Assume that a cross is made between two organisms that are both heterozygous for a gene that shows incomplete dominance. What genotypic ratio is expected in the offspring?

Answers

Answer 1

Answer:

Genotypic ratio: 1 homozygous for parental trait : 2 heterozygous : 1 homozygous for other parental trait

Explanation:

Incomplete dominance is a non-mendelian type of inheritance in which one allele of a gene portrays Incomplete dominance over the other allele, hence, they combine to form a third phenotype that is a blending of both parental phenotypes. A very good example of incomplete dominance gene is that of flower colour in snapdragon plants.

One allele of the flower colour gene codes for RED while the other codes for WHITE. However, an heterozygous intermediate phenotype (PINK) is formed as F1 offspring when both parents are crossed.

If the F1 heterozygous offsprings are self-crossed i.e. two heterozygous offsprings, four possible offsprings will be produced with the genotypic ratio 1:2:1, which means that 1/4 of the offsprings will be homozygous for one of the parental traits, 2/4 will be heterozygous for both alleles, 1/4 will be homozygous for the other parental trait.

Answer 2

The branch of biology which deals with the study of genes and inheritance is called genetics. In genetics, there are two types of alleles present in an organism. These alleles are as follows:-

Dominant Recessive

Gregor John Mendel is the father of genetics and describes the functioning of the gene in an organism.

Let's assume the gene of the parents, The genes are as follows:-

Mother - TtFather - Tt

These different type of genes is known as Heterozygous genes. If we cross both the parents together, the gamete formed is "T" and "t".

After crossing the genes, the genotype of the offspring will be "TT", "Tt", "tt".

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Related Questions

If the ease of oxygen pickup depends on oxygen concentration, would respiration be easier in air or in water?

Answers

Answer:

Air

Explanation:

depending on the concentration of Oxygen in any of the  medium- air or water, Oxygen pick up will be easier in air than water. This is because the energy barrier for the exchange of gases is higher in a liquid medium than in a gaseous medium.

Respiration would be easier in air than water

What is Respiration ?

Respiration is the process of taking in oxygen and giving out CO₂ through the oxidation of organic substances in the body. The process of respiration leads to the production of energy in the body.

Respiration would be easier in a gaseous medium ( air ) than it would be in water because air molecules are more loosely packed and therefore allows the free movement of oxygen ( lesser energy barrier ) .

Hence we can conclude that Respiration would be easier in air than water.

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Maria is a 121-lb endurance athlete who is planning to employ carbohydrate loading before her next race. Which strategy correctly follows the guidelines for carbohydrate loading?
Choose the statement below that correctly describes a good carbohydrate loading strategy for Maria:
O Four to six days prior to the event, Maria should consume 550 g of carbohydrate daily and decrease to 220 to 275 g of carbohydrate daily 1 to 3 days prior to the event.O One to three days before the event, 550 g of carbohydrate daily is recommended.O One to three days before the event, 1210 g of carbohydrate daily is recommended.O Six days before the event, her carbohydrate intake should be 550 g per day.O Two days before the event, an intake of 33 g of protein daily is adequate.

Answers

Final answer:

The correct carbohydrate loading strategy for Maria, a 121-lb endurance athlete, is to consume 550 g of carbohydrate daily for 1 to 3 days before her event to maximize glycogen storage for energy.

Explanation:

The correct carbohydrate loading strategy for Maria—an endurance athlete who weighs 121 lbs—would be to consume 550 g of carbohydrate daily during the 1 to 3 days prior to her event. This is because carbohydrate loading is aimed at maximizing the storage of glycogen in the muscles, which is used for energy during prolonged periods of intense exercise. Ingesting 550 g of carbohydrates daily, following the guidance, should increase her glycogen stores, providing an energy reservoir that can be tapped into during her endurance event.

Concerning the options given, the second option correctly follows the guidelines for carbohydrate loading: "One to three days before the event, 550 g of carbohydrate daily is recommended." This is based on the principle that consuming high amounts of carbohydrates can boost glycogen stores within muscles, effectively increasing the athlete's endurance capacity during the race.

16.Globular proteins fold up into compact, spherical structures that have uneven surfaces. They tend to form

multisubunit complexes, which also have a rounded shape. Fibrous proteins, in contrast, span relatively large

distances within the cell and in the extracellular space. Which protein is not classified as a fibrous protein?

(a)elastase (b)collagen (c)keratin (d)elastin

Answers

Answer:(a)elastase

Explanation:

Elastase is a protease enzyme the function involves the cleavage of peptide bonds after amino acids with small side chains. This is responsible for cleaving the peptide bonds between the elastin fibers and aids in digestion of the elastic protein.

