You are running a nursery for garden peas at Baton Rouge, and you have two pure-bred garden pea strains: Yellow Wrinkled [YYrr] and Green Round [yyRR]. You know that Yellow is dominant to Green, and also that Round is dominant to Wrinkled. One customer dropped by and asked about one particular pure-bred strain, Yellow Round [YYRR]. You promised to that customer that you could generate that strain next year through the following breeding experiments.
I. Crossing pure-bred Yellow Wrinkled with Green Round
II. Self-crossing F1 progeny
III. Screening pure-bred Yellow Round among the F2 progeny

What is the genotype for F1 progeny?

A. YYrr
B. rrYY
C. YyRr
D. yyrr
E. none of the above

Answer:

195.2 × 10⁻¹⁹ J.

Explanation:

The relation between the work done, potential difference and charge is as follows;

V = w / q.

Here, the potential difference, V = 122mV and q is the charge = 1.6 × 10⁻¹⁹ C

Substitute these values to find the amount of work done

W = v × q

W = 122 × 1.6 × 10⁻¹⁹

W = 195.2 × 10⁻¹⁹ J

Thus, the answer is 195.2 × 10⁻¹⁹ J.

Answer:

[tex]W = 1.95 * 10^{-20}[/tex] J

Explanation:

The potential on the inner side of the membrane is [tex]0[/tex]mV

And the potential  on the outer side of the membrane is [tex]122[/tex] mV

So the potential difference across the inner and outer membrane is equal to [tex]122[/tex] mV

We know that work done is equal to

[tex]W = q . V\\[/tex]

Where, q represents the charge of the particle and

V represents the potential difference across the inner and outer membrane

Substituting the given values in above equation, we get -

[tex]W = 1.6 * 10^{-19} * 122 * 10^{-3}\\W = 195.2 * 10^ {-22}[/tex]

[tex]W = 1.95 * 10^{-20}[/tex] J

Answer:

a) diversity

b) natural selection

c) living species

d) ancestral species

e) heritable variations in a population

f) greater reproductive success

g) an evolutionary adaptation

Darwin proposed a mechanism for evolution: natural selection, in which heritable traits that help organisms survive and reproduce become more common in a population over time.

See attached picture

Evolutionary adaptations in a population lead to greater reproductive success, increasing living species diversity.

At the core of this process is the presence of heritable variations within a population. These variations, arising from genetic diversity, form the basis for natural selection. Natural selection, in turn, is a mechanism by which certain traits or adaptations confer a reproductive advantage to individuals possessing them.

The phrase "greater reproductive success" reflects the idea that individuals with advantageous traits are more likely to survive, reproduce, and pass on these beneficial traits to their offspring. Evolutionary adaptations refer to the changes in traits over generations that enhance an organism's ability to survive and reproduce in its environment.

As these adaptations accumulate in a population over time, they contribute to the emergence of new species and the overall diversity of living organisms. Therefore, the interplay of heritable variations, natural selection, and the accumulation of evolutionary adaptations is central to the understanding of how ancestral species transform into diverse living species through the process of evolution.

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Answer:

all +bt +vg heterozygotes.

Explanation:

The mutation changes the chromosome structure or might change the gene sequence of the organisms. The mutation might result in the defective phenotype.

The mutation in Drosophila wings causes  bent wings (bt) or vestigial wings (vg). The homozygous bent wing (+bt +bt) are crossed with homozygous vestigial wings ( +vg+vg). Their cross is as follows:

parents : +bt +bt  and +vg+vg

Gametes:   +bt                   +vg

F1 cross:       +bt +vg.

Thus, the correct answer is option (a).

In the case of mating a homozygous bent wing and a homozygous vestigial wing Drosophila in genetics, all offspring are expected to be heterozygotes carrying one allele each of bent and vestigial wings.

When two homozygous parents with different wing mutations are crossed, we can use Punnett squares to determine the expected offspring. In this case, the parents are homozygous bent wing (bt/bt) and homozygous vestigial wing (vg/vg).

Using "bt" to represent the bent wing allele and "vg" to represent the vestigial wing allele, the Punnett square would look like this:

    bt      bt

  +bt/bt  +bt/bt

vg +bt/vg  +bt/vg

In this cross, all the offspring will be heterozygous for wing shape (+bt/vg). Therefore, the expected offspring are:

a. All +bt +vg heterozygotes.

So, the correct answer is (a).

