If thymine makes up 25% of the DNA nucleotides in the genome of a plant species, what are the percentages of the other nucleotides in the genome?

Answers

Answer 1

Answer: Adenine, Guanine and Cytosine takes 25% each

Explanation:

Recall that the sum of all nitrogenous bases in the DNA nucleotide is equal to 100%. And specific base pairings of Adenine to Thymine (A=T), and Cytosine to Guanine (C=G) must be equal.

So, the percentage of Adenine equal thymine, and that of cytosine equals guanine.

Now, A + T + C + G = 100%

So, if thymine makes up 25% of the DNA nucleotides, Adenine is also 25%.

Then, 25% + 25% + C + G = 100%

50% + C + G = 100%

C + G = 100% - 50% = 50%

Thus, divide 50% by 2 to obtain the individual percentage of cytosine and guanine. Each will take 25%

Answer 2

Final answer:

Using Chargaff's rules, if thymine makes up 25% of the DNA nucleotides in a plant species' genome, adenine also represents 25%, with guanine and cytosine each constituting 25% as well, ensuring all nucleotide percentages total 100%.

Explanation:

If thymine makes up 25% of the DNA nucleotides in the genome of a plant species, the percentages of the other nucleotides in the genome can be deduced using Chargaff's rules. According to these rules, in a DNA molecule, the amount of adenine (A) equals the amount of thymine (T), and the amount of guanine (G) equals the amount of cytosine (C). This principle of base pairing dictates that if 25% of the nucleotides are thymine, then adenine also makes up 25% of the nucleotides due to the A=T rule.

Since thymine and adenine together account for 50% of the nucleotides, the remaining 50% must be divided equally between guanine and cytosine, following the G=C rule. Thus, both guanine and cytosine make up 25% of the nucleotides each.

In summary, if thymine constitutes 25% of the nucleotides in the DNA of a plant species, then adenine, guanine, and cytosine also represent 25% each respectively, ensuring the total percentages add up to 100%.


Related Questions

Which of the following is a true statement about antibiotic resistant bacte



they attack the immune system



they can spread through the air



they are not affected by antibiotics



they only attack human, not animals

Answers

Answer: They are not affected by antibiotics.

Explanation:

Antibiotics resistant bacteria are bacteria that are resistant or not affected by antibiotics rather they develop adaptations to antibiotics by living and surviving in the presence of antibiotics.

They resist to antibiotics by changing in a way that reduces the effectiveness of antibiotics. Examples include penicillin resistant enterococcus, methicillin resistant staphylococcus aureus e.t.c.

Answer:

they are not affected by antibiotics

Explanation:

Resistance Bacteria prevent antibiotics from affecting them. They build immune against the activity of the antibiotics in there system by preventing its function.

This can be done by pushing the antibiotic out of the system reducing its concentration. They also change membrane system and preventing permeability of the antibiotic in the membrane. Some bacterial also destroy the active ingredient in the antibiotics thereby preventing the activities.

Hence antibiotic resistance bacteria are not affected by the antibiotic by inhibiting there activities.

A researcher decides to investigate whether there are any significant changes in personality from early adulthood through late adulthood. The researcher begins with a sample of 21-year-olds and plans to follow this group until they are 80 years old. The type of research design is:

Answers

Answer:

Longitudinal research design

Explanation:

Longitudinal research design is an observational design where data is collected for the same subjects over a long period of time which can range from months to years.

Therefore, in our case the researcher begins with a sample of 21 year olds and is planning to follow this group until they are 80 years in order to investigate whether there are any significant changes in personality from early adulthood through late adulthood.

This makes his research design to be Longitudinal research.

Mutations in which of the following genes lead to transformations in the identity of entire body parts?A) morphogensB) segmentation genesC) egg-polarity genesD) homeotic genesE) inducers

Answers

Answer: Option A

Explanation:

A morphogen gradient is an important concept in case of the developmental biology. It produces the signals from the part of embryo which decides the fate of development of the parts.

The differentiation of the cell takes place by the help of the signals produced by the embryo in order to decide where will which part of the embryo grow.

The surrounding cells receive the signals and the development of entire body parts takes place.

So, the mutation in this gene will transform the entire body part.

Answer: Option A.

Explanation:

Morphogens are signaling molecules that originate from a specific location that spread from their origin via diffussion to form a concentration gradients acting at long distances to induce responses in cells base on the gradients the cells interact with. It determines the pattern of tissue development and morphogens gradients cause the differentiation of unspecialized stem cells into different cell types.