Thus the elastase cleaves the protein fibers this can be said that the elastase is a non-fibrous protein.

Answer:

The answer is A: elastase

Explanation:

Fibrous proteins are insoluble, long, rod-like structured proteins that have high α-helix or β-sheet content, and are made up of a repeating motif. They are mechanically strong, cross-linked and play structural roles in the extracellular matrix. Examples are: collagen, keratin and elastin.

You are studying a new variant of a eukaryotic cell. The variant cell has mutated so that it no longer attaches well to surfaces or initiates the formation of a biofilm. The mutation in this cell has most likely affected the _____.

Answers

You are studying a new variant of a eukaryotic cell. The variant cell has mutated so that it no longer attaches well to surfaces or initiates the formation of a biofilm. The mutation in this cell has most likely affected the _____.

Mitochondria

Flagellum

Cell membrane

Glycocalyx

Answer:

Glycocalyx

Explanation:

Biofilms refer to the irregular layers of cells. These are formed when a group of cells arranges themselves at a solid or liquid surface.  The glycocalyx forms a layer around some animal cells. It has glycolipids and glycoproteins.   These substances allow the cells to recognize one another and to make contact with one another. The formation of adhesive communications that hold the cells in the biofilms is mediated by the components of the glycocalyx. Therefore, any mutation in the genes coding for components of glycocalyx would affect the ability of the cells to form biofilms.  

Final answer:

In a eukaryotic cell, mutations hindering attachment and biofilm formation likely affect the cell adhesion mechanisms, particularly the functioning of adhesion molecules such as cadherins, selectins, and integrins.

Explanation:

The mutation in the eukaryotic cell described has most likely affected the cell adhesion mechanisms, specifically the functioning of proteins known as adhesion molecules. These molecules help cells stick to each other and to their surroundings, a crucial process for the formation of biofilms.

Depending on the organism and environment, various types of adhesion molecules might be present. Nonetheless, some universally key ones would include cadherins, selectins, and integrins. If a mutation affects these proteins, the cell may lose its ability to properly adhere to its environment or form biofilms.

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The PCR reaction for Lab 2 is listed below. For each component besides water, briefly describe the component’s role in the PCR reaction. PCR amplification recipe (stock concentrations):

14.4 μL high quality nuclease‐free water
6.0 μL 5x Phusion HF buffer
3.0 μL dNTPs (2.5 mM each of four dNTPs – dATP, dCTP, dGTP, dTTP)
0.3 μL Phusion DNA polymerase
0.3 μL pCD122 (a plasmid that serves as template DNA)
3.0 μL sense primer CD474 (5 μM)
3.0 μL antisense primer CD475 (5 μM)
30.0 μL total reaction volume

Answers

Answer:

PCR is known as polymerase chain reaction used to amplify DNA sequence of our interest into multiple copies. This technique is been commonly used by researchers to study in depth about the gene of interest during their research work.

Explanation:

PCR:- It is known as polymerase chain reaction, used to amplify DNA sequence of our interest into multiple copies mainlty to millions or trillions

Components

Phusion HF buffer:- This buffer Create optimal reaction conditions for high fidelity amplification of DNA

dNTPs:- They are the building blocks in the synthesis of new copies of DNA.There are four dNTPs used in the amplification process that is dATP, dGTP, dCTP, dTTP. these building blocks are added in equal proportion during the PCR reaction

Phusion DNA polymerase:- This is the enzyme used in DNA amplification, it is generally a fusion of  DNA binding domain to a portion of pyrococcus like proof reading polymerase. This enzyme is tolerant to various inhibitors, allowing strong amplification of DNA of interest with minimal optimization

pCD122:- Plasmid that serves as a template DNA for the amplification of desired DNA sequence of our interest.

sense primer CD474:- Sense primer is also known a reverse primer, it attaches to the stop codon of the complementary strand of DNA

antisense primer CD475:- This primer is also known as forward primer, it attaches to the start codon of the template DNA

All these components is added in such a way that the total mixture should have a reaction volume of 30.0μl

Final answer:

In the described PCR setup, specific roles include the buffer for maintaining an optimal enzyme environment, dNTPs as building blocks for new DNA, Phusion DNA polymerase for synthesizing DNA, plasmid template DNA for target amplification, and primers for initiating synthesis. The final concentration of each primer in the PCR reaction will be 0.5 µM.

Explanation:

In the PCR (Polymerase Chain Reaction) described, each component serves a crucial role:

5x Phusion HF buffer provides the optimal pH and ionic environment for the activity of the DNA polymerase.

dNTPs (deoxyribonucleotide triphosphates - dATP, dCTP, dGTP, dTTP) are the building blocks needed for the synthesis of new DNA strands.