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Answer:

Amoeba;

Explanation:

Amoeba:

Amoeba will contain nucleus (contain the genetic information). Nucleus normally appears as dense circular mass under microscope.

Organelles are present. Organelles appears like distinct masses that are rounded in shape and smaller than nucleus.

Plasma membrane is present.

Bacteria;

Nucleus will be absent in bacteria. The nucleoid normally appear lighter in color under microscope.

Ribosomes will be present. Ribosomes appears like black dots within the cytoplasm.

No plasma membrane will be present.

Answer:

Amoeba is Eukaryotic

Explanation:

Under microscope Amoeba like all Eukaryotas has   membrane bound organelles; food vacuole, mitochondria,  nucleus contractile vacuole etc. this is a distinguishing feature.

the cell membrane can be seen under microscope bounding the cytoplasm with stain

its amoeboid movement is additional  identification feature.

Bacterial  like all prokaryotes lack membrane bound nucleus., therefore this distinguished them from Amoeba.

Answer:

D preventing a lake from freezing over

Answer: Option D) preventing a lake from freezing solid

Explanation:

This is based on the difference in density of water in solid form as against it's liquid form

Ice, which is water in solid form is less dense to water as liquid water. Hence, ice floats on liquid water, thereby preventing the total freezing of the lake which would lead to death of aquatic organisms living in polar regions

Answer:B

Explanation:

In C3 plant fixation of carbon occurs rubisco the Calvin cycle enzyme that add CO2 to ribulose bisphosphate to produce a three carbon compound 3-phosphoglycerate.

C4 plants have an alternate mode of carbon fixation that forms a four-carbon compound as its first products. In C4 plants there are two distinct types of photosynthetic cells; bundle-sheath cells and mesophyll cells. The Calvin cycle is confined in the chloroplasts of the bundle-sheath cell.

In the first step of this mechanisms an enzyme present only in the mesophyll PEP carboxylase adds CO2 to PEP to form oxaloacetate a four-carbon products.

The four-carbon is exported to the bundle-shealth cells where it releases CO2, which is reassimilated into organic material by rubisco and the Calvin cycle. The same reaction regenerate pyruvate. ATP is used to convert pyruvate to PEP, allowing the reaction cycle to continue.

In C4 plants ATP is the price for concentrating CO2 in the bundle-shealth. C4 photosynthesis has higher ATP requirements than the C3 pathway.

C4 plants are not universal because C4 photosynthesis consumes more energy than C3 photosynthesis due to the regeneration of PEP consuming ATP, despite being more water-efficient and advantageous in hot, dry conditions.

The primary reason that not all plants are C4 plants despite their higher rates of photosynthesis under normal atmospheric conditions is that C4 photosynthesis requires a significant amount of additional energy. The correct answer to the question is: b. Because the regeneration of PEP consumes ATP, C4 photosynthesis consumes more energy than C3 photosynthesis.

C4 photosynthesis involves an additional set of steps before the C3 cycle, with the first carboxylation happening in the mesophyll cells, forming a four-carbon compound. This compound is then transported to the bundle sheath cells, where it is decarboxylated to release CO2, which then enters the Calvin cycle. The requirement for additional ATP to regenerate phosphoenolpyruvate (PEP) makes C4 photosynthesis less energy efficient compared to C3 photosynthesis. However, C4 plants have adaptations that make them more water-efficient, allowing them to thrive in drier and hotter conditions where C3 plants might struggle.

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Answer:

It will take 1.7 hours for the colony to contain 2,000 bacteria.

Explanation:

One bacteria divides into two by the process of binary fission.

Initial bacteria population = 200

Growth factor = 2

It doubles in size every 30 minutes.

Time = t/30

The exponential growth function is

[tex]y=ab^x[/tex]

where, a is initial value, b is growth factor and x is time.

Substitute a=200, b=2 and [tex]x=\frac{t}{30}[/tex] in the above function.

[tex]y=200(2)^{\frac{t}{30}[/tex]

We need to find the time taken by bacteria to reach 2,000 bacteria.

Substitute y=2000 in the above equation.

[tex]2000=200(2)^{\frac{t}{30}[/tex]

Divide both sides by 200.

[tex]10=(2)^{\frac{t}{30}}[/tex]

Taking log both sides.

[tex]\log 10=\log (2)^{\frac{t}{30}}[/tex]

[tex]1=\frac{t}{30}\log (2)[/tex]

[tex]30=\log 2(t)[/tex]

Divide both sides by log 2.

