In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are intercrossed to produce the F2.
(a) What will be the phenotypic ratio in the F2?
(b) If an F1 plant is backcrossed to the bitter, yellow-spotted parent, what phenotypes and
proportions are expected in the offspring?
(c) If an F1 plant is backcrossed to the sweet, non-spotted parent, what phenotypes and proportions
are expected in the offspring?
(d) If you testcross the F1 plant, what phenotypes and proportions are expected in the offspring?

Answer: Elongation

Explanation:

Elongation (in the context of DNA) is not a form of post translational modification. It is basically an increase in the length of a particular growing nucleotide chain during either replication, translation or transcription. Post translation modifications are basically changes that are most of the times chemical that can take place in a protein after translation and some of these occur to activate or to cleave some specific regions. Examples including phosphorylation, glycosylation (conjugation of carbohydrate by enzymes) etc

Answer:

A homologous structure can be described as structures present in similar and different species. These structures might not perform the same functions but are similar because they might have a common ancestor in the past.

An example of a homologous structure is the arms of a human and the wings of a bat. Although the arms of a human and the wings of a bat do not perform the same function yet they do have similarities because they have a common ancestor in the past.

Answer:

Explanation:

One of the earlier experiments that confirmed DNA and not proteins as genetic material was conducted by Alfred Hershey and Martha Chase

Hershey and Chase experiment showed the genetic material of virus is DNA not protein. Their experiments demonstrated that Phage are composed of DNA and protein, when phage infect bacteria, their DNA enters the host bacterial cell but not protein before infection

Hershey and Chase experiment prove that DNA is the hereditary material.

From chargaff base pairing rule, the ratio of purine to pyrimidine is a cell is 1:1. Since adenine pair with thymine and cytosine pair with quanine, this means the percentage of adenine will equal the percentage of thymine.

%A = %T and %G = %C.

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

If a DNA contains 8% adenine, therefore the percentage of thymine is 8%

8% + 8% + %G +%C = 100%

16% + %G +%C = 100%

%G +%C = 100%-16%

%G +%C = 84% = 84%/2= 42%

Since %G = %C

Then %G = 42%, %C = 42

By Chargaff's law, the percentage of a base is always equal to the base from which it pairs.

So the percentage of adenine is 8%, thymine is 8%, cytosine is 42%, and guanine is 42%.

According to the law :

Thus, the percentage of adenine is 8%, thymine is 8%, cytosine is 42%, and guanine is 42%.

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Answer: All of the answers are correct

Explanation:

Lipids refers to a group of small biomolecules that do not dissolve in water, but dissolve readily in nonpolar solvents and contain fatty acid, sterols.

They have several:

- phospholipids forms the structural components of cells and biological membranes.

- Triacylglycerides serves as energy reserve (release energy during starvation) in the adipose tissues of animals.

- lipids in the adipose tissues of animals also provide insulation against cold, thereby maintaining body temperature

- lipids such as phospholipids are found in the biological membranes of many organs like the heart, brain etc, where they help to absorb shock or damage during severe hit to such organs.

So, all the answers are correct

Lipids are crucially important as they provide a significant energy reserve, form essential structural components of cells, help maintain body temperature and cushion organs against shocks. All the options are correct.

Lipids are a class of organic molecules. These lipids perform several essential functions in the body. Firstly, they serve as a significant energy reserve, meaning that during times when other energy sources like carbohydrates are not available, the body can break down lipids to generate energy.

Second, lipids form essential structural components of cells. They are integral parts of the cell membrane and provide barriers in cellular transportation. Third, they help to maintain body temperature by providing insulation. Lastly, they protect our internal organs by cushioning them against shocks, acting almost like shock absorbers in the body.

