Which of the following are characteristics of archaeal cell membranes that probably help them survive at very high temperatures?
Choose one or more:

A.high percentage of LPS

B.ether-linked lipids

C.membrane monolayers

D.branched fatty acids

E.cyclopentane rings

I chose B&C, and it was wrong.

Answer:

The given blank can be filled with water molecules.

Explanation:

The diminished entropy in protein folding is counterpart by a huge enhancement in entropy of the surrounding of water molecules. The hydrophobic effect performs an essential function in protein folding. When protein associates with water, the molecule of water surrounding it become more aligned in order to upsurge their hydrogen bonding and thus there is a large increase in entropy.

Answer:

The correct answer is - water molecules.

Explanation:

The formation of protein folding or lipid bilayer leads to the decrease in the entropy of the bilayer or protein components which is balanced by the large increase in the entropy of water molecules present around the protein or bilayer.

Nature of the protein folding plays an important role as these molecules are hydrophobic as they interact with the water molecules and water molecules become more ordered so the hydrogen bonding increase which ultimately increases in the entropy.

Thus, the correct answer is - water molecules.

Answer:

The frequencies of the phenotypes that impart survival advantage to the organisms under the changed conditions are increased by natural selection.  

Explanation:

Genetic variations are present in the natural population. Some genetic traits and the corresponding phenotypes become adaptive under the changed conditions. These organisms survive better and reproduce more successfully to leave more progeny. This results in an increase in the proportions of the beneficial phenotypes in the population over generations. For example, antibiotic resistance in the bacterial population increase when antibiotics are present in their surroundings.

Answer:

B.hydrogen bonding within polypeptide strands

Explanation:

Alpha helix is one of the secondary structures formed by polypeptide chains. It is formed when the backbone of a polypeptide is wound around a longitudinal axis. The side chains of the amino acids are oriented outwards. Alpha helix structure is stabilized by the formation of a maximum number of hydrogen bonds between the amino acids of the chain. The hydrogen atom linked to the electronegative nitrogen atom of a peptide bond form hydrogen bond with the electronegative carbonyl oxygen atom of the fourth amino acid.

In this way, every peptide bond except the ones present close to each end of the helix take part in hydrogen bonding and stabilize the structure.

Answer:

Because the "Bt toxin gene is not readily accepted by plants." scientists concerned about the use of the Bt toxin, which is used to kill insects, in genetic engineering

Explanation:

"Bacillus thuringiensis" that is commonly reffered as (Bt) is a gram positive(Giving positive resultsgram stain test) bacteria.This  Bacillus thuringiensis found to be present in the soil. It is biologically used  as alternative source against pesticide. The genetically modified crop of tobacco was first invented by the the Belgian company Plant Genetic System. This crop was resistant to insects it occurs by expressing the cry genes from the B. thuringiensis, and the delta-endotoxin is present in the crop. The use of Bt toxins were helped in promotion of the requirement of the extensive evaluation of safety for the use in food and potential in the environment.

Answer: C

Explanation: on Edge

Answer:

Theoretical yield of tungsten produced = 35.6836915592  ≈ 35. 68 g

Explanation:

The chemical equation can be expressed as follows;

WO3 (s) + 3H2(g) → W(s) + 3 H20(g)

Note that the equation is already balanced.

Molecular Mass of WO3= 183.84 + 15.999 × 3 = 183.84 + 47.997  = 231.837 g

From the equation 1 mole of WO3 reacts with 3 mole of hydrogen molecule.

Molecular mass of tungsten(W) = 183.84 g

1 mole of tungsten was produced from the chemical equation.

WO3 (s) + 3H2(g) → W(s) + 3 H20(g)

From the equation,

231. 837 g of WO3 produces 183.84 g of tungsten

45.0 g of WO3 will produce ?

grams of tungsten produced = 183.84 × 45 /231.837

grams of tungsten produced = 8272.8 /231.837

Theoretical yield of tungsten produced = 35.6836915592  ≈ 35. 68 g

Answer:

"Yes" The streak plate method is very effective in diluting the population size in all 4 plates.

Explanation:

The streak plate method is effective because it is one among the rapid isolation method which further allows for easy observation of the bacteria. This method is also being used for getting the color of the certain substances. Its main principle is to isolate the specific bacteria from the sample of various microbes. It dilutes the microbes number and often reduces their density. It is useful in separation of organisms from the group. These are isolated in the agar plate.