[tex]\dfrac{30}{\log 2}=t[/tex]

[tex]\dfrac{30}{0.301}=t[/tex]

[tex]99.667774=t[/tex]

It will take 99.66774 minutes for the colony to contain 2,000 bacteria.

1 hour = 60 minute

[tex]t=\dfrac{99.667774}{60}=1.66112\approx 1.7[/tex]

Therefore, it will take 1.7 hours for the colony to contain 2,000 bacteria.

It will take approximately 1.66 hours for the original bacterial population of 200 to grow to 2000 if the population doubles every 30 minutes. The calculation involves finding the number of doubling times needed and multiplying that by the length of each doubling time.

The subject of this question is exponential growth, a concept in mathematics that is often exemplified with the growth of bacterial colonies. Recognizing that the number of bacteria doubles every 30 minutes, we can calculate the time it would take for a population of bacteria to increase from 200 to 2000.

Our starting number of bacteria is 200, and we know that this number doubles (that is, grows by a factor of 2) every 30 minutes. Thus, we need to find out how many times we need to double 200 to get 2000. To do this, we divide 2000 by our initial number 200, which equals 10. The number 10 is the same as 2^3.322, meaning roughly 3.322 doubling times are needed.

Since each doubling time is 30 minutes, we multiply the number of doubling times (3.322) by the length of each doubling time (30 minutes) to get approximately 99.66 minutes. Converting this time to hours by dividing by 60, we find that it takes approximately 1.66 hours for the population to grow from 200 to 2000 bacteria.

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The question is incomplete. The complete question is as ollows:

For DNA, a plot of the change in absorbance versus temperature would be ________, which indicates ________.

Variable in shape (hyperbolic, linear or sigmoidal); the dependence of the denaturing process on the sequence of the DNA

Linear; the breaking of bonds between the base pairs is an independent process

Hyperbolic; a simple equilibrium between the native conformation and denatured DNA.

Sigmoidal; cooperativity where disruptions in the base stacking at one position destabilizes the stacking in neighboring base pairs

Answer:

Sigmoidal; cooperativity where disruptions in the base stacking at one position destabilizes the stacking in neighboring base pairs

Explanation:

DNA is the main genetic material of all the living organisms except in case of few viruses. DNA contains the nitrogenous bases ( adenine, guanine, thymine and cytosine), pentose sugar and phosphate bond.

The DNA can absorb light due to the presence of the absorbance property of nitrogenous bases. The shape of the plot of the change in absorbance versus temperature is sigmoidal. This is because the temperature denatures the DNA and breaks the hydrogen bond present between the nitrogenous bases. The breaking of the bonds between the nitrogenous base will disrupt the bonding at the neighboring base pair also.

Thus, the correct answer is option (4).

In DNA, a sigmoidal plot of the change in absorbance versus temperature indicates the melting temperature of the DNA. This is measured using a spectrophotometer and represents the tempearture at which 50% of the DNA molecules are denatured.

For DNA, a plot of the change in absorbance versus temperature would be Sigmoidal, which indicates Melting temperature (Tm) of the DNA. The absorbance of DNA increases with temperature due to the unstacking of the base pairs (denaturation), and this can be measured using a spectrophotometer. The temperature at which 50% of the DNA molecules are denatured is called the melting temperature. Therefore, the sigmoidal curve represents the denaturing process. The steep portion of the sigmoidal curve represents the melting temperature, Tm, of the DNA.

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Answer:

Temperature will increase

Explanation:

As we know

[tex]KE = \frac{2}{3} RT[/tex]

Where KE represents the Kinetic Energy, R represents the gas constant and T represents the temperature.

As the kinetic energy increases, the velocity of the gas molecules increases due to which the gas molecule start moving rapidly and hence exerts higher amount of force on the wall of the container in which it is kept.

The higher force leads to generation of higher pressure and hence the higher temperature.

When an object or entity is in a moving state then the energy in it is called kinetic energy.

The formula of kinetic energy (KE) is given by:

[tex]\text{KE} = \dfrac {2} {3} RT[/tex]

Where, [tex]R[/tex] = gas constant and [tex]T[/tex] = temperature

Based on the formula, the temperature will increase.

This can be explained as:

Therefore, kinetic energy is directly proportionate to the temperature.

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Answer:

B - increases (deplolarization.)

Explanation:

the influx of sodium ion into the cell increases the positive charge of the axoplasm, cause charges reversal of   the   axon  membrane and it is called  depolarization.