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Answer: 55

The expected chromosome number in the hybrid would be 55

Explanation:

To obtain the hybrid organism:

- the diploid (2n) cell of horse will undergo meiosis to produce a gamete with halved chromosome. This gamete is with haploid number (n)

Thus, 64 / 2 = 32

- the diploid (2n) cell of Zebra will undergo the same pattern as the horse

Thus, 46 / 2 = 23

Then, the gametes of the two animals will be fused to form a zygote of the hybrid.

32 + 23 = 55

Hence, the zygote will form a diploid hybrid organism with an expected chromosome number of 55

Answers: Two populations involved in the study. 1) Root system 2) Watering regimens

Explanation: The experiment to investigate the effect of watering on the life span of certain type of root system using three watering regimens are determined based on the life span of the root system. Therefore, two populations involved in the experiment. One is the population of watering regimens and the other is the root system.

The model of the cell cycle provides details about how cell division promotes genetic continuity, including DNA duplication, accurate segregation of chromosomes, and cytokinesis.

The model of the cell cycle provides several details that support the argument that cell division promotes genetic continuity:

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Answer:

Centromeric DNA

Explanation:

Genes are responsible for formation of proteins. A codon produces a single amino acid and it comprises of three base pairs. The whole DNA of organism does not contribute to encode protein. There are sequences in DNA that does not contribute in coding known as non-coding DNA and perform different functions. These may include the O O O 5-end, DNA coding for a signal sequence, noncoding intron and O promoter sequence .

The centromere is the part of DNA that links the sister chromatids. The centromere is not converted to mRNA or proteins.  It is also not involved in regulation of genes. So it did not contribute to 864 bp found in the yfg gene.

Answer:

Soon after the island of Hawaii rose above the sea surface (somewhat less than one million years ago), the evolution of life on this new island should have been most strongly influenced by habitat differentiation

Explanation:

Habitat differentiation has to do with when new environment is carved out of old one to form a new one, just as the description that occurs in island of Hawai Rose.

Answer:

The result would differ in the sense that the reproductive success would not decrease at the same rate at 20% as it would at 100%. At 20% it would decrease faster.

Final answer:

A 20% negative selection pressure results in a slower rate of allele elimination from the population, leading to a more gradual evolutionary change compared to 100% negative selection pressure. This allows for greater genetic diversity and adaptability in facing future environmental changes.

Explanation:

Naturally occurring selective pressures, such as predators, diseases, or environmental conditions, tend to reduce an organism's reproductive capacity rather than completely negate it. If there was only 20% negative selection pressure, as opposed to 100%, this would mean that the affected individuals still retain a significant proportion (80%) of their reproductive capability. Consequently, the alleles or genetic traits subject to this pressure would be eliminated from the population at a slower rate compared to a scenario with 100% negative selection pressure. Over time, this could lead to a more gradual change in the allele frequencies within the population, affecting the pace at which evolution occurs under these selective pressures.

The evolutionary impact of differing levels of selective pressure illustrates how natural selection can shape populations over time. A 20% selective pressure allows for a greater diversity of genetic material to remain within the population, potentially enabling it to adapt to future changes in the environment or new selective pressures. Thus, understanding the nuances of selective pressure helps us grasp the complex dynamics of evolution and the survival of species in changing environments.

Answer:

G

Explanation:

NADPH (Nicotinamide Adenine Dinucleotide Phosphate) and ATP (Adenosine Triphosphate) are forms in which energy is stored in metabolic reactions

Final answer:

The question is about identifying carrier molecules among ATP, NADPH, FAD, and AMP in cells. ATP, NADPH, and FAD act as carrier molecules, facilitating energy transfer and electron shuttling in cellular processes. Therefore, the correct answer is that all of the mentioned molecules serve as carrier molecules, with a clarification on the unique roles of NADPH, FAD, and ATP.