Phusion DNA polymerase is the enzyme that synthesizes the new DNA strands by adding dNTPs to the primed DNA template.

pCD122 (plasmid template DNA) contains the specific region of DNA that we aim to amplify.

Sense and antisense primers (CD474 and CD475) are short pieces of single-stranded DNA that mark the starting point of DNA synthesis on each strand of the DNA template.

Regarding your specific query, mixing 5.0 µL of each 2.0 µM primer into a 20 µL PCR reaction will result in 0.5 µM final concentration of each primer in the reaction. This is because the primers are diluted by a factor of four (10 µL combined primers in a total volume of 40 µL).

What phenotypic ratio is expected in the offspring? Write your answer as numeric values and use a colon to separate numbers (e.g., 1:1)

Answers

Answer:

According to the statement of the question, this problem is solved by analyzing the sex chromosomes for the determination of sex, that is, as it does not present the problem (genotype and phenotype) of the parents, the indication that refers to the determination of sex, in women and his sex chromosomes are "XX" and the man who determines sex is "XY".

The result of sex in the offspring corresponding to the man, if the chromosome in the sperm fertilizes the ovule is "X", then the sex or phenotype of the individual will be "XX" (female) and if the fertilizing sperm is "Y" then the sex will be "XY" of male, the probability of being born female or male will be 50% and 50% and the ratio is 1: 1

Describe the biological roots of Behavioral Neuroscience.

Answers

Answer:

Explanation:

The biological perspective or roots, a way of looking at neuroscience is by studying the physical basis for animal and human behavior. It involves such things as studying the brain, immune system, nervous system, and genetics.

The biological perspective tends to stress the importance of nature.

The study of physiology and biological processes has played a significant role in neuroscience since its earliest beginnings. Charles Darwin first introduced the idea that evolution and genetics has roles to play in human behavior. Natural selection influences if certain behavior patterns will be passed down to future generations. Behaviors that help survival skills most likely are passed down than those that prove dangerous.

The biological perspective is a way of looking at human problems and actions. For instance in aggression, someone using the psychoanalytic perspective might view aggression as the result of childhood experiences, another might take a behavioral perspective to show how the behavior was shaped by association and punishment. A neuroscientist with a social perspective might look at the group dynamics and pressures that contribute to such behavior. The biological viewpoint, on the other hand would look at the biological roots that lie behind aggressive behaviors by considering how certain types of brain injury might lead to aggressive actions. Or might consider genetic factors that can contribute to such displays of behavior.

Which set of details correctly identifies a series of events in a sympathetic pathway?

a. thoracolumbar origin, long preganglionic fiber, NE release at ganglion, short postganglionic fiber, NE release at effector
b. craniosacral origin, short preganglionic fiber, ACh release at ganglion, long postganglionic fiber, ACh release at effector
c. thoracolumbar origin, short preganglionic fiber, ACh release at ganglion, long postganglionic fiber, NE release at effector
d.craniosacral origin, long preganglionic fiber, ACh release at ganglion, short postganglionic fiber, ACh release at effector

Answers

Answer:

Answer is C.

Explanation:

The sympathetic pathway is the pathway of the sympathetic nervous system. Sympathetic nervous system is a part of the autonomic nervous system , that  comprises of neurons regulating the body's involuntary actions. It is also regarded as the pathway by which organisms respond to stress, because it prepares the body for stress.

The sympathetic nervous system help the body to respond to stress by increasing the rate of the heart beat, activating the release of adrenaline among others.

The other part of the autonomic nervous system is the parasympathetic nervous system, which brings the body to the state of calmness and relaxed feeling after a stressful event. This is done by slowing down the heart rate and increasing the rate of digestion.

Which of the following answers describes the most direct consequence of tropomyosin binding to F-actin in muscle cells? a. The (-) end of F-actin is stabilizedb. Loss of actin-ADP subunits from the (+) end is preventedc. Myosin binding to F-actin is blockedd. ATP hydrolysis by myosin is blockede. The movement of myosin towards the (+) end is blocked

Answers

Answer: A

Explanation:

The "-" end of F actin is stabilized. The myosin head bind to actin and makes the actin filament to slide

Which primates are included among the prosimians? Why is this taxonomic group problematic?

Answers

Answer:

The primates that are included among the prosimians are all the existing or extinct species of strepsirrhines. The taxonomic groups are problematic because they fail to mark specific distinctions between the species.

Explanation:

It is from 85 to 55 million years that the primates have been existing on the Earth. They emerged as small terrestrial animals and developed into some large ones through the continued evolution of species. Daubentoniidae, Tarsiidae, Lemuridae are some examples of primate families.

