Answer:
Post reproductive isolation → Hybrid sterility
Explanation:
The biological concept of species states that individuals of a species can not mate and reproduce with individuals of another species. But if they get to reproduce, the progeny will not be viable or fertile. There will not be any reproductive success.
There are different reproductive isolation mechanisms, which are barriers that inhibit or interrupt the genetic flow between different species.
Reproductive barriers are isolation mechanisms that prevent mating between two or more species. The prezygotic mechanism avoids fertilization between individuals of different species, while the postzygotic mechanism impedes the zygote to develop and reach the adult stage.
Postzygotic mechanisms or barriers include
Hybrid inviability, Hybrid sterility, Hybrid reduced viability or fertility, Cytoplasmic interactions.In the exposed example, sympatric speciation occurs. It seems that the mountains separating the snails´ populations made a place for speciation and the development of postzygotic barriers, specifically hybrid sterility. After the erosion process, both populations got to meet again. Snails from one population get to mate and produce offspring with the snails of the other population, but their progeny is sterile.
what type of stream valley forms oxbow lakes ?
1. Youthful stream valley
2.mature stream valley
3. Old age stream valley
Answer:
I thik answer is 2 number
Match the terms in column B to the descriptions in column A.
Column A:
1. Connects the larynx to the main bronchi
2. Includes terminal and respiratory as subtypes
3. Food passageway posterior to the trachea
4. Covers the glottis during swallowing of food
5. Contains the vocal cords
6. Indentation on the lung where the lung root structures enter and exit
7. Pleural layer lining the walls of the thorax
8. Site from which oxygen enters the pulmonary blood
9. Connects the middle ear to the nasopharynx
10. Pleural layer in contact with the surface of the lung
11. Increases air turbulence in the nasal cavity
12. Separates the oral cavity from the nasal cavity
Column B:
a. alveolus
b. bronchiole
c. conchae
d. epiglottis
e. esophagus
f. hilum
g. larynx
h. palate
i. pharyngotympanic tube
j. parietal pleura
k. trachea
l. visceral pleura
Answer:
1. Connects the larynx to the main bronchi k. trachea
2. Includes terminal and respiratory as subtypes b. bronchiole
3. Food passageway posterior to the trachea. e. esophagus
4. Covers the glottis during swallowing of food d. epiglottis
5. Contains the vocal cords g. larynx
6. Indentation on the lung where the lung root structures enter and exit f. hilum
7. Pleural layer lining the walls of the thorax j. parietal pleura
8. Site from which oxygen enters the pulmonary blood a. alveolus
9. Connects the middle ear to the nasopharynx. i. pharyngotympanic tube
10. Pleural layer in contact with the surface of the lung l. visceral pleura
11. Increases air turbulence in the nasal cavity c. conchae
12. Separates the oral cavity from the nasal cavity h. palate
Explanation:
1. The trachea is between the main bronchi and the larynx. It has semicircular rings of cartilage. The function of this organ is to conduct the air from the larynx to the primary/main bronchi.
2. After the primary (main), secondary, and tertiary bronchi, the terminal and the respiratory bronchioles come. They also conduct the air towards the alveoli so that the oxygen can enter the blood. The diameter of the bronchi terminal is smaller than the bronchi, and the diameter of the respiratory bronchi is smaller than the previous ones.
3 and 4. The esophagus is not part of the respiratory system. It is a tube that belongs to the digestive system since food has to pass through it to go to the stomach. The esophagus is posterior to the trachea, and the epiglottis closes the entrance to the larynx when we swallow to stop food from going to the lungs.
5. The vocal folds are in the larynx. This one is between the trachea and the pharynx. There are two types of vocal folds, the true and the false vocal folds. Both of them vibrate when air passes through them, allowing us to speak and make different tones.
6. The hilum is the lungs section where the bronchus, the pulmonary artery, and the pulmonary vein enter the lung. It is an indention that is in the middle part of the lungs.