You fill a shallow pan with water and place a drop of red ink in one end of the pan and a drop of green ink in the other end. Which of the following is true at equilibrium? a. The red ink is uniformly distributed in one-half of the pan, and the green ink is uniformly distributed in the other half of the pan. b. The red and green inks are uniformly distributed throughout the pan. c. Each ink is moving down its concentration gradient. d. The concentration of each ink is higher at one end of the pan than at the other end.

Answers

Answer is (B)

Explanation:

When we fill a shallow pan with water and place a drop of red ink in one end of the pan and a drop of green ink in the other end we see that the red nd green inks are uniformly distributed throughout the pan. This is due to the Diffusion of red ink and green ink in the water. Diffusion is the process where the movement of particles takes place from the region of higher concentrations to the region of lower concentartions. Until the concentration of substances is uniform throughout, this phenomena continues.

Final answer:

At equilibrium, the red ink is uniformly distributed in one-half of the pan, and the green ink is uniformly distributed in the other half of the pan.

Explanation:

At equilibrium, the correct statement is:

a. The red ink is uniformly distributed in one-half of the pan, and the green ink is uniformly distributed in the other half of the pan.

Equilibrium is reached when the concentration of each ink becomes uniform throughout the pan. This happens because the molecules of the red and green ink are constantly moving due to Brownian motion. Over time, they will mix and spread evenly, resulting in a uniform distribution of the inks in the pan.

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"A parakeet with green plumage is crossed to a white parakeet. The progeny are all green. Crossing these progeny together gave the following offspring: 3 white, 8 blue, 29 green, and 9 yellow. Propose a genetic hypothesis to explain the results. -g

Answers

Answer: This is due to the theory of independent assortment of gene.

This means that gene separate and assort with each other, independently. There a gene that is recessive in the parents may not be recessive in the offspring.

A recessive gene is the gene that makes up the physical appearance of the offspring. While a dominant gene is the gene that exist in the offspring, but is not among the gene that makes up the physical appearance of the offspring.

For the two pure stock parent plants, their have yellow and blue as the dominant gene, while green and white are the recessive gene.

That's is why the yellow and blue gene has a tendency of occurrence in future generations.

For their first progeny which appears to be all green, is because the green gene has sorted themselves independently to be recessive, while others become dominant.

For the second progeny, the recessive gene from the parent plants which was carried by the offspring as dominant gene, now became recessive in some plants.

The genetic pattern observed in parakeets suggests the presence of two gene pairs determining plumage color, influenced by dominance and epistasis. A Punnett square could be used to predict offspring ratios, reflecting complex inheritance patterns.

The question describes a genetic phenomenon observed in parakeets when crossed, which showcases inheritance patterns akin to those discovered by Gregor Mendel. A green parakeet was crossed with a white parakeet, and all progeny were green. When these progeny were crossed, various colors were observed, suggesting a relationship between different gene pairs influencing plumage color. Using Mendelian genetics principles, we can hypothesize that two gene pairs are interacting, where one gene pair determines the presence of color (G for green dominant over white g), while the other pair determines the color type (Y for yellow and B for blue, both dominant over white).

To align with the observed progeny, we might consider that the green color is dominant and that the white is recessive, masked by the presence of the green trait. We possibly analyze this using a Punnett square to predict the offspring ratios. Additionally, the appearance of blue and yellow parakeets indicates a pattern where other genetic factors or allele interactions impact the final phenotype. If we consider epistasis, where one gene affects the expression of another, we may infer that the yellow gene could be epistatic to the white color, and the green could be showing a typical Mendelian 3:1 ratio when the y allele is not present. This hypothesis aligns with the observation that crossing the progeny results in multiple plumage colors, adhering to complex inheritance patterns beyond simple dominance.

Catecholamines mediate their effects throughI. α-adrenoreceptors.II. β-adrenoreceptors. III. receptor tyrosine kinases.IV. monomeric G protein receptors

Answers

Answer is I and II

Explanation:

Catecholamines are hormones made by your adrenal organs, which are situated on your kidneys. Models incorporate dopamine; norepinephrine; and epinephrine (this used to be called adrenalin or adrenaline).  Our adrenal organs send catecholamines into our blood when you're genuinely or sincerely focused. They cause you to inhale quicker, raise your pulse, and send more blood to significant organs, similar to your cerebrum, heart, and kidneys.

1. _______________________ is a process of gaining and testing knowledge & must be proven through experimentation. 2.___________________or non-science claims to be scientific but does not use experimentation. Cryptozoology is an example of this.

Answers

Answer:

Sciencepseudoscience

Explanation:

Science is the study of nature and behavior of everything around us. It is the knowledge on the physical or material world obtained through observation and experimentation.