Which of the following statements is FALSE? A. Biological systems are highly ordered so entropy changes are not relevant.B. The entropy of a biological system can decrease. C. Entropy is a measure of disorder. D. The entropy of an isolated system will tend to increase to a maximum value

Answers

Answer: Option A is false.

Biological system is highly ordered, entropy changes is irrelevant.

Explanation:

Biological systems is the network of complex important biological individual like cells, organelles, ordans, macromolecules. Biological systems obey the second law of thermodynamics in that entropy changes with gain or loss of energy and for this to happen, it must increase the entropy of the universe. Living organisms taking in food to decrease their entropy yet the overall entropy is increased when entropy within the organism decreased.

Final answer:

The false statement is that entropy changes are not relevant in highly ordered biological systems. Entropy is certainly relevant as it can decrease with an input of energy to the system, and it is a measure of disorder, with the entropy of an isolated system tending to increase over time.

Explanation:

The statement that is FALSE among the provided options is: "Biological systems are highly ordered so entropy changes are not relevant." This statement contradicts the principles of thermodynamics which apply to all systems, including biological ones. Biological systems, while highly ordered, still adhere to the laws of thermodynamics, and thus changes in entropy are indeed relevant.

Entropy can decrease in a biological system but only if there is an input of energy. This is because biological systems are not closed systems; they exchange energy with their surroundings. The second statement that says the entropy of a biological system can decrease is true given that energy is constantly being input into these systems to maintain their order and function. Thus, entropy in a local system can decrease if work is done on it, although the overall entropy of the universe still increases.

Entropy is a measure of disorder; a higher level of entropy corresponds to a higher state of disorder in the system. Finally, the entropy of an isolated system will tend to increase to a maximum value, which aligns with the second law of thermodynamics.

Cholesterol is an important component of animal cell membranes. Cholesterol molecules are often delivered to body cells by the blood. which transports the molecules in the form of cholesterol- protein complexes. The complexes must be moved into the body cells before the cholesterol molecules can be incorporated into the phospholipid bilayers of cell membranes. Based on the information presented, which of the following is the most likely explanation for a buildup of cholesterol molecules in the blood of an animal? The animal's body cells are defective in exocytosis. The animal's body cells are defective in endocytosisy c) The animal's body cells are defective in cholesterol synthesis. (D) The animal's body cells are defective in phospholipid synthesis.

Answers

Answer:

The correct answer is - option B. The animal's body cells are defective in endocytosis.

Explanation:

Endocytosis is the process that internalized the substances inside the cells by the area surrounded by the membrane. This is the process that helps in the internalize cholesterol-protein complexes.

If the cells are not able to perform this then cholesterol-protein would not be internalized by the cell and it will remain in the blood and buildup cholesterol molecule.

Thus, the correct answer is - option B.

The most likely explanation for a buildup of cholesterol molecules in the blood of an animal is that 'the animal's body cells are DEFECTIVE in endocytosis'.

Endocytosis refers to the movement of molecules from the surrounding cellular environment to the interior of the cell by transport vesicles.

This process (endocytosis) is caused by the invagination of the cell membrane to form an intracellular vesicle that contains extracellular fluid.

Cells absorb cholesterol from their surrounding extracellular environment by receptor-mediated endocytosis.

In conclusion, the most likely explanation for a buildup of cholesterol molecules in the blood of an animal is that 'the animal's body cells are DEFECTIVE in endocytosis'.

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Consider a population of 425 diploid giant Sequoia trees. In this population, you observe the following genotypic counts: 100 homozygous dominant genotypes, 250 heterozygous genotypes, and 75 homozygous recessive genotypes. What is the allele frequency of the recessive allele in this population (use two decimal places and the usual rounding conventions)

Answers

Answer:

q = 0.42

Explanation:

This question is an example of Hardy-Weinberg question and there are two equations necessary to carry out this question;

p + q = 1

p² + 2pq + q²  = 1

where;

p = the frequency of the dominant allele

q =  the frequency of the recessive allele

= the frequency of individuals with homozygous dominant genotype

2pq  = the frequency of individuals with heterozygous genotype

= frequency of individuals with the homozygous recessive genotype

Since the total  population = 425

q² = [tex]\frac{individuals with recessive genotype}{Total Population}[/tex]

= [tex]\frac{75}{425}[/tex]

q² = 0.1765            

To find q; we need to square root both side to           eliminate the square from  q².

∴ [tex]\sqrt{q^2}=\sqrt{0.1765}[/tex]

q = 0.4201

q = 0.42       (to two decimal places)

What type of protist is heterotrophic and includes species whose cells can come together to form a slug that moves to a new habitat?