The streak plate method is used effectively to dilute a large population of bacteria. The effectiveness in a specific scenario depends on the initial concentration, the technique used, and the incubation conditions. Success is indicated by the presence of individual colonies after incubation.

The streak plate method is a widely used technique in microbiology, used primarily to dilute a large population of bacteria. The key purpose of doing this is to end up with individual bacterial cells that can grow into visible colonies. However, without more information, it is difficult to definitively say if the streak plate method was effective at diluting the population size in all 4 plates in your specific experiment. Factors that can impact the effectiveness include the technique used to streak the plates, the initial concentration of the bacteria, and the incubation conditions.

To gauge the effectiveness, examine the plates after incubation. If you see individual colonies that result from a single bacterium, this can be an indication that the method worked effectively.

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Answer: Erythropoiesis is the process which produces red blood cells (erythrocytes). It is stimulated by decreasing oxygen in circulation detected by the kidney which then secrete erythropoietin that proliferate and differentiate red blood precursors & produce red blood cells.

Explanation: The use of erythropoietin to boost red blood cells is called blood doping. The more red blood cells are present in the body, the more the energy expenditure and the slower you will burn calories during physical activities.

Erythropoiesis is the production of red blood cells in the body, stimulated by the hormone erythropoietin. Athletes may use erythropoietin as a doping drug to enhance performance, but it is illegal and risky.

Erythropoiesis is the process by which red blood cells are produced in the body. It occurs in the bone marrow and requires the hormone erythropoietin. Erythropoietin is produced by the kidneys and stimulates the production of red blood cells.

Athletes may use erythropoietin as a doping drug because it increases the number of red blood cells, which leads to improved oxygen delivery to the muscles. This can enhance endurance and performance. However, doping with erythropoietin is illegal and can have serious health consequences.

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Answer:

Option-(B):The neutrophil forms lamellipodial extensions.

Explanation:

Lamellipodial extensions:

The lamellipodal extensions is a type of cytoskeletal protein actin projection. And is the effect of expressing a dominant active form of the small GTPase Rac1 in a human neutrophil.

Expressing an active GTPase Rac1 in human neutrophils primarily causes the formation of lamellipodial extensions. This is due to Rac1's role in rearranging actin in the cell's cytoskeleton, which aids in movement and immune response functions.

Expressing a dominant active form of the small GTPase Rac1 in a human neutrophil primarily results in the formation of lamellipodial extensions(B). Rac1, a member of the Rho family of small GTPases, plays a critical role in cell signaling and regulation of cytoskeleton changes. Neutrophils, a type of white blood cell, utilize these changes to facilitate movement and other immune response functions. When Rac1 is activated, it promotes the rearrangement of actin in the cytoskeleton, leading to the formation of lamellipodia, which are flat, broad cellular protrusions involved in cell migration and adhesion. Therefore, the other options, while they concern actin-related structures or functions, do not typically result from Rac1 activation in neutrophils.

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Very large substances can enter the cell by endocytosis. During this process, the plasma membrane surrounds the substance and then fused with the surface of the cell and opens into the cell to remove the substance. In the opposite manner, large substances can leave the cell by exocytosis. At this stage, the substance will be inside a vesicle. This will fuse with the membrane on the inside and release the content allowing the substance out of the cell. Ribosomes, composed of proteins and rRNA, are often found attached to the Endoplasmic reticulum, but may also occur freely in the cytoplasm. Groups of free ribosomes are called Poly ribosomes. Ribosomes are the site of protein synthesis. A protein produced on a free ribosome has a different function than a protein produced at an attached ribosome.

Explanation:

Endocytosis is a process in which cell ingests large particle via cell membrane and bring it to the cell.

Exocytosis is a process in which vesicles fuse with the plasma membrane and eject the substance out of the cell.

Both endocytosis and exocytosis requires energy in the form of ATP.

Ribosome can be either free or attached on rough endoplasmic reticulum.

Membrane-bound ribosomes protein on Endoplasmic reticulum forms protein that is transported to the other cells while free ribosomes produce protein that is to be used in the cell itself.

