Answer: Mycorrhizae
Explanation:
Mycorrhizae is a mutualistic symbiotic relationships that form between fungi and and the root of vascular plants.
The Mycorrhizae relationship that exist between fungi and root of plants, show how fungi colonize the root system of a their host plant, they make use of their extensive network of hyphae and large surface area which is in contact with the soil to channel water and minerals from the soil into the plant, thereby increasing nutrients absorption in plants.
The plant in return provide food in form of carbohydrates formed from photosynthesis to the fungi
Fungi of the phylum ascomycota are recognized on the basis of their production of _____ during sexual reproduction.
Answer:
The correct answer will be- sac like structure called ascus.
Explanation:
Ascomycota is the largest division of the fungi with about 64000 species which reproduces by both the ways of reproduction.
The ascomycetes are named so because of their sac-like microscopic structure called ascus formed during the sexual reproductive phase of the fungi. A single ascus produces eight ascospores through meiosis under optimal conditions which form the fungi.
Thus, sac-like structure called ascus is the correct.
Answer:
Asci
Explanation:
Ascomycota is a phylum of the fungi which is recognized based on characteristics sac-like structures known as ascus produced by them.These asci contain the ascospores during the sexual stage which may vary in number from four to eight. Due to the production of these sacs containing ascospores the fungi is also popularly known as sac fungi. The sac that is produced by the sac fungi is known as the ascus which is a sexual structure and is microscopic.Thus, the name of the phylum Ascomycota has been derived from the word ascus that is the sacs that are produced by these fungi.Describe a landscape design technique that benefits the environment.
Answer:
Green landscape technique
Explanation:
Land scape design technique is a technique used to design, create, and maintain your landscape to save time, money, and energy.
The main purpose of land scape is to minimize both the input of resources and the output of waste in our various yards and gardens.
Some of the best plants for land scaping are; Dogwood, flowering cherry tree, hedge Holly, Leyland Cypress and so on.
Green landscape technique nurture wildlife and also to reduce pollution (air, water and soil pollution).
Naturally obtained soft water collected from rainwater runoff from the roof in rain barrels can be used freely.
Conservation landscaping uses biological diversity, resource conservation, long-term planning, low impact, and water conservation to benefit the environment. It minimizes resource usage and supports local wildlife, reducing maintenance needs.
Landscape Design Technique Benefiting the Environment
One effective landscape design technique that benefits the environment is Conservation Landscaping. Here are its key features and benefits:
Biological Diversity: Utilizes a variety of plant species to promote beneficial insects, provide food and habitat for animals, and reduce pest and weather impacts.Resource Conservation: Uses existing topography, water, plants, and views in the design to minimize environmental disturbance and reduce costs. Sourcing locally supports the local economy.Long-Term Planning: Considers the mature sizes and future appearance of plants to reduce maintenance and labor costs.Low Impact: Minimizes site disturbance and uses plants appropriate for the site, requiring less maintenance, pesticides, and fertilizers.Water Conservation: Groups plants by water needs and employs tools like rain sensors, rain chains, barrels, and gardens to manage runoff effectively, conserving valuable water resources.Implementing these strategies not only enhances the aesthetic appeal and functionality of landscapes but also supports the larger community and watershed health.
Which type of genetic exchange occurs among bacteria in which dna is carried into a bacterial cell by means of a virus?
Answer:
Transduction
Explanation:
Among the three types of gene transfer in bacteria, transduction does not require physical contact between the two bacteria cell. Viruses that infect bacteria accidentally move pieces of chromosomal DNA from one bacterium to another. These viruses that infects bacteria are called bacteriophages.
Simply put, transduction is a process of genetic recombination in bacteria which involves the incorporation of the genes of a host cell (bacterium) into the genome of a viral cell (bacteriophage) and then conveyed to another host cell when the bacteriophage infects it. The process of transduction is of two types; generalized and specific depending on whether any gene or specific gene is transduced respectively.
The aim of these viruses when they infect bacteria cell is to harness the ability of bacteria to replicate, transcribe and translate their genetic material and use it to procreate into many virons or complete their viral particle.
A 30-year-old male who manages his type 1 diabetes with glyburide presents at the emergency room reporting headache, confusion, and tachycardia. He has come from a party at which he drank two beers to celebrate running his first half-marathon. Which of the following is likely to be the cause of his complaints?A)Blood glucose levels are primarily a result of the timing, quantity, and character of food intake.B)Ingested glucose that is not needed for cellular metabolism circulates in the blood until it is taken up to meet cellular needs.C)Blood glucose levels are kept in a steady state by selective excretion and reuptake by the kidneys.D)Glucose that exceeds metabolic needs is converted and stored by the liver.
