foot

Answer: Options B and E are correct

Explanation: molluscs are found in freshwater, marine and terrestrial habitat. Some small species such as the land snail lives in dampy and dark places in terrestrial habitats. Molluscs are animal, are multicellular, have tissue, a digestive tract which includes mouth, stomach, anus and they also possess an oral structure called the radula. Some cephalopods do not have shells as they have lost their shells at one point in evolution and examples include octopuses. Lobsters are crustaceans belonging to the phylum arthropoda. Molluscs do have three main body parts head, visceral mass and the muscular foot.

Answer:

A. ability to infect a particular type of cell within the host.

Explanation:

In plants, tropism refers to the movement of a plant part in a particular direction in response to any stimulus. In microbiology, tropism refers to the ability of pathogen to infect a particular type of cell in the host.  

The tropism in which a particular cell is infected is called cell tropism. The tropism in which a particular tissue is infected then it is cell tissue tropism. Host tropism is also seen in pathogen when they infect specific host. Stimulus is required for any tropism. So the right answer is A.

Answer:

A) ability to infect a particular type of cell within the host.

Explanation:

Viral tropism refers to the ability of a virus to infect a particular cell, tissue and a host species.

The viruses have to interact with a variety of negative and positive factors in the cell so the viral tropism is determined at different levels of the viral-host interaction that is during replication and viral progeny production.

The viral tropism occurs at two-step and based on this is of two types- the receptor-independent tropism which takes place intracellularly and the receptor-dependent tropism which takes place at the cell surface.

Thus, Option-A is the correct answer.

Answer:

Biomass: can be defined as the total number of viable organisms, such as plants, animals in a given area.

While,

Standing crop: can be defined as the amount of biomass at a particular time or it is the amount of above-ground plant biomass.

Answer:

The correct answer is option D) "Ethanol / industrial microbiologists".

Explanation:

Ethanol is one of the most common components of biofuels in the United States. Ethanol is used in biofuels for its property of producing oxygen when burned, which makes car's engine to burn fuel more efficiently. Ethanol is produced by microbial fermentation of glucose from cellulose or cornstarch, and industrial microbiologists are the professionals responsible of making this product commercially available.

Answer:

D) Ethanol / industrial microbiologists

Explanation:

In this case of bio-fuels we have already prepared bio ethanol that is being produced after a process called as fermentation run by utilization of specially designed strains of micro-organisms. This simple fermentation is on a small scale and is not feasible for let say the whole country. Now, the industrial microbiologists are the only expert people that shall advance in some kind of already present hyper fermentation techniques that'll eventually give us enough quantity of ethanol to be used commercially.

The given question is incomplete. The complete question is:

Sort the molecules in the glycolysis pathway based on whether they are intermediates or products in the first half of the pathway that requires energy, or are intermediates in the second half of the pathway that produces energy. Drag the appropriate items to their respective bins.

glucose-6-phosphatefructose-6-phosphatefructose-1,6-bisphosphateglyceraldehyde-3-phosphatedihydroxyacetone phosphate1,3-bisphosphoglycerate3-phosphoglycerate2-phosphoglyceratephosphoenolpyruvate

Answer:

The procedure in which glucose gets dissociated into pyruvate and generates energy in the form of ATP is known as glycolysis. The process takes place in the cytosol and can get differentiated into two phases, that is, the preparatory phase that required energy in the form of ATP and the pay-off phase in which the production of ATP takes place.  

Post glycolysis, the process of cellular respiration takes place in which the eventual outcomes, that is, two pyruvates are further used. Of the mentioned intermediates, the intermediates that are used in the first half of the pathway are glucose-6-phosphate, fructose-6-phosphate, fructose-1,6-bisphosphate, dihydroxyacetone phosphate and glyceraldehyde-3-phosphate.  

The intermediates used in the second half of the pathway are 1,3-bisphosphoglycerate, 3-phosphoglycerate, 2-phosphoglycerate, and phosphoenolpyruvate.  

Answer: It will take 1.7 hours. If it isn't correct then please feel free to delete this. If it is however please mark me brainliest. Thanks!

Answer: Option D) the climate of the community in which the species mentioned above inhabit

Explanation:

Abiotic refers to non-living factors. Thus, the abiotic factors in an habitat where living organisms dwells include the atmospheric conditions such as climate, weather; inorganic nutrients such as phosphorus, carbon compounds, water etc present in the soil of such habitat.