Answers

Answer: Fungus like protists are heterotrophic and feed on organic matter.

Cellular slime mould is a species of fungus like protists that form slug.

Explanation:

Fungus like protists are heterotrophic and the feed on organic matter and mostly unicellular.

Cellular slime mould are protists belonging to class Dictyostelia. They are heterotrophic and decomposers that live on organic matter. When there is deterioration condition, the cells migrates together to form slugs and move to form new habitat. Some of the cells form stalk and others form spores.

Bird feathers are modified scales.
a. True
b. False​

Answers

Answer:

A. True.

Explanation:

True. b is wrong

The statement Bird feathers are modified scales is a. True.

Bird feathers and reptile scales share a common evolutionary origin. Feathers are considered to be highly modified scales that have evolved over millions of years.

This evolutionary relationship is supported by both developmental and genetic evidence.

Feathers and scales are made of the same protein, beta-keratin, which is a protein unique to reptiles and birds.

The fundamental structure of feathers and scales is quite similar, with both composed of layers of beta-keratinized cells.

This similarity in composition and structure strongly suggests a common ancestry.

During embryonic development, both feathers and scales start as simple epithelial cells.

As development progresses, these cells differentiate and form either feathers in birds or scales in reptiles, depending on the genetic instructions within the organism.

The evolutionary transition from scales to feathers is thought to have provided advantages such as improved insulation, enhanced aerodynamics for flight, and eventually elaborate displays for courtship. Feathers underwent various modifications and specialized into diverse forms serving different functions, from flight feathers for flying to down feathers for insulation and display feathers for attracting mates.

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Glucose moves from the plasma into a skeletal muscle cell, where it is used for energy. Through which fluid compartment does glucose move between the plasma and the skeletal muscle cell?

(A) Intracellular fluid
(B) Interstitial fluid
(C) Extracellular fluid inside of blood vessels
(D) Cytosol

Answers

Answer:

(B) Interstitial fluid

Explanation:

The interstitial fluid and blood plasma together make the extracellular fluid. The extracellular fluids are present outside the cells. The extracellular fluid that is present in the narrow spaces between cells of tissues is known as interstitial fluid. When a substance moves from blood plasma into the cells of a tissue, it crosses the interstitial fluid present between its cells. Therefore, when a skeletal muscle cell picks glucose molecules from blood plasma, it moves from plasma to the interstitial fluid to enter the cell.

Final answer:

Glucose moves from the plasma, through the interstitial fluid, which surrounds cells, to reach the skeletal muscle cells where it's used for energy. The correct answer is (B).

Explanation:

The movement of glucose from the plasma into a skeletal muscle cell follows a specific path through different fluid compartments in the body. After digesting carbohydrates, glucose is absorbed into the bloodstream and circulates in the blood plasma. From here, glucose must then pass through the interstitial fluid that surrounds the cells. This movement is facilitated by a concentration gradient, where glucose levels are higher in the blood compared to inside the cells, and by glucose transport proteins present in the cell membrane. Thus, the correct answer is (B) Interstitial fluid, as the glucose moves from the plasma, through the interstitial space, and finally into the skeletal muscle cells where it can be utilized for energy production.

It is important to note that insulin plays a significant role in this process by stimulating the uptake of glucose into cells, particularly into liver and muscle cells for storage and energy use. Additionally, the sodium-potassium pump and facilitated diffusion mechanisms are involved in maintaining proper electrolyte balance and glucose transport, respectively.

Inspired by Louis Pasteur's swan-neck flask experiment, a microbiology student was interested in repeating this experiment using hay (dried grass) infusion instead of nutrient broth. He placed a few strands of dried grass into a sugar solution in an open neck flask, drew the neck of the flask into a swan-neck shape using a flame and then boiled the contents of the flask for 30 minutes. After 3 days of incubation at room temperature he was surprised to see bacterial growth in the flask.Does this experiment support spontaneous generation? If not, how would you explain the growth bacteria in the flask?

Answers

Answer:

Bacterial endospores.  Some bacteria, specially from the phylum Firmicutes, produce endospores. An endospore is a resistance form of the bacteria, which is not reproductive. This form of the bacteria is resistant to heat, UV radiation, drought, cold, and might remain dormant for  long periods. The formation of the endospores is triggered by starvation. In the experiment mentioned, instead of nutrient broth, which production is controled, a hay infusion is used. Since this is a material picked up from the soil, there is a high possibility of the presence of the endospores, could be from Bacillus, since this genera of bacteria is widely present in the soil. Once the endospores are in a suitable envoronment (culture media) they will stop dormancy and became metabolically active. It is mentioned that the culture media was boiled for 30 minutes, and endospores are resistant at 100 °C for several hours.