Glucose that exceeds metabolic needs is converted and stored by the liver
Explanation:
The patient is suffering from type 1 diabetes and takes glyburide which is an oral hypoglycemic.
In the glucose pathway, excess glucose will be converted to glycogen and will be stored in the liver. Under normal conditions, when the body needs excess glucose, like in the above case after a marathon run, glycogen converts back to glucose by the processes of glycogenolysis and gluconeogenesis.
Glyburide acts against increasing blood glucose levels by inhibiting glycogen conversion to glucose in the liver.
Hence, the blood lacks the necessary amount of glucose leading to hypoglycemic conditions. Further, an imbalance of glucose efflux and influx also lead to hypoglycemia.
Hyoglycemia can cause neurogenic and sympathoadrenal symptoms like headache, confusion, tachycardia. or postural orthostatic tachycardia syndrome.
In a plant, the allele for tall stems (T) is completely dominant to the allele for short stems (t). A second gene located on a separate chromosome determines flower color. The allele for purple flowers (E) is completely dominant to the allele for white flowers (e). A tall plant with white flowers is crossed with a tall plant whose flower color is unknown. The resulting F1 offspring population showed the following phenotype frequencies: Phenotype Frequency Tall stems, purple flowers 3/8 Tall stems, white flowers 3/8 Short stems, purple flowers 1/8 Short stems, white flowers 1/8 What are the genotypes of both parent plants?
Answer:
Genotype of parents is TtEe and TtEe
Explanation:
Given –
Allele for tall stem is “T” and allele for short stem is “t”
Trait of tall stem is dominant over trait of short stem
Also given allele for purple color is represented by “E” and allele for white flower is “e”
Trait of purple color is dominant over trait of white color
Genotype of Tall stems, purple flowers – TtEe or TTEE
Genotype of Tall stems, white flowers – Ttee or TTee
Genotype of Short stems, purple flowers – ttEe or ttEE
Genotype of Short stems, white flowers – ttee
This makes it clear the genotype of parents would be
TtEe and TtEe
Final answer:
The most likely genotypes for the parent plants in the question are Tt ee for the tall, white-flowered plant and Tt Ee for the tall plant with unknown flower color.
Explanation:
In a plant, the allele for tall stems (T) is completely dominant to the allele for short stems (t). A second gene located on a separate chromosome determines flower color, with the allele for purple flowers (E) being completely dominant to the allele for white flowers (e). Given the phenotype frequencies observed in the F1 offspring, we can deduce that one parent is a tall plant with white flowers. Since it has white flowers, it must be homozygous recessive for flower color (ee). As the plant is tall, it can be either heterozygous (Tt) or homozygous dominant (TT) for stem height.
The second parent plant has an unknown flower color, but because there are both purple and white flowering offspring, it must have at least one purple allele (E) passed onto the offspring. Since there are short stems in the offspring and shortness is recessive, the second parent must also have at least one recessive allele for stem height (t).
To produce the observed 3:8 ratio for each tall phenotype (purple flowers and white flowers) and the 1:8 ratio for each short phenotype, the most likely genotypes for the parents are: Tall, white-flowered plant: Tt ee (tall and homozygous recessive for flower color) and Tall, unknown flowered plant: Tt Ee (tall and heterozygous for both traits). These genotype combinations explain the observed offspring phenotypes when utilizing the principles of Mendelian inheritance.
Zortronian scientists have noticed that Bellvoices in the cities use a different range of frequencies than Bellvoices in the forests do. Assuming the sound characteristics of Zortronian cities and forests are like those on Earth, what frequencies do Bellvoices use in the city and why?
Answer:
low frequency
Explanation:
Bellvoices in the city use lower frequencies because the back ground noise is low pitched and drowns out any high pitched sounds they make. Frequency and pitch are similar characteristics. SO, when the frequency is increase the pitch also increases.
The adaptation that made possible the colonization of dry land environments by seed plants is most likely the result of the evolution of _____.
A. pollen
B. heterospory
C. cones
D. ovules
Answer:
The correct answer is A. pollen.
Explanation:
Pollen is a very efficient way to colonize and reach further. It's believed that this adaptation is the main reason why land plants could colonize so many environments. The fact that pollen is little and easily transported also was very important.