Pseudoscience incorporates convictions, speculations, or practices that have been or are viewed as logical, however have no premise in scientific fact. This could mean they were invalidated experimentally, can't be tried logically, or need proof to help them. The term commonly has a negative implication. At the point when utilized, somebody's intimating the theme did not depend on logical discoveries and is, in this manner, ailing in truth.

Which term is not correctly matched with its proper definition? Prototroph: possesses the same metabolic/growth capabilities as the parent, wild type strain. The prefixes chemo- and photo-: define the electron source Macronutrient: phosphorous and sulfur are in this category Chemoorganotroph: a heterotroph could be considered this Growth factor: components, either defined or undefined, that promote/support growth

Answers

Answer:

The prefixes chemo- and photo-: define the electron source

Explanation:

The prefix 'chemo' indicates something that is a “chemical” or “chemically induced” while the prefix "photo" relates to light. An electron source is that which produces electrons. The prefixx chemo and photo do not relate to sources of electron production.

Final answer:

The term 'Prototroph' is incorrectly matched with its description. A Prototroph is a microorganism that can synthesize all compounds needed for its growth and does not specifically relate to maintaining the same metabolic/growth capabilities as the parent, wild type strain. Other terms are correctly matched with their definitions.

Explanation:

In the list of terms and their corresponding definitions, the term that isn't aptly matched is Prototroph. The definition of Prototroph usually refers to a microorganism that, unlike an auxotroph, does not require any additional nutrient other than a source of organic carbon for growth. It does not specifically relate to maintaining the same metabolic/growth capabilities as the parent, wild type strain.

Specifically, a Prototroph is a strain of microorganism that possesses a complete set of genes, enabling it to synthesize all compounds needed for its growth, thereby not requiring any growth factors. Similarly, a wild strain usually refers to the 'normal' or 'natural' condition of an organism, but it may or may not always have the same growth capabilities as its parent strain.

On the other hand, the terms and definitions of Macronutrient, Chemoorganotroph, Growth factor, and terms with the prefixes chemo- and photo- are correctly matched. These terms are standard in the field of Microbiology.

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Suppose that the resistance between the walls of a biological cell is 7.0 × 109 Ω. (a) What is the current when the potential difference between the walls is 80 mV? (b) If the current is composed of Na+ ions (q = +e), how many such ions flow in 0.85 s?

Answers

Answer:

Explanation:

Using Ohm's law

V ( voltage) = I (current A) × Resistance R in ohms

R = 7.0 × 10⁹Ω

V = 80 mV = 80 / 1000 = 0.08 V

0.08 V = I × 7.0 × 10⁹Ω

a) I = 0.08 V / 7.0 × 10⁹Ω = 1.142857 × 10 ⁻¹¹ A

b) quantity of charge = I × t = 1.142857 × 10 ⁻¹¹ A × 0.85 s = 9.7142857 × 10⁻¹² C

number of Na⁺ ions ( q = +e) = 9.7142857 × 10⁻¹² C / 1.6 × 10⁻¹⁹ C = 60714285.714 Na⁺ ions

Make a hypothesis about how you think the two species of Paramecium sp. will grow alone and how they will grow when they are grown together.

Answers

I hypothesize that Paramecium species will grow faster alone due to reduced competition. When together, resource competition might slow growth.

I expect that when each species of Paramecium is cultured alone, their growth rates will be relatively higher due to the absence of interspecific competition for resources. In a mixed culture, where both species coexist, I anticipate a potential decrease in growth rates compared to individual cultures.

This could be attributed to the sharing of limited resources, leading to heightened competition between the two species. Consequently, the presence of interspecific interactions might result in a more constrained growth environment, affecting the growth trajectories of both Paramecium species when compared to their growth in isolation.

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Final answer:

When grown separately, both Paramecium aurelia and Paramecium caudatum will probably proliferate successfully. However, when they share the same environment and compete for resources, Paramecium aurelia is likely to outcompete Paramecium caudatum. This hypothesis relies on previously observed behaviors, but must still be validated by experimental testing.

Explanation:

A hypothesis about the growth of two species of Paramecium, specifically Paramecium aurelia and Paramecium caudatum, could be: When each species is grown separately, they will both proliferate successfully. However, if they are grown together and forced to compete for the same resources, Paramecium aurelia is likely to outcompete Paramecium caudatum. This hypothesis is based on the observations shared in figures 45.24, 36.25, and 19.19 that reveal this competitive dynamic.

Beyond competition, these species might engage in symbiosis, which represents different types of close, long-term interactions between species. Depending on the conditions and specific interactions, these could be commensal (one species benefits, the other is unaffected), mutualistic (both species benefit), or parasitic (one species benefits at the expense of the other).