Thus, Option D represent the abiotic factor

Answer: This moth has evolved

Explanation:

Evolutionary patterns show different patterns. One of the patterns is that: one species gradually transform into another species.

The above mentioned pattern is responsible for the moth's ability to lay more eggs and have more surviving offspring than other moths in the area. The moth is gradually transforming, so also its offspring will inherit these new traits, and it would further demonstrate evolution of one species.

Thus, it can be concluded that the unique ability of this moth over other moths is because the moth has evolved.

Answer:

II. There is an electronegativity difference between the bonded atoms.

Explanation:

In a compound that has a number of individual dipoles(weaker force of attractions between compounds). There exist a weaker intermolecular force of attraction ( Vander waal's forces), these force of attraction arises from fluctuating dipoles in atom molecules brought by the movement of electrons around the atomic nucleus. They possess weak electronegativity difference between the bonded atoms because they have less energy to break these weaker bonds.

Also, individual dipoles can interact with each other in an way which will also increase the boiling point and as boiling point increases there is decrease in electronegativity difference between the bonded atoms.

On the other-hand, A polar compound are ionic compounds and the bonding between them are electrovalent bonding because of their high melting and boiling points. They possess high electronegativity.

Answer:

The correct answer is 27 residues.

Explanation:

In an antiparallel beta sheet, there is a repeat for every 7 Angstrom units and the number of residues is two for every repeat. In an antiparallel sheet that exhibits a length of 94.5 Angstrom, the number of residues can be determined by using the following formula:  

2 residues per every 7 Angstrom,  

Length of the given segment is 94.5 Angstrom

Number of residues in this segment are:  

= (number of segments) * (residues per segment)

= (94.5/7) * 2

= 13.5 * 2

= 27 residues.  

Thus, the number of residues in this segment is 27.  

Beta sheets are a type of secondary structure in proteins where segments of the polypeptide chain are aligned side by side. The number of residues in a segment of an antiparallel beta sheet can be calculated using the formula number of residues = length / distance between adjacent residues.

Beta sheets are one of the two major types of secondary structure in proteins, along with alpha helices. In a beta sheet, segments of the polypeptide chain are aligned side by side, forming a sheet-like structure. The length of a segment in an antiparallel beta sheet can be calculated using the formula length = number of residues * distance between adjacent residues. In this case, the length of the segment is given as 94.5 Å, so we can rearrange the formula to find the number of residues: number of residues = length / distance between adjacent residues.

It's important to note that the distance between adjacent residues in a beta sheet can vary, but a common value used for calculations is 3.5 Å. So, using this value, we can calculate the number of residues: number of residues = 94.5 Å / 3.5 Å = 27 residues.

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Answer:

Borneo Island was once home to one of the most majestic forests in the world. During the 80s and 90s, Borneo underwent a profound transformation. Their forests were demolished at an unprecedented speed. Its rainforests ended in countries like Japan and the United States in the form of garden furniture, paper pulp and chopsticks. Today, the remaining forests are threatened by the new biofuel market, especially palm oil. As a result, large tracts of land are being transformed to oil palm plantations.

Explanation:

The jungles of Borneo were considered one of the wildest and pristine jungles of the planet, home to nomadic tribes and important populations of orangutans, pygmy elephants and rhinos. Currently, the traditions of these tribes have disappeared, rhinos are almost extinct, and orangutans and elephants are in danger. Otherwise, the rainforests of Borneo have gone from being a net carbon sink, which absorbed greenhouse gases, to being a carbon source, thus contributing to climate change along with deforestation and fires.

Conservation is a priority in Borneo, especially in biologically diverse regions that have escaped intensive logging and fires. The initiative called "Heart of Borneo" is an example of what can be achieved. It is essential that forests be restored. The use of native tree species should be encouraged through financial incentives and education programs, especially with the help of external governments, NGOs and private foundations. In addition, there is a possibility that under future climate agreements, reforestation could pay direct dividends stimulating the local economy and entrepreneurship in the villages.

Answer:

a) mRNA:        AUG  ACC  GGC  AAU  CAA  CUA  UAU  UGA

b) Protein: Methionine, Threonine, Glycine, Asparagine, Glutamine, Leucine, Tyrosine,  Stop.  

Explanation:

The Question was missing the DNA Strand. I have looked up the question and found this to be the DNA strand related to this question. A picture for converting codons to amino acids is also attached. A protein is simply a strand of Amino Acids.