Explanation:

The question asks which of the following act as carrier molecules in cells: AMP, NADPH, FAD, or ATP. In cellular metabolism, activated carrier molecules are crucial for transferring energy and electrons to drive endergonic reactions, which are reactions that require an input of energy. Notably, Adenosine triphosphate (ATP) is the primary energy currency of cells, acting as a direct source of energy for various cellular processes. NADPH and FAD are key electron carriers involved in redox reactions, including those in photosynthesis and cellular respiration, respectively. AMP, while related to ATP, does not serve in the same capacity as a direct energy carrier or electron shuttle in cells. Therefore, the correct answer is E. All of the Above, acknowledge the roles of NADPH, FAD, and ATP as activated carrier molecules in cellular processes.

In oogenesis, the secondary oocyte indeed inherits most cytoplasm and organelles from the oogonium, making statement 1 correct.

The question revolves around oogenesis, the process of ovum (egg) formation, and includes three statements for validation. Let's address each:

The secondary oocyte contains most of the cytoplasm and organelles from the oogonium. This statement is correct. During oogenesis, the secondary oocyte indeed inherits most of the cytoplasm, nutrients, and organelles to ensure the zygote has sufficient resources following fertilization.

The secondary oocyte contains the diploid chromosome complement. This statement is incorrect. The secondary oocyte does not contain a diploid set of chromosomes, but rather a haploid set. After the first meiotic division, the chromosome number is halved.

The polar body forms from discarded DNA during oogenesis. This statement is correct. Polar bodies are formed from uneven cell division, where they receive a minimal amount of cytoplasm and are mostly composed of unneeded DNA, eventually degenerating.

In summary, statements 1 and 3 are correct, while statement 2 is incorrect within the context of oogenesis.

Answer:

Plants are set up in a greenhouse. 1/3 of potted plants are grown under red light, 1/3 under green light, and 1/3 under natural light. Plant biomass is measured at the beginning of the experiment and after a 28 day period.

In this study, is the amount of plant growth the independent or dependent variable?

Dependent variable

Explanation:

Plant is the independent variable as the light varies in color which makes the plant to depend on them for growth

Answer:

The atmosphere

Explanation:

The major reservoir of carbon in the abiotic environment is the atmosphere where it occurs as carbon dioxide. Atmospheric carbon dioxide is used by green plants for the synthesis of organic compounds through photosynthesis.

Exocytosis is a cellular transport process that allows materials to move across the plasma membrane of a cell. It is the primary method of transporting large molecules out of the cell and uses transport vesicles to export materials from the cell. Option E

Exocytosis is a cellular transport process that allows materials to move across the plasma membrane of a cell.

It is the primary method of transporting large molecules out of the cell and uses transport vesicles to export materials from the cell.

Exocytosis is used by cells to remove waste products and release chemical signals or substances, such as digestive enzymes and hormones, into the extracellular fluid.

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Answer:

Carbon convert into methane, Nitrogen converts into Ammonia and Sulfur converts into hydrogen sulfide gas

Explanation:

As the flow time in a wastewater treatment plant increase the color of waste water converts from grey to black as the condition becomes more anaerobic and flow becomes more septic. In such scenario if more waste water is added the sulfur and nitrogen extracts from the waste are converted into ammonia and hydrogen sulfide. Carbon is present in the form of organic waste and in the absence of oxygen it gets converted into methane gas. However , in presence of oxygen this carbon is released as carbon dioxide.

Answer:

time (t) = 2.69 sec

Explanation:

Given that;

The final top speed of 3.96 m/s ( final velocity V) = 3.96 m/s .

The initial speed is 0; &

The acceleration (a) = 1.47 m/s²

We can determine how much time does it take the snake to reach its top speed by using the equation of motion;

V = U +at

3.96 = 0 + (1.47×t)

3.96 = 1.47t

t = [tex]\frac{3.96}{1.47}[/tex]

time (t) = 2.69 sec

∴ It took the snake  2.69 sec to reach the its top speed.