The interaction between these Paramecium species could potentially mirror other observed species interactions, such as between different groups of minnows in a shared environment or different mussel cultures in a lab. It's important to note however that any hypothesis must be tested experimentally to verify its accuracy and be subjected to further refinement based on findings.

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A particular tissue was discovered to have cells that lie in small chambers, separated by a matrix that is solid yet flexible, and fibers that could be stretched while still being able to return to their original length. Name this tissue.

Answers

Answer: Fibrous Tissue

Explanation:

A simple strong or repair tissue which consists of the twisted strands made of collagen and laid down by cells called as fibroblast.

Fibrous tissues are abundant in the body and are found almost everywhere. This tissue has the ability to repair after any scar or wound.

These tissue also gets repaired once it is damaged even in the case of some disease. This is a type of connective tissue.

An inbred strain of plants has a mean height of 74cm. A second strain of the same species from a different geographical region also has a mean height of 66cm. When plants from the two strain are crossed together, the F1 plants are a uniform height of 70cm. However, the F2 generation shows a wide range of heights: the majority are like the P1 and F1 plants, but approximately 1/1024 are only 50cm tall, and about 1/1024 that are 90cm tall. A.) What mode of inheritance is occuring here? B.) How many gene pairs are involved? C.) How much does each gene contribute to plant height? D.) Indicate one possible set of genotypes for the original P1 parents and the F1 plants that could account for these results? PLEASE EXPLAIN your reasoning. E.) Indicate three possible genotypes that could account for F2 plants that are 58cm high and 3 that account for F2 plants that are 74cm high? PLEASE EXPLAIN you reasoning.

Answers

Answer:

Explanation: see attachment

Final answer:

The plant height variation in the F2 generation indicates polygenic inheritance, with four gene pairs involved. Each gene pair contributes 2.5cm to the plant's height. The P1 parents could have been AABBCCDD and aabbccdd, and F2 genotypes causing specific heights would have different combinations of alleles.

Explanation:

The scenario described suggests a polygenic inheritance pattern, where multiple genes contribute to the phenotypic expression of height in these plants. Since the F2 generation exhibits a wide range of heights and specifically shows extreme phenotypes (50cm and 90cm tall plants) at a frequency of 1/1024, we can infer that several gene pairs are involved.

To determine the number of gene pairs involved, we must consider that each gene pair segregates independently and follows Mendel's laws of inheritance. With a 1/1024 occurrence for the extreme phenotypes, this points towards a 4th power of 1/4 (since 1/4 is the probability of a homozygous recessive phenotype from a heterozygote), indicating that four gene pairs are involved. Each gene pair contributes equally to the overall height phenotype, so to find the contribution of each gene, we examine the difference in height between the extreme phenotypes (90cm - 50cm = 40cm) and divide it by the number of gene pair combinations that result in the extreme phenotypes (2^4, where 2 is for dominant or recessive allele, and 4 is the number of gene pairs), therefore each gene contributes 2.5cm to the plant height.

The P1 parental genotypes could have been AABBCCDD (for the 74cm strain) and aabbccdd (for the 66cm strain), resulting in all F1 offspring having the genotype AaBbCcDd (70cm tall, intermediate height). The F2 generation would exhibit a wide range of phenotypic variation due to different combinations of alleles.

For F2 plants that are 58cm in height, with each gene contributing 2.5cm, these individuals are 16cm shorter than the F1 intermediate height, suggesting they have 6 recessive alleles. Therefore, three possible genotypes could be Aabbccdd, aaBbccdd, or aabbCcdd. For F2 plants that are 74cm high, the genotype must be the same as the original 74cm P1 plant, or include combinations that also result in the same height due to the same number of dominant alleles; possible genotypes include AABBccdd, AABbCcDd, or AaBBCcDd.

Information transfer is fundamental to all living organisms. For the following examples, explain in detail how the transfer of information is accomplished.
1. The genetic material from one bacterial cell enters another via transformation, transduction, OR conjugation?
2. The genetic material in one eukaryotic cell is copied and distributed to two identical daughter cells.
3. A gene in a eukaryotic cell is transcribed and translated to produce a protein.

Answers

Answer:

Explanation:

We are to explain in details how

genetic material from one bacterial cell enters another via transformation, transduction, OR conjugation.

For transformation, here, there is genetic alteration of a cell. It is a process of horizontal gene transfer where bacteria take up foreign genetic material from the environment.

Transduction is genetic recombination in bacteria where genes from a host cell are carried into the genome of a bacterial virus and then carried to another host during infection cycle.