DNA (3 to 5 sequence):

TAC    TGG   CCG    TTA    GTT   GAT    ATA    ACT

a).  Transcribing to mRNA (5 to 3 sequence):

AUG   ACC   GGC   AAU   CAA   CUA   UAU   UGA

b).  Related Amino Acids:

 1st Codon:  Methionine

2nd Codon:  Threonine

3rd Codon:  Glycine  

4th Codon:  Asparagine  

5th Codon:  Glutamine  

6th Codon:  Leucine  

7th Codon:  Tyrosine  

8th Codon:  Stop Codon

c.) DNA with Transition Mutation at Position 11 (5 to 3 sequence):

In genetics, Transition is a point mutation which converts a purine nucleotide to another i.e. A ↔ G, or a pyrimidine nucleotide to another pyrimidine i.e. C ↔ T. Hence, after transition at 11 position, T will be changed to C.

TAC    TGG   CCG    TCA    GTT   GAT    ATA    ACT

Transcribing to mRNA (5 to 3 sequence):

AUG   ACC   GGC   AGU   CAA   CUA   UAU   UGA

Related Amino Acids:

 1st Codon:  Methionine

2nd Codon:  Threonine

3rd Codon:  Glycine  

4th Codon:  Serine  

5th Codon:  Glutamine  

6th Codon:  Leucine  

7th Codon:  Tyrosine  

8th Codon:  Stop Codon

Answer:

Part A

[tex]0.108[/tex]

Part B

B.1

Fifty Percent

B.2

[tex]16.4[/tex] %

Explanation:

Part A

Given -

Both the parents are heterozygous for the sickle cell anemia

Let the genotype of the parents be

Ss

where S is the allele for presence of sickle cell anemia

and s is the allele absence of sickle cell anemia

If the above two parents are crossed, following offspring are produced

Ss * Ss

SS, Ss, Ss, ss

The probability of children with sickle cell anemia is equal to

[tex]\frac{3}{4}[/tex]

The probability of children with sickle cell anemia is equal to

[tex]\frac{1}{4}[/tex]

Probability that three of the children have sickle cell anemia and four of the children are healthy

[tex]\frac{7!}{3! * 4!} * \frac{3}{4}^4 * \frac{1}{4}^3\\ = 35 * \frac{81}{256 *64} \\= 0.108[/tex]

Part B

B.1

Hh * hh

Hh, Hh, hh, hh

The first child of the parent has a probability of fifty percent to have Huntington\'s disease

B.2

[tex]\frac{7!}{2!*5!} * \frac{1}{2}^2 * \frac{1}{2}^5[/tex]

[tex]= 16.4[/tex]

Answer:

Communication can be complicated in any organization. This is due to the fact that discussing peer to peer in a corporate office, of the institution, is going to be a lot unlike discussing peer to peer in a low-level office setting.

Also, cultural diversity can take a form in how efficient communication is. It is also very important that varying forms of communication are employed. Businesses that consists of many varying departments won't be dependent wholly on face to face communication. Therefore, it is necessary to have other forms of communication being utilized often. Examples of such Forms: email, phone, radio, etc.

In order to establish an effective route of communication that each sector of the organization feeds from, skilled communication professionals is necessary to be placed at every sector or place of business. This will permit corporate to possess “eyes on the target” per say. In other words, they can possess actual people all through every department or unit that can convey messages back and forth. International seems to do a good job at this. Their communications directors are involved with the relay of messages, permitting corporate’s suggestions which is then tailored to fit their particular department

Answer:

The sandy variation could be as a result of a homozygous recessive allele at one of two varying genes in these two different types of sandy pigs. Let's refer to them as genes A and B.

A kind of sandy pig may be aaBB and the variety AAbb. The F1 generation in this cross produces heterozygotes for both genes to give all red. This indicates that the A and B alleles are dominant. In the F2 generation, 6 out of 16 give us homozygous for either the aa or bb alleles and produces sandy.

Out of 16 will we have one that would be doubly homozygous and be white. The other 9 will possess at least one dominant allele for both genes.

Pedigree attached

Answer:

1)  2/3

2) 1/2

3) 1/12

Explanation:

1) Individual III.1 is affected. He inherited one r allele from his carrier mother, but must also have inherited another allele from his father, who's genotype we are not told. His father is unaffected, but because III.1 is affected, must be a carrier with the genotype Rr. Therefore, the parents of III.3 have are Rr x Rr. We can carry out a punnett square to assess probabilities

           R     r

 R     RR    Rr

 r        Rr     rr

The probability of him inheriting an r from either parent is 1/2 (as there is 50% chance of being Rr in the punnet square). However, we already know that he is not rr (he is unaffected and has a R allele), so that means 2/3 of the options involve him being a carrier

2) The probability III.4 is a carrier can be calculated because we have the genotypes of her parents, RR x Rr

           R     R

 R     RR    RR

 r        Rr     Rr

The probability III.4 is a carrier is 1/2 but with no possibility that she is affected.