The significance of checkpoints can be demonstrated by considering what happens when they are impaired. What would occur if there was a gain-of-function mutation in the promoter for the cyclin E gene such that cyclin E protein was always made at high levels even under conditions in which cyclin E would not normally be made?

a. Cells will pass through the G2/M checkpoint with damaged DNA.

b. Cells will skip the S phase and go directly to G2 phase and not complete DNA replication.

c. Cells will pass through the G1/S checkpoint even if conditions are not ideal for cell division.

d. Cells will pass the M checkpoint with chromosomes unattached to spindles.

Answer:

c. Cells will pass through the G1/S checkpoint even if conditions are not ideal for cell division.

Explanation:

Cyclins are the regulatory proteins that are formed during different stages of the cell cycle. A specific cyclin during each stage bind to corresponding cyclin-dependent kinases (CDKS). The activated CDKs phosphorylate the proteins required for the progression of a cell through a specific stage of the cell cycle.

For example, Cyclin E is synthesized at the peak near the G1 -S phase transition or G1/S checkpoint. They activate the CDK-2 and allow the cells to progress from G1 to the S phase. If the cell is not prepared for DNA replication, cyclin E is not formed or is inhibited by specific protein kinases which in turn does not allow the cell to enter the S phase.

Any gain of function mutation resulting in the constitutive synthesis of cyclin E would allow the cell to pass through the G1/S checkpoint irrespective of the conditions.

Answer:

The correct terms are as - 1. virus, 2. capsid, 3. lytic cycle, 4. lysogenic cycle, 5. prophage.

Explanation:

A virus is considered as nonliving by various scientists as it required a living host to divide, however, it is studied with the microbes such as the bacteria and others. The virus is made up of nucleic acid surrounded by the protein coat called a capsid.

Viruses normally show two types of life cycles inside the host cell that are the lysogenic cycle and the lytic cycle.

The lytic cycle is characterized as the life cycle that ends with the destruction or bursting of the host cell. The lysogenic cycle includes being dormant inside the host genome until the favorable condition reestablished. Prophage is a virus that inserts itself into the host genome on its own.

Thus, the correct answer is - 1. virus, 2. capsid, 3. lytic cycle, 4. lysogenic cycle, 5. prophage.

The questions are about different aspects of virology including viral structure, and the lytic and lysogenic cycles. A virus, although not considered truly alive, is studied as a microbe. Its structure includes a protein coat (capsid) and its genetic content or nucleic acid. The lytic and lysogenic cycles describe how a virus interacts with its host.

1. Although generally not considered to be alive, a virus is studied alongside other microbes such as bacteria. 2. The protein coat that surrounds the nucleic acid of a virus is known as a capsid. 3. A viral life cycle that results in bursting of the host cell is referred to as the lytic cycle. 4. A viral life cycle in which the virus inserts its genome into the genome of its host, where it may remain dormant for long periods is known as the lysogenic cycle. 5. A viral genome that has inserted itself into the genome of its host is referred to as a provirus or prophage.

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Answer:

B

Explanation:

The evidence that prompt this conclusion is the reported amount of  nucleotide present which is varied because double stranded DNA genome which is usually heavier, more stable usually contains Adenine to thymine ratio of 1 to 1 and guanine to cytosine ratio of 1 to 1 and an over all of 1 to 1 ratio of purines to pyrimidines ( Double-stranded genomes have equal amounts of A to C )compared to what is reported which has a varied ratio ( A to T is 1 : 3 and G to C is about 0.77 for ss stranded DNA).

Answer:

Option b. Lowering the pH to kill pathogenic bacteria is correct answer.

Explanation:

bacterial motors are sensitive to pH.  By decreasing the pH bacterial motors stops working. This was identified in a new research.  But, with the weak acids and a lower internal pH they slow and ultimately stop moving (became dead).

Reference: Powell, K. Acid stops bacteria swimming. Nature (2003).