7 and 10. The pleura has two sides, the parietal one and the visceral one. The first one is in contact with the lungs and the second one with the thorax's walls. Between them, there is a space called the pleural cavity. The cavity has fluid that allows the movement of the two pleurae. As a result, the lungs can move and fill with air.
8. After the respiratory bronchioles, we have the alveolar duct. The alveolar ducts lead to alveolar sacs. The alveolar sacs has the alveolus. They are thin walls that are in contact with capillaries. When the air is there, the oxygen passes through the thin walls. Then it goes through the capillaries' walls and into the blood.
9. The pharyngotympanic tube is also known as the Eustachian tube. It connects the middle ear to the nasopharynx. Its function is to regulate the pressure in the ear.
11. The conchae are in the nasal cavity. They are three projections in the nasal cavity, the inferior one, the middle one, and the superior. Their function is to increase the surface of the nasal cavity so that more air can enter with every inspiration. As they are projections, they modify the laminar airflow producing a turbulent flow.
12. The palate is between the nasal cavity and the oral cavity. It has two parts, the soft palate, and the hard palate. The palate helps in the production of certain sounds and divides the nasal cavity from the mouth.
An increase in the biodiversity of an ecosystem leads to an increase in its productivity.
True
False
Answer:
true
Explanation:
Which life process is vital for the survival of all organisms
Answer:
oxygen
Explanation:
Which of the following would provide the best evidence that species A and species B have a common ancestor?
O A. The limbs of species A perform a function that is similar to the limbs of species B, but they have a different structure.
OB. Species A has limbs, and species B does not have limbs.
O c.
The limbs of species A and species B have a different structure and function.
OD
The limbs of species A have a similar structure to the limbs of species B, but they perform a different function.
Answer:
I believe the answer is D
Explanation:
Answer:
B
Explanation:
The answer is B on Plato. Just took the test and got it right.
Transcription and Translation Questions Note: Answer the following questions without taking start codon into consideration. 6. The template DNA sequence is 5’ ACCTGAGTC 3’, what is the transcribed mRNA sequence from 5’ to 3’?
explain how the tissue of the esophagus and tissue of the trachea can be differentiate
Answer:
Trachea: It is the wind pipe — making it a part of the respiratory system
Esophagus: It is the food pipe — making it a part of the digestive system
Trachea: It is shorter, 10–11 centimeters. It connects upper airway to the lungs
Esophagus: It is longer, 25 centimeters. It connects mouth to the stomach
Trachea: It is cartilaginous, made of C-shaped semicircular cartilages. They give it structural stability and prevents it from collapsing
Esophagus: It is muscular. It contracts in a wave-like motion through it’s length to propel food from mouth to stomach a.k.a swallowing
Trachea: It’s opening is protected by Epiglottis, a flap like structure, to prevent food from accidental entering the air passage. It prevents choking
Esophagus: It’s opening is protected by two sphincters. They are muscular rings that constrict to close the esophagus off when food is not being swallowed.
Trachea: It has 2 portions — cervical and thoracic i.e. neck and chest portions.
Esophagus: It has 3 portions — cervical, thoracic and abdominal i.e. neck, chest and stomach portions.
Lainey is looking for a new apartment and her realtor keeps calling her with new listings. The calls only take a few minutes, but a few minutes here and there are really starting to add up. She's having trouble concentrating on her work. What should Lainey do? a) Tell her realtor she can only receive text messages O b) Limit the time spent on each call O c) Turn off her phone until she is on a break O di Call her realtor back when customers won't see her on the phone
Answer:
c) Turn off her phone until she is on a break
Explanation:
If she does option "a" then her phone will still keep ringing with text message alerts. Option " b" will still be consuming her work time and option "d" can't be because it'll be too late by then so only option c makes sense.