3) If both parents were carriers, there would be a 1/4 chance that individual IV.1 would be affected, see punnet square below:

           R     r

 R     RR    Rr

 r        Rr     rr

Only rr genotype is affected, so the probability is 1:4. However, there is only 1/2 probability that their mother is affected and 1/2 that their father is. To work out their probability based on this we have to multiply the probabilities: 2/3 x 1/2 x 1/4 = 1/12

The probabilities that the parents III-3 and III-4 are carriers of a recessive genetic trait are both 1/2, while the probability of their child inheriting the condition is 1/4.

This question is asking about the probability of individuals III-3 and III-4 being carriers of a recessive genetic condition and the probability of their child being affected. Since we're dealing with a recessive genetic trait, if both parents are heterozygous carriers (Rr), the chance that any one of their children will be a carrier is 50% or 1/2. This applies to the probabilities that III-3 and III-4 are carriers. The chance that their child IV-1 will be affected (rr) is 1/4, as in classical Mendelian inheritance, when both parents are heterozygous carriers, there is a 25% chance for the child to inherit two recessive alleles.

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Answer:. Nondisjunction during meiosis I in either the male and female gamete

The failure of   homologous chromosome to separate at mitosis,  or failure of sister chromatids  to separate at Anaphase of  meiosis I or ii is called  non-disjunction.

 In this  cell division anomaly the  unsegregated chromosomes  are  separated  into  cells as single chromosomes, therefore extra copies are  therefore inherited.This  result is one of the cell having an extra copies of  chromosomes than others( chromosomes 21.

If this extra  chromosome copies  is segregated into a sperm or egg, the resulting zygote cells will   have extra  copy as  chromosome 21, and hence Down syndrome.

→It is more common in Meiosis 1 than ii,because of longer duration of the first stage of division than meiosis ii,given rooms to errors.

→Most common in women(with age increase)  than men because of exchange of telomeres in   them,which  worsen with age. Their Meiotic mechanism  which   weaken with age, and  more prone to errors compare to men,

Explanation:

Answer: Fluoroscopy is used as positioning device prior to obtaining film radiographs. It is used for examination where the type of projection and habitus if the patient present difficulties

It is not an acceptable practice because it exposes the skin and underlying tissues to radiation induced injuries and there is a possibility of future occurances of radiation induced cancer in the future

Explanation:

Structures does not exist as a pair d. The urethra Therefore , d. The urethra is correct .

Explanation:

a. The Kidney:

The kidneys are paired organs in the human body, each individual having two kidneys.

They are essential in the urinary system, primarily responsible for filtering waste products and excess substances from the blood to form urine.

b. The Sphincter Muscle:

Sphincter muscles are circular muscles that surround various openings in the body, including the digestive and urinary tracts.

While there are multiple sphincter muscles, they exist as pairs, serving to regulate the flow of substances through the associated openings.

c. The Ureter:

The ureters are a pair of tubes that carry urine from the kidneys to the urinary bladder.

Humans typically have two ureters, one connected to each kidney, allowing the flow of urine from both kidneys to the bladder.

d. The Urethra:

Unlike the kidneys and ureters, the urethra is a single tubular structure. It is the tube that carries urine from the urinary bladder to the exterior of the body.

For such a more question on urethra

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d). The urethra is the correct answer because it is a singular structure, unlike kidneys and ureters, which exist in pairs. Sphincter muscles are specialized muscles and not paired organs.

The question asks us to identify which of the following structures does not exist as a pair: The kidney, the sphincter muscle, the ureter, or the urethra.

Thus, the correct answer is d. The urethra, as it is a singular structure.

Answer:

Tetany

Explanation

The refractory period in the heart refers to the shortest interval between consecutive contraction conducted impulses out of the cardiac muscle.

It is important because it allows adjustments to stimuli and limits amount of action potential sent per minute.

Without a plateau extending the refractory period in the cardiac muscles.

Sarcomeres might be stimulated to contract instead of relaxing leading to sustained contraction also called Tetany.

Tetany is an involuntary muscle cramp or contraction caused by overly stimulated nerves.

Answer:22.180

Explanation:divide the two together then use the first three number after the decimal.

.