Answer:

When the pKa is 6.0, we can determine the fraction of protonated H is by:

pH = pKa + log [A]/[HA]

Where

A = Deprotonated imidazole side

HA = Protonated side

Given, pH = 5.0

5 = 6 + log [A]/[HA]

log [A]/[HA] = -1 (take antilog of both side)

[A]/[HA] = 0.1

The ratio of the deprotonated imidazole side chain to the protonated side chain at pH 5.0 = 0.1

Given, pH = 7.5

7.5 = 6 + log [A]/[HA]

log [A]/[HA] = 1.5 (take antilog of both sides)

[A]/[HA] = 31.62

The ratio of the deprotonated imidazole side chain to the protonated side chain at pH 5.0 = 31.62

Final answer:

The ratio of deprotonated to protonated histidine side chain at pH 5.0 is 0.1, and at pH 7.5 is 31.62, calculated using the Henderson-Hasselbalch equation.

Explanation:

The ratio of deprotonated to protonated histidine side chain can be calculated using the Henderson-Hasselbalch equation:

pH = pKₐ + log ([A⁻]/[HA])

where pKₐ is the acidity constant of the side chain, pH is the environment's pH, [A⁻] is the concentration of the deprotonated form, and [HA] is the concentration of the protonated form.

Calculation at pH 5.0:

pH = pKₐ + log ([A⁻]/[HA])
5.0 = 6.0 + log ([A⁻]/[HA])
log ([A⁻]/[HA]) = -1
[A⁻]/[HA] = 10⁻¹
[A⁻]/[HA] = 0.1

Calculation at pH 7.5:

pH = pKₐ + log ([A⁻]/[HA])
7.5 = 6.0 + log ([A⁻]/[HA])
log ([A⁻]/[HA]) = 1.5
[A⁻]/[HA] = 10^1.5
[A⁻]/[HA] = 31.62

At pH 5.0, the ratio of deprotonated to protonated histidine side chain is 0.1, meaning there is 10 times more protonated form present. At pH 7.5, the ratio is 31.62, indicating there is significantly more deprotonated form present.

Answer:

The answer is option B. autotrophs

Explanation:

Decomposers release nutrients like nitrogen and carbon dioxide as well as water when they feed on dead plants and animals. Decomposers include fungi, bacteria and earthworms.

These nutrients are released into air, soil and water and serve as nutrients for autotrophs. Autotrophs produce their own food from sunlight energy, water and carbon dioxide. Example include green plants; which by the process of photosynthesis produce their own food using carbon dioxide and soil nutrients.

Answer:

Option b. top carnivores is correct.

Explanation:

organism that found at the top of the PYRAMID OF NUMBERS that feed (preys) on other organisms but which is not itself preyed on are called top carnivore. They actually regulate terrestrial ecosystems. For example, tiger. (See attached Image)

In a terrestrial food web, producers have the lowest energy content and biomass.

In a terrestrial food web, the energy content and biomass is lowest in producers. Producers, such as plants, convert solar energy into chemical energy through photosynthesis. They are the primary source of energy in a food web, and as energy flows through the system, some energy is lost at each trophic level, resulting in a decrease in energy content and biomass.

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Answer:

A. Both microbial genomes include a gene that encodes for streptokinase, an enzyme and virulence factor that activates human plasminogen.

Explanation:

Streptokinase (Ska) is a single-chain 414 amino acid protein, produced by certain Streptococci species (group A, C, and G streptococci), which possess an activity closer to two host proteins (urokinase-type and tissue-type plasminogen activators), as it non-enzymatically process inactive plasminogen to proteolytically active plasmin.

Streptokinase is a species-specific virulence factor. If both microbial genomes produces streptokinase, then these microorganisms can both be grouped as Streptococci since they are closely related.

All other choices do not certainly ease the relationship beyond the domain level.

Answer:

A. Both microbial genomes include a gene that encodes for streptokinase, an enzyme and virulence factor that activates human plasminogen.