Suggest how whooping cough spreads from person to person
Answer:
People with pertussis usually spread the disease to another person by coughing or sneezing or when spending a lot of time near one another where you share breathing space. Many babies who get pertussis are infected by older siblings, parents, or caregivers who might not even know they have the disease.
Explanation:
Answer:
Explanation:People with pertussis usually spread the disease to another person by coughing or sneezing or when spending a lot of time near one another where you share breathing space. Many babies who get pertussis are infected by older siblings, parents, or caregivers who might not even know they have the disease
5. Imagine that you encounter two color morphs of S. marcescens in natural environment. If you do not have any prior knowledge on the prodigiosin synthesis pathway of these Serratia species, will you consider the red and white Serratia colonies as one species or two species? Do you think we should designate species based on morphological traits or genetic differentiation? Without limiting to the S. marcescens experiment, provide evidence and example to support your claim.
Answer:
According to the given question that without prior knowledge on the prodigiosin synthesis pathway of these Serratia species, and we experience two different color morphs of the species. So, there is given or known preliminary morphological basis we will consider them different two different colonies. These can be identify only after genetically identify or study.
The morphological characterstics of the organism and the genetic identification both are equally important. So Giving any one more importance over other is not logical.
Morphologically we can differentiate the color of the colonies of mutants. The mutant serratia sp. are known for eficient biosynthesis of prodigiosin, Thus on the basis of color characteristics of their colonies when grown on peptone glycerol medium we can identify.
The difference between active transport and passive transport is that a. concentration gradients are involved in one and not in the other. b. glycolipids play a role in one and not in the other. c. one requires expenditure of energy by the cell and the other does not. d. ions are transported into and out of the cell by one process and not by the other.
Answer:
D) ions are transported into and out of the cell by one process and not by the other
what is The Catalys?
Answer:
A catalyst is a chemical substance that speeds up the rate of a chemical reaction at any given conditions.
WHAT IS PHOTOSYNTHESIS?
Answer:
Photosynthesis is the process in which plants make their own food from inorganic substances like carbondioxide and water to organic substances in the presence of sunlight and chlorophyll
Explanation:
Have a nice day
what are the effects of under secretion of steroid hormones
Answer:
Anabolic steroids are related to testosterone, the major male hormone. Abusing the hormone can lead to physical and psychological side effects. Problems include breast development and hair loss among men, and facial hair growth, menstrual problems, and a deepened voice in women.
Give reason. Mosquitoes and housefly are placed in the phylum arthropoda
Mosquitoes and Housefly are grouped under the phylum Arthropoda because they both posses jointed legs (or appendages).
Animals that have jointed appendages are called Arthropods (i.e animals with jointed legs). They are triploblastic (i.e they have 3 germ layers) and exhibit bilateral symmetry. They are both six-legged arthropods sub-grouped under the class Hexapoda (coined from the word "hexa" meaning six and "Poda" meaning leg). Their body are divided into three parts; head, thorax and abdomen.
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Bacterial strain A (met , his-, arg ) is mixed with bacterial strain B (met-, his , arg-), grown in complete media, then plated on minimal media supplemented with histidine. Bacteria grows on the plate, which indicates that ________.
Answer:
Plate has all the essential nutrients that is required by the bacteria.
Explanation:
Bacteria grows on the plate, which indicates that the plate has all the essential nutrients that is required by the bacteria. Bacteria needs a type of media for its growth which contains all the necessary items which plays a key role in its growth and development. This plate has a media that contains food which the bacteria use for its growth and development so that's why we can say that plate has all the essential nutrients that is required by the bacteria.
A 36 years old female came to see the doctor with the complaint of amenorrhea (absence of menstrual period) for past three months. She is worried if she is pregnant although she never missed her Contraceptive Pills. She recently noticed some milky discharge from her left breast and abnormal facial hairs. She also said that her period was irregular and less in amount for past one year. She also gained 15 pounds during last year but denies any cold intolerance, dry skin, depression, fatigue or hot flushes.