Answer:

Explanation:

Let's start by stating that the work done here would be determined by taking into account the following:

-External forces

-Change in kinetic energy

Work Energy theorem is fully going to guide the solution to this problem.

Let's go now:

Mass of car = 1475kg

Velocity of car= 66.5km/h

Velocity = 66.5km/h * 0.278

V = 18.472 m/s

Kinetic Energy (K.E) = 1/2(mv^2)

K.E = 0.5 * 1475 * 18.472^2

K.E = 737.5 * 341.214

K.E = 251645.90 kgm^2/s^2

1kgm^2/s^2 = 1 Joule

So therefore,

K.E = 251645.90 Joules

K.E = 60.144KCals

The 'lub-dub' sound we hear is the sound of the heart valves closing, not the contraction of the heart muscle. The 'lub' is the sound of the mitral and tricuspid valves closing, while the 'dub' is the sound of the aortic and pulmonic valves closing.

The statement 'a. The "lub-dub" sound originates from the physical act of the heart contracting' is NOT true regarding the sounds associated with the heart. The "lub-dub" sound we hear is actually made by the closing of heart valves, not the contractions of the heart muscle itself. Specifically, the "lub" is the sound of the mitral and tricuspid valves closing at the beginning of systole (an active phase when your heart pumps blood), while the "dub" is the sound of the aortic and pulmonic valves closing at the beginning of diastole (a resting phase when your heart fills with blood).

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Final answer:

The incorrect statement is that b. the "lub-dub" sound originates from blood turbulence  through the heart. These sounds are actually the result of the closing of the AV valves (for "lub") and the semilunar valves (for "dub"), which can be heard with a stethoscope.

Explanation:

The question is asking which statement is NOT true about the origins of the heart sounds. The answer to this is b, "The "lub-dub" sound originates from blood turbulence through the heart." It is not correct because the "lub-dub" sounds, known medically as S1 and S2, are primarily caused by the closing of the heart valves, not by blood turbulence.

The "lub" or the first heart sound, S1, occurs when the atrioventricular (AV) valves close at the beginning of ventricular systole, while the "dub" or the second heart sound, S2, happens when the semilunar (SL) valves close at the end of ventricular systole. When a medical practitioner uses a stethoscope to listen to the heart, these sounds provide critical information regarding the heart's condition.

Final answer:

A nonsense mutation in the mRNA sequence can introduce a stop codon, prematurely ending the translation process and therefore shortening the protein. The mutated mRNA sequence for the given DNA would have a point mutation turning the second codon into a stop codon, thus producing a truncated and likely non-functional protein.

Explanation:

To induce a point mutation that shortens the protein, you would create a nonsense mutation in the mRNA sequence. For the original DNA sequence TAC CGA ACC TTA, the corresponding mRNA sequence is AUG GCU UGG AAU. If we introduce a point mutation in the second codon to change it to a stop codon, we can use the stop codon UAA. The mutated mRNA sequence would be AUG UAA UGG AAU. Here, UAA is a stop codon and will cause the translation process to terminate prematurely, therefore shortening the resulting protein.

Nonsense mutations convert a codon that can specify an amino acid into a stop codon (UAA, UAG, UGA), leading to premature termination of translation. This type of mutation can have a profound effect on the protein, as it will be truncated, likely destroying its normal function. This is because the ribosome will cease to translate the mRNA into protein upon reaching the stop codon, resulting in an incomplete and non-functional protein product.

Mutations can profoundly impact an organism by altering its proteins. Depending on where the mutation occurs, the consequences may be mild or severe. Inducing such mutations intentionally can be a valuable tool for research but can be detrimental when they occur naturally and disrupt the organism's biological functions.

The genotype ratio for a monohybrid cross of heterozygous pea plants with round (R) and yellow (Y) seeds is 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green. This ratio comes from the dihybrid cross of RrYy plants, reflecting the principle of independent assortment.

In a monohybrid cross of heterozygous pea plants with round (R) and yellow (Y) seeds, the genotype ratio you are asking about is derived from the principle of independent assortment and dominance, leading to a phenotypic ratio of 9:3:3:1 in a dihybrid cross. When focusing on one trait, a monohybrid cross, the ratio simplifies to a 3:1 ratio because each trait segregates independently.

For the dihybrid cross of RrYy (round, yellow) × RrYy, the ratios of expected phenotypes are 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green, which corresponds to the same proportions in genotype.