Conjugation. In this, DNA plasmid is transferred from one bacterium to another through pilus.

2. We explain also in details how genetic material in one eukaryotic is copied and distributed.

Genetic material is copied during DNA replication and genetic material is distributed during mitosis.

The followings take place during DNA replication.

• DNA is copied during interphase, S phase of cell cycle.

• Site origin of DNA is known.

• Unwinding of DNA at the origin

• Synthesis of new strands with enzyme called helicase.

• Formation of replication forks.

• Proteins associated with replication forks help in the initiation and continuation.

• DNA polymerase synthesizes new strands by adding nucleotides that complement each strand.

Mitosis.

• Chromosomes are doubled

• Prophase: phase of condensation of chromosomes, formation of spindle.

• Metaphase: Alignment of chromosomes

• Anaphase: Chromatids are separated

• Telophase: reformation of nuclear membrane, division of cells, cell cycle control.

3. We also explain how transcription and translation of gene in eukaryotic.

Transcription

• Here we have DNA sequence to RNA sequence.

• Occurs in the nucleus

• DNA is carried into nucleosomes.

• RNA polymerase: add complementary RNA nucleotides to DNA template.

• Growth of new strands

• Leads to mRNA processing.

Translation

• Here we have mRNA base sequences to amino acids initiation

• Sequence of events such as complexes, small unit of ribosomes.

• Also we have the first tRNA

• Structure of ribosomes formed.

• Complete description of two subunits, 2 action sites, rRNA and proteins.

Saturated fatty acids are degraded by the stepwise reactions of beta-oxidation, producing acetyl-CoA. Under aerobic conditions, how many ATP molecules would be produced as a consequence of removal of each acetyl-CoA?

Answers

Answer:

4 ATP molecules

Explanation:

Normally, about 11 ATP molecules are generated as the result of Beta-Oxidation of saturated fatty acid in Kreb's Cycle. But the total removal of acetyl-CoA under certain aerobic condition decreases the overall yield and approximately 4 ATP molecules comes out from each removal of acetyl-CoA.

Final answer:

Each acetyl-CoA molecule produced from the stepwise degradation of saturated fatty acids through beta-oxidation can yield around 10 ATP molecules.

Explanation:

Saturated fatty acids are degraded through beta-oxidation, which produces acetyl-CoA. Each acetyl-CoA molecule can yield a certain number of ATP molecules when it enters the citric acid cycle and oxidative phosphorylation. One mole of acetyl-CoA metabolized by the citric acid cycle yields about 10 ATP molecules. Therefore, the energy produced by the removal of each acetyl-CoA molecule from the beta-oxidation of saturated fatty acids can be estimated to be around 10 ATP molecules.

Which of the following forms of cellular transport is being illustrated below?
Wy-laigh molecule
outside
CELL MEMBRANE
Bit formas
Inside
membrane
encloses
yeslcſó forma
D
A,
facilitated diffusion
(В.
exocytosis
C. pinocytosis
OD. endocytasis

Answers

Final answer:

It sounds like the described process is facilitated diffusion, which is a type of cellular transport that helps molecules cross the cell membrane from areas of high concentration to lower concentration using transport proteins. This process does not require energy as it's a passive form of transport.

Explanation:

Without a visual, it's difficult to definitively identify which form of cellular transport is being described here. However, if we have a scenario where molecules are moving from the outside to the inside of the cell through the membrane, it sounds like facilitated diffusion. This process involves specific transport proteins that act as channels or carriers, helping molecules that cannot cross the cell membrane alone to do so. Facilitated diffusion does not require energy as it's a type of passive transport, and the movement is down the concentration gradient, from a high concentration outside the cell to a lower concentration inside.

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in describing sound, what is a more common name for propagation velocity?

Answers

Answer: It is called Velocity factor.

Explanation:

Velocity factor or propagation velocity of a transmission medium is the ratio of speed of electromagnetic signal or radio signal move through the transmission medium to the speed of light in a vaccum. It means a ration of speeds is confined in sound cables or computer network

The genetic condition Xeroderma pigmentosum, which can lead to skin cancer, results from A. inability to correct UV induced lesions B. inability to process phenylalanine. C. inability to produce functional hemoglobin. D. inability to correct transitions E. breaks in the X chromosome

Answers

Answer:

Option A

Explanation:

Xeroderma pigmentosum arises as a result of the cell being unable to correct lesions induced by UV. This can be as a result of mutations in the enzymes which include XP A-E needed for correction of the lesions. Failure to correct these lesions leads to their accumulation and then damage to the cell.