The key adaptation facilitating the colonization of dry land by seed plants was the evolution of pollen, enabling reproduction without the need for water.
Explanation:The adaptation that made possible the colonization of dry land environments by seed plants is most likely the result of the evolution of pollen. Pollen and seeds provided evolutionary advantages that enabled plants to reproduce and develop independently of water, which was crucial for their success on dry land. Pollen, as the male gametophyte, is encased in a protective coat to prevent desiccation and can be transported far from the parent plant, ensuring gene spread. This adaptation, along with the development of seeds, allowed seed plants to become widespread and diverse by being less reliant on moist environments for reproduction.
An organism is discovered that thrives in both the presence and absence of oxygen in the air. Curiously, the consumption of sugar increases as oxygen is removed from the organism's environment, even though the organism does not gain much weight. This organism _____.
Answer: Facultative anaerobe
Explanation: Facultative anaerobes are organisms that can live with or without oxygen supply. this organism processes carbohydrates and proteins as oxygen is taken away from its enviroment in order to derive oxygen.
The organism is a facultative anaerobe, which prefers oxygen for aerobic respiration but shifts to anaerobic respiration or fermentation when oxygen is scarce, increasing sugar consumption without significant weight gain.
Explanation:The organism described is a facultative anaerobe, which is an organism that grows better in the presence of oxygen but can also proliferate in its absence. The increased consumption of sugar as oxygen is removed from the organism's environment is typical for facultative anaerobes as they switch from aerobic respiration, which is more efficient and generates more ATP, to fermentation or anaerobic respiration when oxygen is unavailable. Despite consuming more sugar via less efficient metabolic pathways in the absence of oxygen, the organism does not gain much weight due to the continuous loss of matter as carbon dioxide and water, products of cellular respiration.
Therefore, the correct answer to the organism's behavior when oxygen is removed is B. The organisms are facultative anaerobes.
Black eyes are dominant to orange eyes, and green skin is dominant to white skin. Sam, a MendAlien with black eyes and green skin, has a parent with orange eyes and white skin. Carole is a MendAlien with orange eyes and white skin. If Sam and Carole were to mate, the predicted phenotypic ratio of their offspring would be _____.
a) 1 black eyes, green skin :
b) 1 black eyes, white skin :
c) 1 orange eyes, green skin :
d) 1 orange eyes, white skin
Answer:
ratio: 25% for each phenotype
Explanation:
See attached document for proper explanation of this case
Compared with rods, cones are__________.a more sensitive to any light and less sensitive to fine detail.b less sensitive to dim light and less sensitive to fine detail.c more sensitive to dim light and less sensitive to fine detail.d more sensitive to dim light and more sensitive to fine detail.e less sensitive to dim light and more sensitive to fine detail.
Answer:
e less sensitive to dim light and more sensitive to fine detail.
Explanation:
Rods and cones are light sensitive receptors located at the back of the eye. The human eye contains between 5 million and 7 million cones and 110000000-130000000 bars.
Function
Cones and canes are essential for vision. Together they are able to detect movement, light and color, and transmit that information back to the brain.
Rods
The rods are highly sensitive cells located in the outer area of the retina (the lining of the back of the eye). They are used in low light areas and are more acute to light, shape and movement changes. Rubs do not detect color.
Cones
The cones are located in the central fovea (central area of the retina). They are less sensitive than rods and require bright lighting. Cones are fundamental to our ability to see color.
The cones have a high resolution and can detect the color. Its maximum sensitivity is located at a wavelength around 555 nm (yellow light). This is the so-called photopic vision, the one used during the day to see things in detail and color, making use of direct vision, the one with the highest definition. During astronomical observation, cones can also be used in the case of bright objects, such as planets or stars, whose light is intense enough to detect color. Sometimes we will also notice it in deep sky objects, such as planetary nebulae with high surface brightness.
Darkness arrives, when the light diminishes, the canes begin to work. This is the so-called scotopic vision. The rods are located outside the optical axis, with their area of greatest sensitivity located at an angle of approximately 20 ° around the fovea, and a maximum sensitivity at 507 nm (green light) The rods are responsible for night vision , also being the most sensitive motion detectors. The well-known peripheral vision technique uses the rods to detect weak objects at the time of observation.