Therefore, the correctly completed sentence would be a genotype ratio of 9 round yellow (RRYY, RRYy, RrYY, RrYy) : 3 round green (RRyy, Rryy) : 3 wrinkled yellow (rrYY, rrYy) : 1 wrinkled green (rryy).

Final answer:

In a mono hybrid cross of heterozygous pea plant with round (R) yellow (Y) seeds, the genotype ratio is 9:3:3:1 for round, yellow seeds : round, green seeds : wrinkled, yellow seeds : wrinkled, green seed.

Explanation:

The mono hybrid cross of a heterozygous pea plant with round (R) yellow (Y) seeds would result in a genotype ratio of 9 round, yellow seeds : 3 round, green seeds : 3 wrinkled, yellow seeds : 1 wrinkled, green seed. This ratio is obtained by applying the product rule to the independent assortment and dominance of the alleles for seed shape and color.

Final answer:

During metaphase of mitosis and metaphase I of meiosis, cells have 46 chromosomes. Following telophase of mitosis, cells also contain 46 chromosomes. However, after telophase I and telophase II of meiosis, cells contain 23 chromosomes.So ,option b) metaphase I of meiosis is correct.

Explanation:

Human cells normally have 46 chromosomes, comprising 23 pairs, with one set from each parent. This diploid state (2n) changes during the process of cell division.

a) metaphase of mitosis: Each cell has 46 chromosomes. The chromosomes align in the center of the cell, but each one still consists of two sister chromatids at this stage.

b) metaphase I of meiosis: Each cell also has 46 chromosomes. However, these are in the form of 23 pairs of homologous chromosomes that will be separated in the next phase.

c) telophase of mitosis: The cell still contains 46 chromosomes, but they are now being divided into two new daughter nuclei.

d) telophase I of meiosis: The two cells created after this phase each have 23 chromosomes, as the homologous pairs have been separated.

e) telophase II of meiosis: After this final phase of meiosis, there are four cells, and each cell has 23 chromosomes—these are the haploid gametes.

Dendrites

hope this helps

Answer: the correct answer is dendrite.

Explanation: Dendrite is actually what receive signals from axon of other neurons, synapse is an area which joins axon and Dendrite. Signals move across the axon in the form of action potential, coming from the cell body.

Answer:

The correct answer is the couple: 0.010 M NaCl | 0.010 M CH₃COOH

Explanation:

A semipermeable membrane allows to pass through solvent molecules but not solute molecules. In this case, all solutions have the same molarity (M) but they do not have the same quantity of solute particles because some of them dissociate in two or more ions (van't Hoff factor i is higher than 1) . Osmotic flow proceeds always from the side with lower concentration of solute (with more solvent molecules) to the side with higher concentration of solute. For each pair of solution, we have to determine the number of particles of solute or the van't Hoff factor (i). If the right side has the lower concentration of solute (higher i), the osmotic flow will proceed from the right to the left.

0.010 M NH3(aq) |  0.010 M NaCl(aq): NH₃ is a nonelectrolyte (i=1) and NaCl has i= 2. Osmotic flow proceeds from left to the right.

0.010 M NaCl(aq) |  0.010 M FeCl3(aq): NaCl has i=2 and FeCl₃ has i=3 (it dissociates in Fe⁺ and 3 Cl⁻). Osmotic flow proceeds from left to the right.

0.010 M NaCl(aq) |  0.010 M CH3COOH(aq): NaCl has i=2 and CH₃COOH is a nonelectrolyte (i=1). The lower concentration is in the right side, so the osmotic flow proceeds from right to left.

0.010 M NaCl(aq) |  0.010 M CaCl2(aq): NaCl has i=2 and CaCl₂ has i=3 (it dissociates in Ca⁺ and 2 Cl⁻). The lower concentration is in the left side so the osmotic flow proceeds from left to right.

0.010 M CaCl2(aq) |  0.010 M FeCl3(aq): CaCl₂ has i=3 and FaCl₃ has i=4 (it dissociates in Fe⁺ and 3 Cl⁻), so the lower concentration is in the left side and osmotic flow proceeds from left to right.

In the given case, the osmotic flow proceeds from the right to the left arm - 0.010 M NaCl(aq) | 0.010 M CH3COOH(aq)

It is the spontaneous flow of molecules of the solvent through a semipermeable membrane. The flow would be for the zone of high concentration in order to achieve equal concentration on both sides.

Since we have to apply the pressure on the left arm of the tube to stop osmosis. It indicates that the solution in the left arm is hypertonic to the one in the right arm of the tube.

Learn more about osmotic pressure:

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