Explanation:

Streptokinase is an enzyme (also synthesize as a medication) produced specifically by streptococci spp. If the two unidentified microbes could produce this enzyme, then   they should contain the gene that codes for streptokinase .Thus   they  share certain genome characteristics,  and are related.  

Other options related the similarities  to the domain levels only

Question was't arranged i have arranged it in  ask for detail section.

Answer:

Option e. homozygous is the correct answer.

Explanation:

A gene which has two identical alleles on homologous chromosomes is called homozygous. It is denoted by XX (capital letters)  for dominant character (alleles) and  xx (lowercase letters) for recessive character (alleles).

An individual possessing two of the same alleles at a gene locus is described as "homozygous."

Homozygous:

Homozygous refers to a genetic condition where an individual carries two identical alleles at a particular gene locus on a pair of homologous chromosomes.

These alleles can be the same for a particular trait, whether dominant or recessive.

For instance, if a person has two identical alleles for brown eyes (let's denote this as "BB"), they are considered homozygous for the eye color gene at that specific locus.

Understanding Alleles and Genetic Loci:

Genetic loci are specific positions on a chromosome where a particular gene is located. Each gene at a given locus can have multiple forms, known as alleles.

Alleles can be either identical (homozygous) or different (heterozygous) for a specific gene.

Contrast with Heterozygous:

In contrast, heterozygous individuals have two different alleles at a specific gene locus.

For example, if an individual has one allele for brown eyes and one allele for blue eyes (denoted as "Bb"), they are considered heterozygous for the eye color gene.

For such a more question on homozygous

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Answer:

Option A,  Excessive heat

Explanation:

A denatured protein is the one whose structure changes with the loss of an activity. Generally, a denatured protein unfolds itself when the 2 D network of water around the protein molecule breaks by the breaking of hydrogen bonds. This primarily happens when the temperature is high. Both hydrogen bonds and hydrophobic interactions are disturbed by the heat thereby breaking hydrogen bonds.

Answer:

excessive heat

extreme pH

Explanation:

Protein denaturation can occur when they are exposed to abruptly increased temperature conditions and extremes of pH. Heat denature proteins by affecting the weak interactions that stabilize their secondary or tertiary structures. It makes the proteins less functional or completely non-functional. Similarly, extremes of pH alter the net charge on the protein. This causes electrostatic repulsion between the amino acids and disrupts some hydrogen bonds. Loss of these weak interactions denatures proteins.

Answer/Explanation:

An important way of selecting for bacteria carrying specific recombinant DNA is to add a resistance gene (such as antibiotic resistance) to your gene of interest. That means that any bacteria that carry your DNA will be resistant to a specific antibiotic (in this case streptomycin). Therefore, you can grow your bacteria on plates with streptomycin, and in theory, the bacteria will be unable to grow if they don't have the resistance gene and your gene of interest.

Here, strain A does not grow on streptomycin plates, but strain B now has 100 colonies. This suggests strain B can grow on streptomycin, meaning it is resistant.

Strain B is therefore the streptomycin resistant master plate strain.

To calculate the percentage of receptors that are bound to the ligand, subtract the concentration of free ligand from the total concentration of receptors (this gives the concentration of receptors that are bound), divide by the total concentration of receptors and multiply by 100. In this case, 25% of the receptors are bound to the ligands.

To calculate the percentage of receptors that are bound to a ligand, you first need to determine how many receptors are not bound to the ligand. This value is the total concentration of receptors minus the concentration of free ligand. In this case, it would be 10 mM - 2.5 mM = 7.5 mM.

To find the percentage of receptors that are bound, we need to divide the receptors that are bound (free receptors) by the total concentration of receptors and then multiply by 100.

Thus, the calculation is as follows: (2.5 mM / 10 mM) x 100 = 25%.

The total concentration of receptors is 10 mM, and the free ligand concentration is 2.5 mM, therefore, 25% of receptors are bound to the ligand.

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