Th influx of sodium ion is due combination of three factors

1, to the opening of voltage gated  sodium channels,

2. increase in chemical gradient for sodium ions, and

3. high concentration of sodium ions outside compare to the axoplasm.

The combined effects of  2 and 3 above is called electrochemical gradients; and this pull sodium ions with the psotive  charges  through the voltage gated sodium channels into the axoplasm.

Few  gated channels were open initially however as the intensity of the stimulus increases, more gated channels of sodium opens, with many Na+ diffused in.And if the voltage generated due to deoplarization is up to the threshold levels,Action potential occurs.

During an action potential, sodium (Na+) levels increase inside the cell due to the activity of the sodium-potassium pump in the cell membrane, which creates an electrochemical gradient. Option B

During an action potential, sodium (Na+) levels increase inside the cell. This is due to the sodium-potassium pump in the cell membrane, which pumps three Na+ ions into the cell for every two K+ ions it pumps out.

This creates a short-term electrochemical gradient, with a higher concentration of Na+ ions inside the cell compared to outside. This change in concentration and the resulting electrical charge difference is what allows the signal to be transmitted down the neuron.

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Answer:

Arranging the events from the oldest to the youngest-

1. Cambrian explosion

2. dominance of amphibians

3. Earth hit by a large carbonaceous meteorite

4. diversification and dominance of mammals

5. extinction of giant mammals like mastodons

Explanation:

The Cambrian explosion took place on earth about 540 million years back, where majority of the animal species appeared and slowly started to evolved, which formed the primitive life forms on earth. This event took place because of the sudden increase in the oxygen concentration on earth that facilitate the appearance of new organisms, and it lasted for nearly 15 to 25 million years, resulting in the production of many phyla (metazoan).

The amphibian were first evolved during the Devonian period that ranges from about 420 to 360 million years back. They slowly became the dominant species during the Carboniferous and the Permian period.

By the end of Permian, there occurred a global mass extinction event, due to the meteoric impact on earth, which is commonly known as the Permian-Triassic (P-T) boundary, about 250 million years back. This meteorite was largely comprised of carbon materials.

After this mass extinction event, there occurred divergence of mammals species even though they existed during the carboniferous period, but they were dominant after the extinction of dinosaurs, about 65 million years back.

The mammals like mastodons were abundant during the Cenozoic and they slowly got extinct about 10,000 years back during the time of Pleistocene period.

Final answer:

The sequence of events from earliest to most recent is the Cambrian explosion, dominance of amphibians, diversification and dominance of mammals, Earth hit by a large carbonaceous meteorite, and extinction of giant mammals like mastodons.

Explanation:

Earth's biodiversity and the course of evolution have been influenced by various environmental changes and catastrophic events. To place the events in chronological order, we must refer to key moments in Earth's geological history where these changes have been evident.

Answer:

There are many points at which eukaryotic gene expression can be controlled, through pretranscriptional control, transcriptional control, and posttranscriptional control

Explanation:

The pretranscriptional control determines the accessibility of chromatin to the transcription machinery. It is affected by supercoiling and methylation. It is also known as epigenetic regulation, and it does not depend on the sequence but on the conformation of the DNA.

While transcriptional control determines the frequency and / or speed of transcription initiation through the accessibility of the start sites, the availability of transcription factors and the effectiveness of promoters.

The post-transcriptional control is the one that is exercised once the transcript has finished synthesizing. It can be of several types:

• Maturation control: As the RNA adjustment can be made.

• Transport control: Most RNA has to go out to the cytoplasm to perform its function. For this they have to cross the pores of the nuclear membrane, where you can select the RNAs that will be transported and those that will not.

• Stability control: The half-life of RNA can be regulated by the expression of RNAs or mRNA stabilizing proteins in the cytoplasm.

• Translational control: It is exercised on the frequency with which the mRNAs begin to be translated. It can also affect the frequency with which proteins mature and the availability of enzymatic effectors.

Regulating the individual steps of the gene expression process can impact the expression of a specific protein. RNA stability, splicing, and translation efficiency are three key factors that can be regulated to control protein expression.

The individual steps of the gene expression process can be regulated to increase or decrease the expression of a particular protein. Here are three hypotheses:

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Answer: Option C) Dichotomous distribution

Explanation:

A simple inherited trait can be

- influenced by environment

- monogenic i.e controlled by one gene

For example, an individual can inherit the trait of dark colour from parent where dark is completely dominant over other skin color

However, a simple inherited trait can not express dichotomous distribution, because

dichotomous distribution involves the control of a trait by two different genes.