Here we can notice that the maximum sensitivity between the cones and the rods is different, 555 nm for the former and 507 nm for the latter. This has the effect that during the transition from vision with cones (photopic) to vision with rods (scoopic) the maximum response moves at shorter wavelengths. This is called the Purkinje effect, and it means that for weak sources the eye is more sensitive to blue, while for bright sources it is more red.
Cones are less sensitive to dim light and more sensitive to fine detail compared to rods. They are responsible for color vision and high visual acuity. Rods, on the other hand, work well in low light conditions and are involved in peripheral vision and motion detection.
Explanation:Compared with rods, cones are less sensitive to dim light and more sensitive to fine detail. Cones are specialized photoreceptors located in the fovea, where images are focused, and are responsible for color vision and high visual acuity. In contrast, rods are distributed throughout the retina, function well in low light conditions, and are involved in peripheral vision and motion detection.
Learn more about Cones vs Rods in Vision here:https://brainly.com/question/29609086
#SPJ5
In the late 1960s, Robert Paine conducted landmark studies on diversity in the rocky intertidal zone, comparing the species diversity in control plots with diversity in experimental plots from which he removed the top predator, sea stars. After 5 years, 15 species of intertidal invertebrates lived in the control plots, while the experimental plots were dominated by only two species, one mussel and one barnacle. The process MOST likely responsible for the loss of species diversity in the experimental plots was:A. mutualism.B. predation.C. competitive exclusion.D. parasitism.E. resource partitioning.
Answer:
C: competitive exclusion
Explanation:
The competitive exclusion principle states that organisms living in the same community while competing for the same resources cannot coexist at a constant population rate.
Once some of the species within the community get a slight competitive edge over other species, they become dominant and this might lead to the extinction of the weaker species in the long run.
In the experimental plot, the removal of sea stars provided mussel and barnacle with a competitive advantage over other species within the community (sea stars are predators of mussels and barnacles). This led to the dominance of mussel and barnacle and the eventual extinction of other species within the experimental plot as compared to the control plot.
The correct answer is C.
A mutation that has risen to high frequency through a selective sweep shows a characteristic pattern in which only one allele is found for other loci that occur nearest the selected mutation. Which of the following contribute(s) to this pattern?
Answer:
Recombination is less likely to separate nearby allele from the favored one contribute(s) to this pattern.
Explanation:
Selective sweep is the reduction/elimination of the difference in the nucleotide (allele) in the mutation. This usually occurs in a rare allele. Mutation is the alteration in the existing gene of a species.Recombination is the exchange of DNA between the chromosomes of the parents. The mutation levels are prone to increase or decrease during recombination.Which of the following statements is correct? a)All animals share a common ancestor. b)Sponges are diploblastic animals. c)Eumetazoans have three embryonic tissue layers. d)Most animal phyla belong to the clade Radiata. e)The origin of all extant animal phyla can be traced to the Cambrian explosion.
All animals share a common ancestor.
Explanation:
According to various phylogenetic gene sequence analysis, there are various evidences that proved all animals originated from a common ancestor.
Initially, it was stated that all organisms descended from a single cell which then gave rise to multicellular organisms. Organisms that descend from a common ancestor are closely related and grouped.
The lineage of the common ancestor can be traced in the neoproterozoic era.
The last common ancestor or the basal animal was sea sponge according to some researchers. The last universal common ancestor is called as the concestor.
What caused metamorphism to occur in the blue ridge district during the alleghanian orogeny?
Answer: Strong differential stress generated by compressional stress was rocks were thrust westward.
Explanation:
Measuring its costs and benefits in terms of energy spent and/or gained, predation:
A. results in a gain for one individual and neither a gain nor a loss for the other.
B. results in a gain for both individuals.
C. results in a gain for one individual and a loss for the other.
D. is a lose-lose interaction.
E. None of the answer options is correct.
Answer:
The correct answer is C. results in a gain for one individual and a loss for the other.
Explanation:
Predation results in one individual gaining energy, the predator, and as the principle of energy conservation says, therefore, the prey represents a loss of energy.
In predation, there is an energy transfer from the prey to the predator. The act of predation can thus be described as a gain for one individual (the predator), and a loss for the other (the prey). So, the correct answer is option C.
Explanation:The act of predation is a biological interaction where a predator, an organism that is hunting, kills and eats its prey, the organism that is attacked. Thus, in terms of energy spent and gained, predation results in a gain for one individual and a loss for the other. The predator uses energy to catch its prey, but in return, gains more energy from the prey itself, which is usually more than what was expended in the hunting process, thus resulting in a net gain. However, the prey loses both its life and any energy it had. So, the correct answer to this question is C.