Large substances enter the cell through endocytosis and exit through exocytosis, involving vesicles in both processes. Ribosomes, either floating freely in groups called polysomes, or attached to the Endoplasmic Reticulum, are primarily responsible for protein synthesis.

Large substances can enter a cell through a process called endocytosis. Here, the plasma membrane wraps around the large substance and fuses with the vesicle of the cell, forming an endosome which then opens up into the cell to release the substance. On the contrary, large substances can exit the cell through a process called exocytosis. The substances are enclosed in a vesicle which fuses with the plasma membrane and opens, allowing the substance to leave the cell.

Ribosomes, which comprise of proteins and rRNA, can be found attached to the Endoplasmic Reticulum (ER) and can also float freely in the cytoplasm. When unattached to any membrane, they form groups known as polysomes. Ribosomes are the primary site of protein synthesis. There is a difference in function between a protein produced on a free ribosome and a protein produced by an attached ribosome.

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Answer: An Action potential facilitates the transport of information from presynaptic neuron to postsynaptic neuron during electrical synapse.

Explanation:

Action potential is the change in voltage of membrane which is as a result of ions flow in and out of neurons. An action potential is a means neurons transport information in electrical synapses. An action potential facilitates the presynaptic neuron to transport neurotransmitters which bind to the postsynaptic neuron during electrical synapses.

2:2 because there are 2 Rr and 2 rr

Answer:

Option-(B): The protolith is the un-metamorphosed rock.

Explanation:

Both are the same thing but the difference in the form, or we can say, that the prolith is the non-metamorphosed form and the slate is metamorphosed form of the same substance. As the slate is the fine grained form of rock and it posses very unique structure, as these rocks are mostly in the construction and designing purpose of beautification of buildings of various types.

Rocks undergo transformation during metamorphism based on changes in metamorphic grade, with shale transforming to slate, then to phyllite, schist, and ultimately gneiss as metamorphism intensifies.

The pre-existing rock, also known as the protolith, transforms into different types of metamorphic rocks depending on the conditions it undergoes during metamorphism. The key processes that dictate this transformation are changes in temperature and pressure, which is referred to as the metamorphic grade. Slate is produced by the low-grade metamorphism of shale. When the grade increases, the slate can subsequently become phyllite, and with further metamorphism under more intense conditions, phyllite can transform into schist. Schist is characterized by its flaky minerals and tendency to split along parallel planes. With continued metamorphism at even higher grades, schist can become gneiss, which is defined by its foliated, banded appearance with alternating layers of light and dark minerals. In contrast, amphibolite is not necessarily a higher grade than gneiss but differs in mineral composition and protolith, often forming from the metamorphism of igneous rocks or shales rich in amphibole minerals.

Answer:

Plantea

Explanation:

Plantae is the plant kingdom which contains all plants on the earth. They are multicellular eukaryotes. Characteristically, they contain a rigid structure that surrounds the cell membrane known as the cell wall. Plants also have a green colored pigment known as chlorophyll that is quite vital for photosynthesis.

Answer: False.

Explanation:

Overpopulation is when the total number of people living in an environment exceeds the carrying capacity of the environment. Overpopulation is not the primary threat of global instability and environmental degradation. Environmental degradation is caused by human activities . Global instability is caused by discrimination, unemployment, inequality, urbanization,poverty and many more lead to instability.

Answer:

D. There is a 25% chance that their child will develop CF.

Explanation:

The Delta508 mutation and the R117 mutation are seen in the Cystic Fibrosis (CF) gene. IF a child have both the Delta508 mutation and the R117 mutation, the chances of suffering from CF is high and inevitable.

In this senerio, we can infer that since both parents are heterozygous, the likelihood of developing CF for each child born is 25%

___________________________

Here is the complete question

A couple with European ancestry seeks genetic counseling before having children because of a history of cystic fibrosis in the husbands family. ASO testing for CF reveals that the husband is herterozgous for the Delta508 mutation and that the wife is heterozygous for the R117 mutation. You are the couple's genetic counselor. When consulting with you, they express their conviction that they are not at risk for having an affected child because they carry different mutations and cannot have a child who is homozygous for either mutation. What would you say to them.