Thus, Dichotomous distribution is the answer.

A characteristic of a simple inherited trait that is NOT true is that it is influenced by the environment.

A characteristic of a simple inherited trait that is NOT true is that it is influenced by the environment. Simple inherited traits are monogenic, meaning they are controlled by a single gene, and they have a dichotomous distribution, which means they have only two possible phenotypes. Therefore, the correct answer is a. Influenced by the environment.

A characteristic of a simple inherited trait is that it is not influenced by the environment. Simple inherited traits are typically determined by genetic factors and are not significantly impacted by environmental factors. These traits are usually controlled by one or a few genes, making their inheritance relatively straightforward.

In contrast, complex traits are influenced by both genetic and environmental factors. Examples of simple inherited traits include certain genetic disorders, blood type, or the ability to taste certain substances. These traits follow predictable patterns of inheritance and are less susceptible to environmental influences compared to complex traits.

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Answer:

heart, get rid of toxins/flush out toxins

Explanation:

Project Renew:

In prokaryotes the 5' UTR is 3-10 nucleotides.

In Eukaryotes the 5'UTR is 100 to many thousand nucleotides long.

Explanation:

Leader sequence or 5' UTR starts at transcription site and ends at the initiation codon just one nucleotide away from it.

It is present in mRNA.

These are GC rich and form secondary structure, helps in protein synthesis.

Shine Dalgarno sequence in prokaryotes is an example of 5'UTR.

It acts as an entry point of ribosome.

The length of the 5' untranslated region (5' UTR) varies for different mRNAs and is not a fixed value. It requires specific gene information to determine the exact length in ribonucleotides.

The length of the 5' untranslated region (5' UTR) differs among mRNAs and is not a fixed value. This region stretches from the 5' cap to the nucleotide just upstream of the start codon. The 5' UTR is a crucial part of the mRNA as it plays roles in translation regulation and mRNA stability. Its length can influence the efficiency with which a ribosome binds and initiates translation, which can affect protein synthesis.

Proteins known as RNA-binding proteins (RBPs) can also bind to the 5' UTR, affecting the stability and lifespan of the mRNA molecule. Understanding the exact number of ribonucleotides in a specific 5' UTR would require sequencing data or detailed annotation from a specific gene. Without such specific information, the exact length in ribonucleotides cannot be determined.

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Answer: c. provide the basis of ecosystems in the polar environments.

Explanation:

Cyanobacteria is found in the freshwater, glacial environments of the polar and alpine regions. The Cyanobacteria mats can be found at the bottom of the ponds, lakes and streams within the melted water habitats in the ice shelves and glaciers.

The Cyanobacteria accounts for the dominant fraction of the total ecosystem production. These are the key producers of the food chains in the polar region and these are also responsible for contributing to the nitrogen and carbon cycles. Their presence was observed in the Arctic and Antarctic polar regions.

Answer:

Option C, Both alleles are recessive, but they must be located at different gene loci.

Explanation:

It is given in the paragraph that when obese mice from two laboratories are crossed with another obese mice of the same group, then all the offspring are obese. This clearly indicates that mice would be having recessive allele.  

For instance let obese gene be represented by O1 for laboratory 1 and as O2 for laboratory 2. Let the normal gene be N1 and N2 for laboratory  1 and 2 respectively.  

Thus, cross between tow homozygous recessive species will produce recessive offspring  

O1 O1 * O1O1

O1O1, O1O1, O1O1, O1O1

O2O2* O2O2

O2O2, O2O2, O2O2, O2O2

However, when two obese mice from different laboratories are crossed, then all the offspring are normal  

i.e  

O1O1 * O2O2

O1O2, O1O2, O1O2, O1O2

Being at two different loci, these obese genes could not express themselves and hence all the offspring are normal.

Hence, option C is correct

Final answer:

c) Both alleles 1 and 2 are recessive and located at different gene loci, resulting in normal offspring when obese mice from different labs are crossed due to the masking effect of the normal alleles at each locus.

Explanation:

The correct answer to why all offspring are normal when crossing two obese mice from different laboratories but all are obese when crossing obese mice from the same laboratory is that c) both alleles 1 and 2 are recessive and located at different gene loci. This scenario is consistent with Mendelian inheritance patterns and suggests that each obese mouse carries a different recessive allele at a separate locus that leads to obesity. Therefore, when mice from the two different laboratories are crossed, the offspring are heterozygous at both loci, carrying one normal allele and one recessive allele for obesity at each locus, which results in a normal phenotype. This is due to the presence of at least one normal allele at each locus, which masks the effect of the recessive obesity alleles.