Her menarche was at 14 and she had regular 21 days cycle till last year when it became irregular and gradually scanty.
Her LMP (last menstrual period) was 3 months back.
She has a 10 years old healthy child (normal vaginal delivery). She has regular Pap test, last test was done 10 months ago which was normal.
Her mother had menopause at age 55.
Her vitals were normal.
Her pregnancy test was done at the office which was Negative.
The doctor reassured and discussed hormonal imbalances with her. She recommended her urine and blood tests for hormone levels and pelvic ultrasound to check her reproductive organs. The doctor advised her to come back with all the test result
1. After reading given case, what will be the most likely diagnosis of this patient?
2. Normal woman does not show and breast discharge, what could be the reason in this patient for breast discharge?
Help me answer to your best of ability
Answer:
The correct answer is -
1. galactorrhea or prolactinoma
2. hormonal imbalance (prolactin imbalance)
Explanation:
The given symptoms experienced by the patient are the absence of menstrual cycles, an increase in weight, some milky discharge from her left breast, and abnormal facial hairs. The pregnancy tests show she is not pregnant and she also does not experience cold intolerance, dry skin, depression, fatigue or hot flushes.
All these suggest that there is a hormonal imbalance in her and due to producing a white substance from the breast it can get that there is too much production of prolactin, which suppresses the effect of estrogen, which is producing by the pituitary gland.
Galactorrhea or prolactinoma are two disorders that affect the production of prolactin that cause milky discharge, it can be caused by overuse of contraceptives or underlying conditions.
difference between chromosomes and DNA
Answer:
The DNA in a human body is organized into many stretches of genes. Proteins attach themselves to these stretches and coil them so that they form chromosomes. These stretches are very important in the formation of an organism. Do you know why?
Macronutrients is the most readily available for energy production
Answer:
The correct answer is - true.
Explanation:
All the macronutrients carbohydrates, fats, and protein all three produce energy. Carbohydrates are the most preferred source of energy for the human body as it is the macronutrient that your system most requires.
The human body easily breaks down most carbohydrates and provides a significant amount of energy. The energy-providing process is called cellular respiration starts with glucose as a substrate. Glucose is the simplest carbohydrate.
Which of the following sequences on a DNA molecule would be complementary to
GCTTATAT?
A.TAGGCGCG
B.COD ATCCGCGC
C.KUCGAATATA
D.TGCCTCTC
E.None of the above
In eukaryotes, miR-5 is a microRNA with sequence complementarity to the 3’UTR of the mRNA transcript encoded by the UCD gene. The UCD gene encodes for a protein that acts as an allosteric activator of the protein p101. The p101 protein is a DNA methyltransferase (DNMT) which is an enzyme that methylates DNA when the protein is in its active form.
A) In a lung cancer cell line, the p101 protein is overly active leading to misregulation of gene expression. In these lung cancer cells which of the following would be a correct prediction in regards to miR-5? [choose ALL that apply, note that incorrect selections will deduct points]
A) The gene encoding miR-5 is in a region of chromatin with hypoacetylation
B) The gene encoding miR-5 is in a region of chromatin with hyperacetylation
C) The gene encoding miR-5 in a region of euchromatin
D) The gene encoding miR-5 in a region of heterochromatin
E) The miR-5 transcript is translated at high levels
F) The miR-5 transcript is activating translation of the UCD gene
B) A different lung cancer cell line is found to have significantly decreased amounts of DNA methylation. Which of the following could explain why there is decreased DNA methylation in this cell line? [choose ALL that apply, note that incorrect selections will deduct points]
A) increased amounts of the p101 protein due to low levels of miR-5
B) increased amounts of the UCD protein due to high levels of miR-5
C) decreased amounts of the UCD protein due to low levels of miR-5
D) the absence of the UCD protein due to miR-5 inhibiting translation of the UCD mRNA transcript
E) the absence of the p101 protein due to miR-5 inhibiting transcription of the p101 mRNA transcript
Answer:
A) A) The gene encoding miR-5 is in a region of chromatin with hypoacetylation.