Explanation:

Individuals with Huntington's disease possess an allele that causes severe symptoms by

a. replacing pairs of chromosomes.
b. halting the process of gene expression.
c. producing certain abnormal proteins.
d. regulating the sequences of nucleotides.

Answers

Answer:

C)Producing certain abnormal protiens

Explanation:

Mutations in HTT gene causes Huntington disease. The HTT gene provides instructions for making a protien called huntingin. This protien plays an important role in the neurons of brain. Thus as the gene coding for this protien is mutated ,abnormal protien is synthesized.

Indicate whether below is True/False.
1. Mammary glands are specialized for milk and hormone production.
2. Breasts contain areolar connective tissue but little dispose tissue.
3. Alveolar glands occur in lobes of mammary glands.

Answers

Answer:

1. True

2. True

3. False

Explanation:

Mammary glands are made up of glandular tissue. These are meant for the secretion of milk after the newborn birth. The milk ejection is done by different hormones, especially by progesterone and prolactin.

Glandular tissue acts as both exocrine and endocrine glands. The mammary glands are a suitable example as it secretes milk which is an exocrine secretion.  The hormone progesterone and prolactin released from the mammary gland are endocrine secretions.

The mammary glands consist of a sac-like structure called alveoli.

These are group alveoli form grape-like appearance called lobules.

The alveolar sacs contain sweat glands. These are the modified sweat glands that secrete milk. It is made up of fibrous connective tissue.

Which lymphatic tissue is associated with mucous membranes and is called mucosa-associated lymphatic tissue, or MALT?

Answers

Answer:

MALT or mucosa-associated lymphoid tissue refers to a bundle of lymphatic cells, known as lymphatic nodules, situated inside the mucous membranes, which envelopes the respiratory, gastrointestinal, urinary, and reproductive tracts. These nodules comprise macrophages and lymphocytes that fight against the entering bacteria and other pathogens, which moves into these pathways along with air, food, or urine. These nodules can be grouped together in clusters or can be present solitary.  

The major clusters of lymphatic nodules comprise adenoids, tonsils, and Peyer's patches.  

A cultured cell line appears to be having trouble surviving. You find that the cells do not appear to have normal chromosome separation. You decide to check for the presence of a mutated protein. Which of the following would your best target for analyzing chromosome separation?
a. Actin
b. Melanin
c. Tubulin
d. Shugoshin

Answers

Answer:

c. Tubulin

Explanation:

Tubulin protein is polymerized to form the cylindrical structures of microtubules. Microtubules form the spindle apparatus during cell division. The spindle microtubules become attached to the kinetochores of chromosomes and mediate the alignment of chromosomes at the equator of cells during metaphase. The shortening of spindle microtubules is responsible for the movement of sister chromatids during anaphase. The same event also moves the homologous chromosomes during anaphase-I.

Any failure in the formation of the spindle apparatus would not allow the proper separation of chromosomes. Therefore, the cell with abnormal chromosome separation might have a faulty or no tubulin.

Clearly cellulose is very abundant on earth, and it is a long-lasting stable substance. Many animals cannot digest cellulose. Given this, what prevents the bodies of dead plants from filling the earth? Something must decompose cellulose. This is where fungal decomposition comes in--fungi digest cellulose, as do many prokaryotes. Name two organisms that consume cellulose and make an educated guess as to whether each breaks down cellulose or simply excretes it as fiber.

Answers

Answer: Termites and herbivores.

Explanation:

The two organism that can digests cellulose are termites and herbivores. The termite contains protists known as mastigophorans carry out digestion of cellulose in the body.

The cellulose in this case is digested and prevents it from getting deposited in the environment.

The other organism are animals like ruminants which can digest cellulose in their gut. They partially digest the cellulose and regurgitate it into the mouth an broken down further.

This process of digestion of cellulose in the gut is anaerobic so methane is released in the environment as a product of digestion.

The human beings cannot digest cellulose and a very less amount of it is considered as fiber and is simply excreted.

Final answer:

Ruminants like cows and fungi have the ability to digest cellulose due to cellulases, allowing them to break down this complex carbohydrate and utilize it as a food source, contributing to nutrient cycles in ecosystems.

Explanation:

Despite the fact that cellulose is a major structural component of plant cell walls and abundantly found on Earth, not all organisms can digest it due to its robust β(1,4) glycosidic bonds. However, certain organisms such as ruminants and termites have symbiotic relationships with microorganisms in their guts that produce enzymes capable of breaking down cellulose. For instance, cows have bacteria in their rumens that secrete cellulases, allowing them to break cellulose down into usable sugars.