Answer:

Asexual lineages, being either unicellular organisms or organisms with a small number of germ-line cell divisions, survive due to the very high fidelity of DNA replication, which is enough for the reliable self-reproduction.

Explanation:

I found this off the internet

I hope it helps.

Answer: d) binding of keratin to cadherine) assembling keratin fibers

Explanation:

The desmos (membrane domains that facilitates the cell-cell contact) like cadherins are able to provide the strong adhesion to the intermediate filaments between the epithelial cells and muscle cells. The binding of the cadherine with the keratin fibers helps in providing the strong connection and adhesion of cell to cell in epithelial cell layers.

Due to mutation of the acidic keratin epithelial cell a fragile epithelial cell layer develops this will lead to disassembly of the keratin epithelial cells or will prevent the assembly of the keratin epithelial cells.

Answer:

The answer is a) gene flow

Explanation:

Gene flow is any displacement of genes from one population to another. The gene flow includes a multitude of different types of events, such as pollen that is transported by air to a new destination or people who move to another city or another country. If genes are transported to a population where those genes did not exist, gene flow can be a very important source of genetic variability.

Gene Flow, Genetic Drift, Natural Selection and Mutation are key forces that maintain genetic diversity in a population and can prevent the permanent fixation of an allele. They cause changes in gene or allele frequencies and variations leading to biodiversity.

In the context of population genetics, both Gene Flow, Genetic Drift, Natural Selection and Mutation are key forces that can maintain genetic diversity and avoid the permanent fixation of an allele.

Gene Flow is the transfer of genetic variation from one population to another. If gene flow occurs, then the gene frequencies in the source population can change and genetic diversity can be maintained.

Genetic Drift is the change in the frequency of an existing gene variant in a population due to random sampling. It can lead to genetic variation within populations.

Natural Selection refers to the process by which traits become more or less common in a population due to consistent effects upon the survival or reproduction. Over time, this process can lead to biodiversity.

Mutation is a change to the sequence of a gene. Mutations provide new genetic variations and are the basis for changes in alleles and therefore prevent their permanent fixation.

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Answer:19:7

Explanation:If we know the standard state free energy change, Go, for a chemical process at some temperature T, we can calculate the equilibrium constant for the process at that temperature using the relationship between Go and K. In this equation: R = 8.314 J mol-1 K-1 or 0.008314 kJ mol-1 K-1.

Answer:

R = 8.314 J mol-1 K-1 or 0.008314 kJ mol-1 K-1.

Explanation:

Answer: e. 2.5 ml

Explanation:

According to the dilution law,

[tex]C_1V_1=C_2V_2[/tex]

where,

[tex]C_1[/tex] = concentration of stock solution = 20X

[tex]V_1[/tex] = volume of stock solution = ?

[tex]C_2[/tex] = concentration of required solution= 1X

[tex]V_2[/tex] = volume of  required solution= 50 ml

[tex]20\times V_1=1\times 50ml[/tex]

[tex]V_2=2.5ml[/tex]

Thus 2.5 ml much of a 20X tricaine stock solution is needed to dilute in order to make your solution.

Answer:

Transport of sodium ions out of the cell

Transport of potassium into the cells

Transport of calcium ions out of cell

Transport of calcium ions OUT OF the cytoplasm of a cell and INTO an organelle that has a high concentration of calcium ions.

Explanation:

In general the concentration of sodium ion is higher outside the cell in the extracellular fluid and the concentration of potassium ion is higher inside the cell.  

During an active transport, a cell uses energy to move molecules against the gradient as a part of primary active transport. In the secondary active transport, the molecules move as per the electrochemical gradient established by the primary active transport.  

In primary active transport, sodium ion moves out of the cell and potassium ion moves into the cell through the assistance of sodium potassium pump. In order to maintain the stability, calcium ion from the sarcoplasmic reticulum is pumped out of the cell

Hence , the correct answers are

Transport of sodium ions out of the cell

Transport of potassium into the cells

Transport of calcium ions out of cell

Transport of calcium ions OUT OF the cytoplasm of a cell and INTO an organelle that has a high concentration of calcium ions.