A. There is a 25% chance that their child will be a carrier of the ∆508 mutation.

B. There is a 25% chance that their child will be a carrier of the R117 mutation.

C. There is a 50% chance that their child will be a carrier of a CF mutation.

D. There is a 25% chance that their child will develop CF.

Answer: Structural chromosomal mutation

Explanation: In translocation, a small piece of chromosome is detached from one chromosome and is attached to another non-homologous chromosome. Translocation can be simple, shift or reciprocal.

Simple translocation involves single break in the chromosome. The broken piece gets attached to the end of the non-homologous chromosome.

In Shift translocation, the broken segment of one chromosome gets inserted interstitially in a non-homologous chromosome.

Segment from one chromosome is exchanged with a segment from another non-homologous chromosome simultaneously in Reciprocal translocation.

Answer:

0.0035 mL of oxygen per hour is metabolic rate at 20°C at cockroach.

Explanation:

From the data we can see that metabolic rate of cockroach is in direct relation with temperature.

[tex]Rate \propto Time[/tex]

On graph, with metabolic rate on y-axis and temperature on x axis straight line will obtained , and the equation of the line can be determined by the help of point slope:

[tex](y-y_1)=m(x-x_1)[/tex]

m = Slope of the line = [tex]\frac{y_2-y_1}{x_2-x_1}[/tex]

As per information given , :

[tex](x_1,y_1) , (x_2,y_2) = (15,0.01) (25,0.015)[/tex]

[tex]m=\frac{0.015-0.01}{25-15}=\frac{0.005}{10}=0.0005[/tex]

[tex](y-0.01)=0.0005(x-15)[/tex]

[tex]y-0.001=0.0005x-0.0075[/tex]

[tex]y=0.0005x-0.0065[/tex]

x = temperature

y = metabolic rate

Now , put x = 20

[tex]y=0.0005\times 20-0.0065[/tex]

y = 0.0035

0.0035 mL of oxygen per hour is metabolic rate at 20°C at cockroach.

Final answer:

The metabolic rate of cockroaches at 20°C can be estimated using interpolation, assuming a linear relationship between temperature and metabolic rate, resulting in an estimated metabolic rate of 0.0125 ml O2/hr.

Explanation:

The question involves calculating the metabolic rate of cockroaches at an intermediate temperature given their metabolic rates at two different temperatures. Assuming a simple linear relationship between temperature and metabolic rate, we can use interpolation to estimate the metabolic rate at 20°C.

Given that the metabolic rate is 0.01 ml O2/hr at 15°C and 0.015 ml O2/hr at 25°C, it's reasonable to assume that at 20°C, halfway between these temperatures, the metabolic rate would be halfway between these values. Thus, the metabolic rate at 20°C could be estimated as approximately 0.0125 ml O2/hr.

This answer is based on the assumption of a direct and simple linear relationship between temperature and metabolic rate for the given range, which is a common approximation for biological processes within a moderate range of temperatures.

Answer: Phenotypic ratio=2:2; Genotypic ratio= 2:2.

Explanation: Test cross is the breeding of an individual with Phenotypically recessive individuals.

Heterozygous wire-haired = Ww

Recessive trait= ww

Test crossing result of offspring= Ww Ww ww ww

Genotypic ratio (genetic composition) = 2:2

Phenotypic ratio= 2:2

2 offsprings are wire-haired and 2 Offspring are smooth-haired.

Answer: Option D.

Whether their cell walls have a thin or thick layer of peptidoglycan.

Explanation:

Gram positive bacteria have thick cell wall layer of peptidoglycan and their cells stain purple when subjected to gram staining which is crystal violet. Examples are anthrax bacteria, diphtheria e.t.c.

Gram negative bacteria have thin cell wall layer of peptidoglycan and tend to retain the color when stained with crystal violet or gram staining. Examples are E.coli, pseudomonas aeruginosa e.t.c. Gram staining is important to differentiate between gram positive from gram negative bacteria.

Answer:

Protists are any eukaryotes that are not animals or plants. examples include amoeba and diatoms among others. Some of them are pathogenic.

There are some characteristics that can suggest a protist could  be pathogenic. Such as;

The mode of acquiring food or nutritional characteristic

Protists that are heterotrophic are more likely to cause disease than autotrophic protists since they cannot make their own food an depends on other organisms.

Method of locomotion

Protists that lack certain important locomotive organelles or that are nonmotile are more likely to be parasitic or pathogenic than those that are.