Suppose a species of tulip has three alleles for the gene that codes for flower color. The C R allele produces red tulips, the C p allele produces purple tulips, and the C w allele produces white tulips. C R is dominant over C p and C w , and C p is dominant over C w . For each cross, determine the expected phenotype ratio of offspring flower color.

Answers

Answer and Explanation:

Available data:

The Cr allele produces red tulipsThe Cp allele produces purple tulipsThe Cw allele produces white tulips. Cr is dominant over Cp and CwCp is dominant over Cw

Cross 1

Parental)        CrCp   x   CpCw

Gametes)    Cr   Cp      Cp   Cw

Punnet Square)     Cr          Cp

                      Cp  CrCp     CpCp

                      Cw  CrCw     CpCw

F1 phenotype ratio)

                    2/4=1/2  Red flowers, CrCp and CrCw

                    2/4=1/2  Purple flowers, CpCp and CpCw

Cross 2

Parental)        CrCw   x   CpCw

Gametes)    Cr   Cw      Cp   Cw

Punnet Square)     Cr          Cw

                      Cp   CrCp     CpCw

                      Cw  CrCw     CwCw

F1 phenotype ratio)

                   2/4=1/2  Red flowers, CrCp and CrCw

                    1/4 Purple flowers, CpCw

                   1/4 White flowers, CwCw

Answer: The expected phenotype ratio of cross between CR and Cp is 1:1:2.

1 for CC,

1 for Cp, purple

2 for CR and Rp(red) all which are offsprings from the cross.

For cross between Cp and Cw, phenotype ratio is 1:1:2.

1 for CC,

1 for Cw, white,

2 for Cp and pw,purple.

Explanation:

Phenotype gives the observable characters of an individual.

Here, the character observed is flower color of tulip species.

The expected phenotype ratio of cross between CR and Cp is 1:1:2. The offsprings from the cross are CC, Cp,CR and Rp

1 for CC,

1 for Cp, purple

2 for CR and Rp red, since red is dominant over purple.

For cross between Cp and Cw, phenotype ratio is 1:1:2. The offsprings from the cross give CC, Cw, Cp and pw

1 for CC,

1 for Cw, white,

2 for Cp and pw,purple since purple is dominant over white.

What is the expected Hardy-Weinberg frequency of heterozygotes in a population of green anoles, where the locus under investigation has two alleles and the frequency of one allele is 0.005 (give your answer to five decimal places)?

Answers

Answer:

0.00995

Explanation:

If the population is in Hardy Weinberg Equilibrium, the sum total of frequencies of the alleles of a locus would be one. The frequency of one allele for a locus is 0.005. So, the frequency of the other allele for the same locus would be=1-0.005= 0.995

According to the given information, there are only two alleles for the locus in the population. So, the frequency of heterozygous genotype in a population= 2 x frequency of one allele x frequency of other allele= 2 x 0.005 x 0.995= 0.00995

10
Select the correct text.
Which of the following is the correct definition of biomass?
the total amount of organic matter from plants and animals in a habitat
all materials that form from the remains of living organisms
hydrocarbons and nitrogen oxides that are released when materials are burned

Answers

Answer: the total amount of

Organic matter from plants and animals in a habitat.

Explanation:

Answer:

the total amount of organic matter from plants and animals in a habitat

Explanation:

Archeological evidence suggests that Neanderthals might have been aware of the medicinal properties of plants over 60,000 years ago. Imagine you are a modern-day ethnobotanist and have identified a compound from fossilized tree pollen that binds to the alpha subunit of the most common G-proteins. To test how whether the compound affects the activity of Gα, you treat liver cells with the compound alongside epinephrine. You observe that the cells fail to produce glucose. Which could be the activity of the compound?

Answers

Answer:

It could be prevent Ga from binding to the epinephrine receptor.

Explanation:

Hindering GTPase action will make G-protein to tie to adenyl cyclase for all time so Glucose is delivered persistently.  

Diminishing the proclivity for GDP will build the opportunity of authoritative of GTP to G protein which thus will initiate the pathway.  

On the off chance that G protein ties with adenyl cyclase it will invigorate it.  

Yet, in the event that the G-protein is kept from official with epinephrine receptor, at that point the receptor can't enact trade the guanine nucleotide to G-protein for initiation. Thus the G-protein stays in latent state.

The activity of the compound should be that  It could prevent Ga from binding to the epinephrine receptor.

How to make G-protein:

In the case when GTPase activity should be inhibiting for binding the adenyl cyclase in a permanent way due to this the glucose should be generated in a continuous way. In the case when it reduced the affinity for GDP it raised the chance of binding for GTP to G protein that turned the activation of the pathway.  