Final answer:

Active transport requires energy in the form of ATP to move substances like ions against a concentration gradient, such as sodium ions out of cells, potassium ions into cells, and calcium ions into organelles with high calcium concentration.

Explanation:

Cells use energy to move substances against a concentration gradient in a process known as active transport. This energy commonly comes in the form of adenosine triphosphate (ATP). In the case of ions, active transport is required when they are moved from an area of lower concentration to an area of higher concentration, such as when cells transport:

Sodium ions OUT OF cells, which counters the higher extracellular sodium concentration.

Potassium ions INTO cells, against the extracellular potassium's lower concentration.

Calcium ions OUT OF the cytoplasm and INTO an organelle with a high calcium concentration.

Conversely, substances that are moved with the concentration gradient do not require energy and are transported via passive transport mechanisms.

Answer:

Mrs. Leonard's child has a severe allergic reaction. Based on the description we can infer that Mrs. Leonard's child has a severe allergic reaction. We can advice Mrs. Leonard to immediately bring the child to the hospital so that anti- allergic such as advil, motrin or  Epi-pen could be given to the child. We can also instruct her to place ice at the area where the bee has stung.

The instructions that should be given is that bring to the child to the hospital for treatment.

It is the reaction where it should be allergy for the skin, nose, eyes. The examples could include the sneezing, watery eyes, blue skin, etc. Here the child should provide the epinephrine that should be used for the anaphylaxis treatment. After that the person should call the ambulance so that the treatment should be started as soon as possible.

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Question

Pedigree attached

a. Provide evidence from the pedigree that conclusively shows that the tune deafness allele is autosomal dominant, not autosomal recessive. Explain your reasoning.

b. Identify the genotypes of individuals 5 and 6, and then draw the Punnett square for the cross of these two individuals.

c. Compare the expected percentage of each phenotype of the offspring from the cross in part (b) with the actual percentage of each phenotype observed in the children of individuals 5 and 6.

Answer/Explanation:

a. Autosomal dominant means that even having one copy of the allele T will produce a tune deaf individual. The evidence of this is that individuals II.3 and 4 have children (III. 8 and 9) with normal tune perception. This suggests that they have inherited the normal allele from both parents. That means II.3 and II4 must be heterozygous, in order to pass on the normal allele. This makes sense if tune deafness is dominant (because 3 and 4 will both be Tt and Tt, so can pass on the t alleles). However, if tune deafness was recessive, that would mean that individual II 3 and 4 both have to carry 2 copies, meaning there is no way their children would have normal tune perception.

b. Since individual 5 is unaffected and we know this is the recessive trait, they must be Tt. Since individual 6 is affected, they must be either Tt or tt. However, they have children who are unaffected (11, 12, 13) who must be tt. This means, individual 6 must have the t allele to pass on, and must be Tt. Therefore, the cross between individuals 5 and 6 is tt x Tt:

         T        t

t        Tt       tt

t       Tt       tt

c) The punnet square shows the children from (b) have a 50:50 chance of being tune deaf vs normal tune perception (50% each). In contrast, the actual ratios are 25% tune deaf to 75% normal tune perception. (1:4). This deviates from the expected ratio, but since all of these are due to chance, it is not an unexpected occurrence. Perhaps if there were 10, 50 or 100 children (!) from this cross, the results would be more like the expected ratio

Answer:

The endomembrane system helps in the protein transport, processing and secretion of protein to the different target. The endomembrane system is absent in prokaryotes.

The localization pattern of sec 61 and sec 63 determines the different roles. The protein processing starts in the endoplasmic reticulum and moves to the golgi. If sec 61 is present in ER, this plays an important role in translocation. If sec 33 is present in golgi it might have a role in processing or transport of protein.

Final answer:

The localization patterns of Sec61 and Sec33 suggest they have different roles. Sec61 localizes in the ER, involved in initial protein modification, while Sec33 localizes in the Golgi, involved in further protein modification and sorting.

Explanation:

The localization patterns of GFP fusion with Sec61 and Sec33 suggest that they have different roles in the cell. GFP fusion with Sec61 localizes in a pattern consistent with the endoplasmic reticulum (ER), which is involved in the initial stages of protein synthesis and modification. On the other hand, Sec33 localizes in the Golgi compartment, which is responsible for further protein modification and sorting.

Since Sec61 is found in the ER, it is likely involved in protein synthesis and initial modification, while Sec33's localization in the Golgi suggests it is involved in the further modification and sorting of proteins before they are transported to their final destinations.