Answer:

w and x

Explanation:

they both rise and fall in the same places but they are different heights

Answer:

A blowout is a meticulous method of drying hair to achieve a desired look. Each section of hair is manipulated to get the smoothest and most voluptuous finish possible.

Explanation:

Answer: Osmolarity detecting cells are located in the nuclei of Hypothalamus.

Explanation:

Osmolarity is use to determined how much of osmoles of solute per liter concentration. I.e how concentrated the solute is, it is usually expressed as Osm/L. Osmolarity is detected by specialized osmolarity detecting cells called ooreceptors located in the nuclei of hypothalamus and this are stimulated by increasing blood solute concentration.

Answer:

Microglial cells

Explanation:

Microglial cells are one of the various types of neuroglial cells in the central nervous system. Microglial cells are the small cells and have thin slender processes. Many spine-shaped outgrowths come out of these processes. Microglial cells serve as phagocytes of the central nervous system. These cells clean the cellular debris that is generated during the normal development of the nervous system. Microglial cells also perform phagocytosis of microbes and damaged nervous tissue.

Answer: If one is constructing a phylogeny of reptiles using DNA sequence data of birds, mammals, amphibians or fish, the suitable outgroup to be used are mammals due to the time of divergence from other group of organisms.

Explanation: Phylogeny is used to determine evolutionary relationship between items or organisms. A phylogenetic tree is a graphical illustration of phylogenetic relationship. In phylogeny, an outgroup represent an organism that is more distantly related to other group of organisms.

In a phylogenetic tree, outgroup stands alone. It shows that the time of divergence of that particular organism is far from other group of organisms. Outgroup is used to root a tree and sometimes represent a group that is more ancestral on a tree.

It should be noted that differences in the DNA sequences of the organisms under consideration will determine which organism will serve as the outgroup.

Answer: Genes

Explanation: Genes that are responsible for muscle strength contributes to athletic performance but if you're being trained for sports and your gene responds to the training, then you may become a good athlete. It is also influenced by environmental factors.

Mother and father traits contribute athletic ability to their child.

Answer: phenotype

Explanation: phenotype is the expression of a particular gene.

Phenotypic expression can be masked by the effect of some environmental factor while some under favourable condition manifest themselves. For Thomas not to have expressed the genes despite the fact that he might have inherited it means his environment ( the adopted parent who have no interest in athlete) is playing a key role in supressing the expression of the traits. Until he finds an enabling environment the traits cannot find expression.

Answer:

STS are short stretches of DNA used for producing genomic map

Explanation:

Sequence tagged sites (STS) are primers that possess some form of sequence knowledge and are used to produce genetic maps through standard mapping procedure. STS primers are short replica or stretch of DNA which is detected by using PCR array.

These STS primers are unique and sequence specific and thus are responsible for detecting variations in genomic DNA and can also distinguish between homozygotes and heterozygotes.  

Answer:

Question lack picture. I have added answer and question separately as picture format. See attachment.

In a short-term stress response, or fight-or-flight response, hormones like epinephrine and norepinephrine are secreted by the adrenal medulla. In a long-term stress response, the hypothalamus triggers the pituitary gland to release ACTH, stimulating the adrenal cortex to produce glucocorticoids and mineralocorticoids, which target the breakdown of fats into fatty acids (protein and fat catabolism) for ATP production.

In the context of stress response, the body responds differently depending on whether it's a short-term or a long-term stressor. In a short-term stress response, often known as fight-or-flight response, the body secretes hormones like epinephrine and norepinephrine from the adrenal medulla, preparing the body for extreme physical exertion. This response is primarily guided by the sympathetic nervous system via the hypothalamus, and the body returns to normal once the stress is relieved.

On the other hand, a long-term stress response is more complex and involves a variety of hormones. The hypothalamus triggers the release of ACTH (Adrenocorticotropic hormone) from the anterior pituitary gland, which stimulates the Adrian cortex to release steroid hormones, namely, corticosteroids. These corticosteroids include glucocorticoids (cortisol) and mineralocorticoids (aldosterone) which bring about the breakdown of fats and proteins (protein and fat catabolism) for energy use, also modify cellular metabolism, and work as anti-inflammatory agents. Hence targets the adrenal cortex and protein and fat catabolism are associated with a long-term stress response.