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The _______________ community has low diversity with microorganisms primarily in the Leptospirillum groups II and III and from the Ferroplasma types I and II.

Answers

Answer:

The Acidophilic Microbial Community has low diversity with microorganisms primarily in Leptospirillum groups II and III and from Ferroplasma types I and II.

Explanation:

An acidophilic microorganism or plant is one which grows best in acidic conditions.

They are also referred to as microorganisms which occur in acidic natural (solfataric fields, sulphuric pools) and man-made (eg. Acid mine drainage) environments.

Acidophilic Microbes otherwise known as Acidophiles are an ecologically and economically important group.

They possess networked cellular adaptations for regulating intracellular pH. Several extracellular enzymes from acidophilic microbes are known to be functional at much lower pH than that inside the cells.

Acid stable enzymes have applications in several industries such as starch, baking, fruit juice processing, animal feed and pharmaceuticals, and some of them have already been commercialized. Acidophiles are widely used in bio-leaching of metals from low grade ores

Recent studies show that acidophiles are currently being considered to be utilized in bio-conversion and bio-remediation, as well as in microbial fuel cells to generate electricity.

Acidophilic microbes of similar characteristics are classifed in groups for ease of study and identification.

Leptospirillum Group II and II as well as Ferroplasma types I and II are groups of acidophilic microorganisms within the Acidophillic Microbial community.

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Imagine a cell whose membrane is permeable to water, but impermeable to solute X. If solute X is more concentrated outside the cell than inside, which way will water move by osmosis?

A. Water will move equally in both directions, in and out of the cell.
B. Water will enter the cell.
C. Water will not move in either direction, in or out of the cell.
D. Water will leave the cell.

Answers

Answer: D

Explanation:osmosis is the movement of molecules from a region of lower concentration to a region of higher concentration through a semi permeable membrane.

Final answer:

Water will move from inside the cell to the outside by osmosis because the concentration of solute X is higher outside the cell, making the outside a region of lower water concentration.

Explanation:

If solute X is more concentrated outside the cell than inside, and the cell's membrane is permeable to water but impermeable to solute X, water will move by osmosis from inside the cell to outside. This is because osmosis is the diffusion of water across a membrane from an area of higher water concentration (or lower solute concentration) to an area of lower water concentration (or higher solute concentration). Therefore, the correct answer is D. Water will leave the cell.

Plant cells form a cleavage furrow or indentation of membrane between new daughter cells.

Animal cells lack centrioles and no spindle forms during cell division.
The cell plate is the final partitioning of plant cells.
Plant cells resort to binary fission.
There is no difference.
Plant cells and animal cells undergo the same cellular processes during mitosis.

Answers

Mitosis Process

Explanation:

Plant cells and animal cells undergo the same cellular processes during mitosis.The centrioles absent in the plant.Animals  cells have centrioles, to what delivers the shaft strands is one of the differences among plant and creature mitosis. The periods of mitosis and goes on in each stage is the equivalent aside from when it comes to how cytokinesis is practiced. Plant cells separate the cytoplasm between the two new cores by framing a cell plate (new cell wall)between the cores. Animals cells achieve this by shaping a cleavage wrinkle which is a squeezing in at the center of the cell until two new cells have been framed with cytoplasm and organelles for each  new nuclei.

Chromatin structure can increase or decrease transcription of a gene according to the:
A. combination of amino acid modifications in the histone tails.
B. coding sequences in the messenger RNAs for histone proteins.
C. combination of histone proteins found within the nucleosome.
D. proportion of arginine and lysine amino acids in the histone proteins.

Answers

Answer:

D. proportion of arginine and lysine amino acids in the histone proteins.

Explanation:

Epigenetic modifications are one of the main reasons of methylation and de-methylation, acylation and de-acylation of histone proteins that trigger the transcriptional process. All of these mentioned modification causes the change in the ration of lysine and arginine residues and these are the main reasons that affect the structure of chromatin as required.

An article in the Los Angeles Times (Dec. 3, 1993) reports that 1 in 200 people carry the defective gene that causes inherited colon cancer. In a sample of 1000 individuals, what is the approximate distribution of the number who carry this gene

Answers

Answer: 5

Explanation:

Since 1 out of 200 people carry the defective gene that causes inherited colon cancer.

In a sample of 1000 individuals, we say

1 = 200

y = 1000

cross multiply

y x 200 = 1000 x 1

200y = 1000

y = 1000/200

y = 5

Thus, in a sample of 1000 individuals, approximately 5 will be carriers

One of Mendel’s Laws says that the probability of inheriting a maternal allele for a gene does not affect the probability of inheriting the maternal allele for another gene. If each gene has two alleles, what is the probability that one individual inherited the paternal allele at three different genes?