D) The gene encoding miR-5 in a region of heterochromatin.
B) D) the absence of the UCD protein due to miR-5 inhibiting translation of the UCD mRNA transcript
Explanation:
MicroRNAs (miRNAs) are small regulatory RNA sequences (approximately 20-24 nucleotides in length) that regulate gene expression by RNA interference (RNAi) mechanisms. These sequences (miRNAs) bind by base complementarity to messenger RNAs and thus inhibit protein translation and/or trigger mRNA degradation. In this case, miR-5 binds to the 3’UTR of the mRNA transcript of the UCD gene, thereby inhibiting/slowing protein UCD synthesis. The UCD protein is an allosteric regulator that binds and activates the expression of the p101 protein, thereby the miR-5 RNAi pathway also indirectly decreases the expression of the p101 gene. Moreover, hypoacetylation is an epigenetic mark generally associated with gene silencing (heterochromatin is a transcriptionally inactive state of chromatin).
Direct counts of cells in liquid samples can be performed using a Petroff-Hausser counting chamber. Research this method and describe how it compares to the viable plate count method of determining the number of CFU in a sample.
The viable plate count is the most commonly used method to estimate the number of viable cells in a sample; while the Petroff-Hausser counting chamber is a direct microscopic count method generally used for counting bacteria and sperm cells.
The Petroff-Hausser counter is a counting chamber used to count microorganisms and cells (e.g., bacteria and spermatic cells) under the microscope.
The Petroff-Hausser counter chamber is divided into 25 large squares which are in turn divided by double lines. Moreover, within each large square, there also are 16 small squares divided by single lines.
Petroff-Hausser counting chambers are often used to directly determine the number of bacteria in a culture or liquid medium. Petroff-Hausser counting chambers are used to count the number of cells in a given volume of culture liquid by observing 10 to 20 microscope fields.
The viable plate count, also known as simply plate count, is the most commonly used procedure to estimate the number of viable cells in a sample.
The viable plate count method consists in obtaining serial dilutions of a sample having viable microorganisms/cells which are plated onto a growth medium.
This method (viable plate count) can be used to determine the number of actively dividing cells.
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Part 1 of 1 -
Question 9 of 10
10 Points
When DNA is copied, sometimes there are mistakes. Approximately how often does this
happen?
O A. There aren't any mistakes.
OB. 1 in a billion bases.
OC. 1 in a million bases.
OD. 1 in a trillion bases.
Reset Selection
Answer:
D. 1 in a trillion bases
Explanation:
A mutagen agent can change the genetic information of organisms increasing mutations over the natural level. Mutagens cause changes in the bases, and pairing bases, that compose DNI strands.
A mistake in the process of DNI copy during cell division might cause genetic changes in daughter cells. Defects DNI replication might be inherited if it occurs in germinal cells. But it can also cause many significant epigenetic changes.
Many of these changes can be detected on time by enzymes such as DNI polymerase. This enzyme can correct these mistakes or at least some of them, moving from 3´to 5´direction, and eliminating the mistakes.
The highly effective replication system, together with the action of enzymes, makes it rare to occur a mistake in DNI replication. Generally speaking, the mistaken rates in DNI replication are very low, meaning that only one in a trillion times occurs a mistaken DNI copy.