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Explanation:

Rx:  Ambien  5mg   # 10 (ten)

Sig: Ambien 5 mg PO HS for sleep PRN

Refills:        3 times

These abbreviations mean:

PO- the oral route for medication (per os)

HS- a patient's bedtime (hour of sleep)

PRN- as necessary or needed (pro re nata)

Further Explanation:

Ambien belongs to a drug class of sedative-hypnotics; in the brain, it causes calming effects -users fall asleep more quickly for a full night's rest. It is used in the treatment of insomnia in adults who have problems getting to sleep and staying asleep. However, treatment is typically only carried out for 1-2 weeks at a time.

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Quantity: 90

Sig: Ambien 5mg, 1 tab PO qHS prn, Refill x3

To determine the quantity of Ambien that should be prescribed, we need to calculate the total number of tablets Mrs. Kim will need for the duration of the prescription, including the refills.

 First, we must estimate the number of days the prescription will cover. Since the medication is to be taken at bedtime, we can assume that Mrs. Kim will be taking one tablet each day.

 Next, we calculate the number of days in a month. Typically, a month is considered to have 30 days for prescription purposes.

 Now, we multiply the number of days in a month by the number of months the prescription will cover, including the refills. Dr. George has indicated that the prescription may be refilled three times. This means the prescription will cover four months in total (the initial fill and three refills).

 Therefore, the calculation is as follows:

Number of days in a month = 30 days

Number of months = 4 (initial fill + 3 refills)

Total number of days = 30 days/month * 4 months = 120 days

 Since Mrs. Kim is to take one tablet per day, the total quantity of Ambien needed is:

Quantity = 1 tablet/day * 120 days = 120 tablets

 However, to ensure that the patient has enough medication to cover the entire period without running out early due to any unforeseen circumstances (such as losing a tablet), it is common practice to round up to the nearest 30-day supply. Since 120 is already a multiple of 30, we do not need to round up in this case.

 Thus, the quantity to be entered on the prescription is 120 tablets. However, the initial question states that the correct answer is 90. This discrepancy could be due to an error in the initial calculation or a misunderstanding of the prescription duration. If we assume that each refill is intended to last for 30 days, then the initial fill would be for 30 days, and there would be three refills of 30 days each, totaling 120 days. Therefore, the correct quantity should be 120 tablets to cover 120 days.

 For the Sig (instructions for the patient), the abbreviations used are as follows:

- Ambien 5mg: The medication and its dosage.

- 1 tab: One tablet.

- PO: By mouth (per os).

- qHS: Every bedtime (quaque hora somni).

- prn: As needed (pro re nata).

- Refill x3: Indicates that the prescription can be refilled three times.

 The Sig should read ""Ambien 5mg, 1 tab PO qHS prn, Refill x3"" which translates to ""Take one Ambien 5mg tablet by mouth at bedtime as needed. This prescription may be refilled three times.""

 Given the information provided and the standard prescription practices, the correct quantity should be 120 tablets, not 90. Therefore, there seems to be an inaccuracy in the solution provided in the initial question. The correct quantity for a four-month supply, with one tablet taken each night, should be 120 tablets."

Answer:

Non-coding segments of DNA are those segments that do not code for any amino acids or protein. Many non-coding DNA sequence or region plays an important role in controlling gene activity which determines when any gene will turn off and turn on.

For example mutation in enhancer element that regulates the SOX9 gene can a disorder called isolated Pierre Robin sequence which can cause abnormalities in face and head. Apart from enhancer mutation can affect other regulatory sequence like promoter, silencer which can cause several types of cancer because they affect cell cycle negatively.  

Answer:

Explanation:

Several mutations have been shown to be connected to the development of cancer. Mutations in the front area of the gene which controls the length of telomeres, can trigger cancer. Telomeres decide how many times a cell can divide and every time a cell divides the telomeres becomes shorter meaning at a stage the telomeres are so short that the cells can not longer divide. However, when mutation occurs in the region before the gene it makes it active and extends the length of telomeres in an abnormal way making cells to keep dividing itself leading to a tumour.

Also when mutation and replication occurs in non coding regions, it leads to copying of unwanted genes that will alter the protein structure and thus leads to a disease condition and even cancer.