Answers

Answer:

There's 2 alleles for every quality and there are 3 qualities. The likelihood of acquiring the fatherly allele of those 3 qualities would be 1/2 x 1/2 x 1/2. This is on the grounds that you have a 50/50 possibility of getting the allele from your mom or your dad for three unique alleles. Since you are making sense of the likelihood of acquiring 3 qualities from your dad you duplicate at that point to get 1/8.

What is one advantage of sexual reproduction over asexual reproduction?


a decrease in genetic variation


an increase in genetic mutation


a continuation of genetic weaknesses


an increase in genetic variation

Answers

Answer:

an increase in genetic variation

Explanation:

The cornea is the transparent outer layer of the human eye. Because it must be transparent to light, it does not normally contain blood vessels. Therefore, it must receive its nutrients via diffusion. Oxygen from the surrounding air diffuses to the cornea through the surface tears whereas other nutrients diffuse to the cornea from the inner parts of the eye, such as the vitreous humor and lens.

During operation, the cornea produces waste in the form of CO2 gas that must be expelled to keep the eye healthy and functioning. This is accomplished by the simultaneous diffusion of CO2 from the cornea to the surrounding atmosphere, which generally features a low CO2 concentration.

It is therefore critical that modern contact lens materials allow sufficient diffusion rates of oxygen and carbon dioxide. Without oxygen, the cornea will warp, loose transparency, and become susceptible to scarring. The body may also react by growing additional blood vessels into the eye, which can damage the cornea.

If an increased steady-state flow rate of O2 (oxygen molecules per second) to the cornea is desired, which of the following contact lens / ambient condition modifications is not likely to be useful?

Note: the flow rate is equal to product of the diffusion flux and an area of interest through which diffusion occurs.

(a) Increase the contact lens thickness
(b) Increase the diffusivity of oxygen gas by increasing the contact lens porosity
(c) Increase the ambient temperature
(d) Increase the ambient partial pressure of oxygen gas
(e) All of the suggestions (a-d) are useful for increasing the flow rate of oxygen

the answer is not c

Answers

Final answer:

Increasing the thickness of a contact lens would decrease the flow rate of oxygen to the cornea, not increase it. Thus, this modification would not be useful if the goal is to increase the steady-state flow rate of oxygen to the cornea.

Explanation:

The best answer is (a) Increase the contact lens thickness. In the context of diffusion, the flow rate of a molecule is inversely proportional to the thickness of the layer it has to go through. Increasing the contact lens thickness would effectively reduce the rate at which oxygen reaches the cornea, thereby reducing the flow rate, not increasing it as desired. On the other hand, solutions (b), (c), and (d) could potentially increase the flow rate of oxygen to the cornea. Increasing the diffusivity, ambient temperature, or ambient partial pressure of oxygen gas would all potentially increase the rate of oxygen diffusion to the cornea.

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At the neuromuscular junction, an electrical signal from the motor neuron is translated to a chemical signal and then back to an electrical signal in the muscle cell. What proteins interact with the released chemical to cause the electrical signal in the muscle cell?
a.Sodium voltage-gated channels
b.Potassium ligand-gated channels
c.Cation ligand-gated channels
d.Calcium mechanical-gated channels

Answers

Answer:

Cation ligand-gated channels

Explanation:

At the neuromuscular junction, the arrival of nerve impulse triggers the release of chemical signals called neurotransmitters in the cleft. Binding of these chemical signals  to the receptor present on the motor endplate results in the opening of a ligand-gated cation ion channel. These channels are present in the receptor. Opening of ion channels allow cations, specifically Na+, to flow across the membrane. The entry of cations makes the inside of the muscle fiber more positively charged and triggers a muscle action potential.

Final answer:

Cation ligand-gated channels on the muscle cells interact with the neurotransmitter (chemical signal) released from the motor neuron to generate an electrical signal in the muscle cell at the neuromuscular junction.

Explanation:

The proteins that interact with the released chemical to cause the electrical signal in the muscle cell at the neuromuscular junction are the cation ligand-gated channels. Here's a brief step-by-step of what happens: An electrical signal or action potential travels down the motor neuron. When this signal reaches the end of the neuron, it causes the release of a neurotransmitter called acetylcholine. The acetylcholine diffuses across the neuromuscular junction and binds to the acetylcholine receptors - which are a type of cation ligand-gated channels - on the muscle cell. This binding causes these channels to open and allow positively charged ions to flow into the muscle cell, creating an electrical impulse in the muscle cell that ultimately leads to muscle contraction.

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