Histones are essentially identical in sequence/structure in all eukaryotic organisms from yeast to plants to animals. What does this say about the biophysical properties of DNA-packaging and the evolution of eukaryotic organisms
Answer:
It indicates that core histone genes were present in the last common ancestor of yeasts, plants, and animals
Explanation:
Histones are highly basic proteins that can strongly interact with DNA, which is packaged into nucleosomes, the basic structural and functional unit of chromatin. Each nucleosome is composed of approximately 147 base pairs of DNA wrapped around a core of eight histone proteins (two copies of four types of histones H3, H4, H2A, H2B). These core histones are evolutionarily conserved across eukaryotic kingdoms in terms of sequence and structure. Therefore, DNA-packaging into nucleosomes is considered a constraint for the evolution of core histones. Moreover, the presence of conserved core histones in eukaryotic kingdoms (e.g., yeast, plant, and animal kingdoms) is strong evidence that histone-mediated DNA packaging was presumably present in the last common ancestor of eukaryotic genomes.
Sophia
Which of the following represents the correct flow of information during
homeostasis?
Home
a.) Integrator - Sensor - Effector
O b.) Sensor - Integrator - Effector
Hi
WHAT'S
oc.) Sensor - Effector - Integrator
Welcome to
allows your
o d.) Effector + Integrator + Sensor
1. Homeo
2 Sensors
Answer:
b.) Sensor - Integrator - Effector
Explanation:
During homeostasis, our body sense change that occur in the external environment through sensors. These sensors sends the information to integrator or the control center where the information is analyzed. After analysis, the instructions are send to the effectors to take action against the change which is occur in the external environment of our body so the correct order of homeostasis is Sensor - Integrator - Effector.
2. At what temperature did the prodigiosin-producing genes express in the S. marcescens culture? From the experiment you conducted in this lab, what evidence can you provide to support your claim?
Answer:
The temperature is the key factor for prodigiosin production. It has been shown that , S. marcescens can produce this kind of pigment at about 25 °C, which however could not produce the pigments at elevated temperatures, especially till 37 °C
Explanation:
Which type of seedless plant has a complex leaf arrangement off a vein?
a. java moss
b. club moss
c. ferns
d. horstails
B. Club moss
Explanation:
This is because club moss is an seedless evergreen plants that have scale-like leaves.
Can you plz mark me as brainliest!!!
answer : club moss
explanation: Because they have vascular tissue, seedless vascular plants
are often larger than nonvascular plants. Vascular tissue is spe-
cialized to transport water to all of the cells in a plant.
The body regulates the amount of hormones are released by using feedback loops. A __ feedback loop increases the response whereas a __ feedback loop decreases the response.
Positive feedback loop increases the response whereas a negative feedback loop decreases the response.
What is positive feedback?Positive feedback is the amplification of a body's response to a stimulus. For example, in childbirth, when the head of the fetus pushes up against the cervix (1) it stimulates a nerve impulse from the cervix to the brain (2).
A feedback mechanism resulting in the inhibition or the slowing down of a process.
Examples of processes that utilise negative feedback loops include homeostatic systems, such as thermoregulation (if body temperature changes, mechanisms are induced to restore normal levels), blood sugar regulation (insulin lowers blood glucose when levels are high ,glucagon raises blood glucose when levels are low).
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Which structure transports urine to the bladder by peristaltic action?
Answer:
The muscular layer of the ureter consists of longitudinal and circular smooth muscles that create the peristaltic contractions to move the urine into the bladder without the aid of gravity.
Answer:
The ureter
Explanation:
The ureter is a long thin tubular structure 10-12 inches long which carries urine produced in the kidney to the bladder. The urine is transported by a process called peristalsis. The ureter actively propels urine from the kidney down into the bladder.
How are oxygen and carbon dioxide exchanged between the alveoli and the
capillaries?
A. Endocytosis
B. Osmosis
C. Simple diffusion
5
D. Active transport
Answer:
B. Osmosis
Explanation:
Osmosis is the process in which oxygen and carbon dioxide exchanged occur between the alveoli and the capillaries because the oxygen enters the body and the carbondioxide gas leaves the cell through a semi-permeable membrane and we know that Osmosis is a process in which smaller molecules moves from higher concentration to lower concentration through semi-permeable membrane of the cell.