Therefore, the different localization patterns of Sec61 and Sec33 indicate that they have distinct roles in the process of secretion.

Answer:

Select all of the carbohydrates from the list below

Cellulose, Sucrose

Explanation:

Cellulose and Sucrose are types of carbohydrate but the simplest form of carbohydrate is glucose.

Cellulose is a polysaccharide that needs to be broken down to its simplest form in order to be used

The carbohydrates from the list are glycogen, cellulose, and sucrose. While antibodies are proteins, cholesterol is a lipid, and estrogen is a hormone.

The carbohydrates from your provided list are glycogen, cellulose, and sucrose. Carbohydrates are biomolecules consisting of carbon, hydrogen, and oxygen atoms, usually with a hydrogen-oxygen atom ratio of 2:1.

In living organisms, they play crucial roles, including serving as an energy source and forming structural components. For clarification, Antibodies are proteins by nature involved in the immune response. Cholesterol is a type of lipid and is part of cell membranes. Estrogen is a hormone, specifically a steroid hormone.

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Answer: To find out specific DNA of interest.

Explanation: By this technique; mutation in DNA can be detected, restriction site can be determined, single strand after separation can be analyzed, genetic disorders can be identified, infections can be identified.

This procedure is also helpful in the forensic.

Answer: The goal of this procedure is to identify the fractionated DNA that would be complimentary to the labelled probe DNA.

Explanation:

The labelled probe DNA contains an already known specific sequence of bases that would be observed if it is complementary to the population of seperated DNA molecules. If it is, then a hybridization of the complimentary DNA sequence occur.

Thus, the goal of this procedure is to identify the fractionated DNA that would be complimentary to the labelled probe DNA.

Answer:

[tex]5 * 10^{10}[/tex]

Explanation:

The question is not complete. Remaining part of the question is as follows - Minimal growth medium for bacteria such as E. coli includes various salts with  characteristic concentrations in the mM range and a carbon source. The carbon source is  typically glucose and it is used at 0.5% (a concentration of 0.5 g/100 mL). For nitrogen,  minimal medium contains ammonium chloride (NH4Cl) with a concentration of 0.1 g/100 mL

How many cells can be grown in a 5 mL culture using minimal medium before the medium exhausts the carbon?

Solution -

We will first find the mass concentration of 0.5 g/100 mL of solution.

[tex]\frac{0.5}{100}[/tex] gram per ml of glucose

The chemical formula of glucose is [tex]C_6H_{12}O_6[/tex]

The molecular weight of glucose molecule is [tex]180[/tex] grams per mole

Now, we will find the number of moles of glucose in a 5 ml medium -

[tex]\frac{\frac{0.5}{100} * 5}{180} \\1.39 * 10^{-4}[/tex] mole

The number of carbon atom in each glucose molecule is equal to six, thus, number of minimal carbon mole is equal to

[tex]1.39 * 10^{-4} * 6\\= 8.34* 10^{-4}[/tex]mole

Number of carbon atoms is equal to

[tex]8.34* 10^{-4} * 6.023 * 10^{23}\\= 5 * 10^{20}\\[/tex] Carbons

One bacteria has [tex]10^{10}[/tex] carbon molecule.Thus, [tex]5[/tex] ml medium will have [tex]5 * 10^{10}[/tex] bacteria

Answer: 1. The genotypes of parent that produced 72 red-winged,24 clear-winged are Rr and Rr.

2. The genotypes of parent that produced 54 red-wing,49 clear-wing are Rr and rr.

3. The genotypes of parent that produced 96 red-winged can either be RR and RR or RR and Rr.

Explanation: To get parent's genotypes from offspring's phenotypes, find the phenotypic ratio among the offsprings produced. That will give insight on the allele combination of the parent.

In the case where 72 red-winged,24 clear-winged were produced, the ratio will be 72:24. Dividing the ratio to the lowest term, we will get 3:1. This implies that the ratio of dominant to recessive alleles in the offspring genotypes is 3:1. Only heterozygous crossing can produce the ratio. We can then assign the parent's genotypes as Rr and Rr.

In the case where 54 red-wing,49 clear-wing were the offspring produced, the ratio will be 54:49. Dividing it to the lowest term, we will get approximately 1:1. This implies that the ratio between dominant and recessive allele is 1:1. A test cross will produce this ratio. We can then assign the parent's genotypes as Rr and rr.

In the cross that produced only 96 red-winged flies, the parent's genotypes can either be RR and Rr or RR and RR.