Fungi also play a crucial role in cellulose degradation in ecosystems. They secrete exoenzymes into the environment to decompose dead plant material, preventing accumulation of undegraded biomass. These enzymes cleave the cellulose into glucose monomers, which they then utilize for energy and growth, contributing to nutrient cycling in the environment. Therefore, both ruminants and fungi do not simply excrete cellulose as fiber but effectively break it down for consumption.

Which of the following is NOTa sign that a survivor may need stabilization?Select one:a. Excessive talkingb. Glassy and vacant eyesc. Strong emotional responsesd. Uncontrollable physical reactionse. Frantic searching behavior

Answers

Answer:

Excessive talking is not a sign that a survivor may need stabilization.

Explanation:

If thymine makes up 19% of the DNA nucleotides in the genome of a plant species, what are the percentages of the other nucleotides in the genome?

Enter your answer in the following order:

a. % of adenine,
b. % of cytosine,
c. % of guanine.

Answers

Answer:

a. 19% of adenine

b. 31% of cytosine

c. 31% of guanine

Explanation:

Chargaff's rule states that in a DNa molecule, adenine pairs with thymine and guanine pairs with cytosine.

For that reason, if 19% of the DNA is made up of T, then 19% as well will be made of A.

Then, 19% + 19% = 38% of the genome is comprised of A+T and the other 62% is comprised of G+C.

Since G and C pair with each other, 31% will correspond to G and 31% will correspond to C.

Final answer:

To determine the percentages of adenine, cytosine, and guanine in the DNA genome of a plant species when thymine makes up 19%, utilize Chargaff's rules. Adenine and guanine percentages can be calculated using the complementary base pairing concept.

Explanation:

a. Percentage of Adenine: Since in DNA, adenine pairs with thymine, if thymine makes up 19%, then adenine also makes up 19% to maintain Chargaff's rules.

b. Percentage of Cytosine: The percentages of adenine and thymine together add up to 38%. As adenine and thymine are complementary base pairs, cytosine pairs with guanine. Therefore, cytosine would make up the remaining 62%.

c. Percentage of Guanine: Following Chargaff's rules and considering the percentages of thymine and cytosine, guanine would also be 31%.

Cindy is 63 years old and at risk for osteoporosis. Which nutrients would be the most important to consume in adequate amounts to preserve bone mass?

Answers

Answer:

Consumption of adequate calcium and vitamin D

Explanation:

Osteoporosis is a condition whereby there is significant amount of porous partition on the surface of the bone,this disease reduces the density and quality of bone. As bones become more porous and fragile, the risk of  getting fractured is greatly increased.

In a bid to overcome or to be resistant to Osteoporosis, Consumption of adequate calcium and vitamin D is required by the body. This because Calcium performs various function which include;building strong bones, regulating heart beat and fluid balance within cells.

Also Vitamin D works in conjunction with calcium to slow down or even reverse osteoporosis.  That is to say, the body cannot absorb calcium at all without some vitamin D. Hence, Vitamin D is vital in assisting the body absorb and use calcium.

Final answer:

The most important nutrients to preserve bone mass and prevent osteoporosis are calcium and vitamin D. Regular exercise, particularly resistance training, is also crucial for maintaining bone density. Other nutrients like vitamin K, magnesium, and omega-3 fatty acids can also support bone health.

Explanation:

The most important nutrients to consume in adequate amounts to preserve bone mass for someone at risk for osteoporosis are calcium and vitamin D. Calcium is a critical component of bone and must be obtained from the diet, while vitamin D is necessary for the absorption of calcium. Both nutrients play crucial roles in bone health and can help prevent bone loss.

Regular exercise, especially resistance training, is also important for preserving bone mass and preventing osteoporosis. Exercise stimulates the deposition of bone tissue and helps improve bone density.

Additionally, other nutrients like vitamin K, magnesium, and omega-3 fatty acids also support bone health.

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You decide to try Alex’s response to glucagon. This test consists of injecting a high dose of glucagonintravenously and then drawing samples of blood periodically and measuring the glucose content ofthe samples. After the glucagon injection, Alex’s blood sugar rises dramatically. Is this the responseyou would expect in a normal person? Explain

Answers

Answer:

YES!

Explanation:

Yes, because the liver is responsible for regulating blood sugar.

What symbiotic partnerships form between plant roots and fungi, and increase water and mineral absorption by the plant?

Answers

Answer: It is a Mutualistic partnership between fungi and roots of plant

Explanation:

Mutualism is a relationship between two organisms in which both of the benefits from the relationship. Fungi live in the roots of plant thereby getting carbohydrates from plants made by photosynthesis and the fungi in turn put mycelia that help plant to increase absorption of minerals and water.

Answer: Symbiosis

Explanation:

The root nodules of legumes contains organisms of the kingdom, fungi that helps the legumes fix atmospheric nitrogen thus meeting it's need for mineral Nitrogen. In turn, the legume plants provides shelter to the fungi.

This codependent partnership is known as Symbiosis

The GART, a gene which is involved with Down syndrome, is found toward the bottom of chromosome 21. This is an example of a ________.

Answers

Answer: It is an example of TRISOMY 21.

Explanation:

Trisomy 21 is a genetic condition in which 21 chromosome fail to separate during the growth of egg cell and sperm which make either of the two to an extra copy of chromosome 21. This is also known as down syndrome. This lead to slow and in appropriate mental development in affected individual.

Phosphoriboglycinamide transformylase(GART) is agene found on chromosome 21, which lead to down syndrome is an example of trisomy 21.

In humans alkaptonuria is a metabolic disorder in whichaffected persons produce black urine. Alkapotonuria results from anallele(a) that is recessive to the allele for normal metabolism(A). Sally has a normal metabolism, but her brother hasalcaptonuria. Sally's father has alcoptonuria, and her mother hasnormal metabolism.
a) Give the genotype of Sally,her mother,father and herbrother.
b) If Sally's parents have another child what is theprobabilty that this child will have alkaptonuria?
c) If Sally marries a man with alkaptonuria, what is theprobability that their child will have alkaptonuria?

Answers

If Sally marries a man with alkaptonuria, there is a 50% chance that  their child will have alkaptonuria.

The genotype of an individual refers to the sum total of genes that the individual received from its parents. Since Sally has a normal metabolism, Sally is Aa. Sally's mother is Aa while Sally's father and brother are aa.

If Sally parents have another child, using the Punnet square method, there is a 50% chance that the child will have alkaptonuria. If Sally marries a man with alkaptonuria, there is a 50% chance that  their child will have alkaptonuria.

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a) Genotypes: : Aa (carrier of the alkaptonuria allele) ,Mother: Aa (carrier of the alkaptonuria allele)Father: aa (has alkaptonuria)Brother: aa (has alkaptonuria)

b) If Sally's parents have another child:The probability of the child having alkaptonuria (aa genotype) is 25%.The probability of the child being a carrier (Aa genotype) is 50%.The probability of the child having normal metabolism (AA genotype) is 25%.

c) If Sally marries a man with alkaptonuria (aa genotype):The probability of their child having alkaptonuria (aa genotype) is 50% (as Sally is a carrier).The probability of the child being a carrier (Aa genotype) is 50%.

The probability of the child having normal metabolism (AA genotype) is 0% (as the husband has alkaptonuria).

Sally: Aa (normal metabolism carrier)

Mother: Aa (normal metabolism carrier)

Father: aa (alkaptonuria)

Brother: aa (alkaptonuria)

b) If Sally's parents have another child:

Probability of the child having alkaptonuria (aa genotype) is 25%.

Probability of the child being a carrier (Aa genotype) is 50%.

Probability of the child having normal metabolism (AA genotype) is 25%.

c) If Sally marries a man with alkaptonuria (aa genotype):

Probability of their child having alkaptonuria (aa genotype) is 50% (as Sally is a carrier).

Probability of the child being a carrier (Aa genotype) is 50%.

Probability of the child having normal metabolism (AA genotype) is 0% (as the husband has alkaptonuria).

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Contrast the genetic content and the origin of sister versus nonsister chromatids during their earliest appearance in prophase I of meiosis. How might the genetic content of these change by the time tetrads have aligned at the equatorial plate during metaphase I?

Answers

Answer:

Sister chromatids are identical  forms of chromatids  of a chromosomes. They are mostly  formed by semi-conservative replication of DNA molecule of a single  chromosome.Thus they  are  like  'photocopies' of  original parent  chromosomes; joined together at the Centromere.

They are exactly similar in all ramification; with the same gene and allele compositions..

However; slight differences  arise between the two identical sisters due to  mutation from errors at replication;and also in  the  length of telomere repeats.

Non-sister chromatids are dissimilar forms of  chromatids of a chromosomes formed  when each half of  a chromosome  at fertilisation from separate   haploid sex-cells, of each parent. fused.They contain different genetic composition;because  they are not on the same homologous chromosomes.Therefore crossing -over ensure variation.

However, they are genetically  similar in composition; if they  are contained in homologous chromosomes. This is because Synapsis of bivalent of these chromosomes allow genetic material to be shared by  chromosomal crossing-over between the non-sister chromatids  on the chromosomes ; therefore identical genetic characteristics are shared .

Explanation:

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