Learn more about Predation here:https://brainly.com/question/12410124
#SPJ11
The reaction-center chlorophyll of photosystem i is known as p700 because
In skeletal muscle, a muscle fiber that generates a larger contractile force has a larger number of:
a. mitochondria present.
b. crossbridges formed.
c. actin molecules present.
d. nebulin and titin molecules present.
e. thick and thin filaments present.
Answer:
b.crossbridges formed
A purebred plant that produces yellow seeds is crossed with a purebred plant that produces green seeds. The F1 plants have yellow seeds. What is the expected phenotypic ratio of seed color of the offspring of an F1 × F1 cross?
Answer:
3 yellow : 1 green
Explanation:
Let yellow seed trait be represented by Y allele and green seed trait by y allele.
Purebred yellow seed plant = YY
Purebred green seed plant = yy
YY x yy = Yy, Yy, Yy, Yy (all yellow)
This thus means that yellow seed trait is dominant over green seed trait.
F1 x F1 = Yy x Yy, resulting in YY, Yy, Yy, yy offspring.
Since Y is dominant over y,
YY, Yy, Yy = Yellow
yy = green
Hence, the expected phenotypic ratio of seed color of the offspring of F1 x F1 cross is 3 yellow : 1 green
In a Mendelian genetics cross where yellow seed color is dominant over green, the expected phenotypic ratio of an F1 x F1 cross is 3 (yellow):1 (green).
Explanation:The question is about predicting the phenotypic ratio of seed color in the F2 generation of a cross between two F1 plants, which came from a purebred yellow seed plant and a purebred green seed plant. In the classic Mendelian genetics framework, yellow and green seed colors are controlled by a gene with two different alleles. In this case, yellow is the dominant trait and green the recessive trait.
For the F1 x F1 cross, each F1 plant is heterozygous, meaning they carry both the yellow (Y) and green (y) allele. If we use a Punnett square to determine the expected outcome from their cross, we would have:
YY, Yy, and yy.
However, because the yellow (Y) is dominant, both YY and Yy will have a yellow phenotype. Thus, the phenotypic ratio expected of seed color in the offspring from F1 x F1 would then be 3 (yellow): 1(green).
Learn more about Phenotypic Ratio here:https://brainly.com/question/29543940
#SPJ6
As animals have evolved large body size, they have also evolved adaptations to improve the exchange of energy and materials with the environment. For example, in many larger organisms, evolution has favored lungs and a digestive tract with ________.
a. increased thickness
b. more branching or folds
c. larger cells
d. decreased blood supply
Answer:
b. more branching or folds
Explanation:
Animal size has increased during evolution on several animals, this has favored enhancements in gas exchange because this enlargement made diffusion too slow to cover internal distances in an efficient time, therefore branches and foldings were developed to cover distances quicker.
I hope you fidn this information sueful and interesting! Good luck!
A nurse is meeting with a woman scheduled to have a modified radical mastectomy to remove an aggressive breast tumor. The woman tells the nurse that she agreed to have the surgery before considering alternative options. Which of the following statements is the nurse's best response?a) "If I were you, I would consider a second opinion."b) "You might want to consider a less invasive surgical procedure."c) "You have a very competent surgeon and you should move forward as planned."d) "Tell me more about your fears and concerns.
Answer:
d) "Tell me more about your fears and concerns."
Explanation:
Statement a) implies the nurse would be imposing her choice on the patient, instead of considering the patient’s concerns and assisting her in deciding a better treatment procedure to undertake.
Statement b) cannot be the best response from the nurse because, the suggested less invasive surgical operation procedure is limited in scope and may not adequately get all the affected tissues removed. The stage of the cancer and how severe it is should inform the type of treatment the patient can be advised to undergo. A combination of treatments might also be needed depending on the stage and nature of the cancer to be treated.
Statement d) is a better response compared to c) and b) and a). Getting to know the fears and concerns of the patient would help enable the nurse to ascertain if the patient has full knowledge about the scheduled procedure as well as address, adequately, the areas that needs further clarification and remove all fears and doubts the patient is having towards the procedure she has decided to undergo.
Cellulose-digesting microorganisms live in the guts of termites and ruminant mammals. The microorganisms have a home and food, and their hosts gain more nutrition from their meals. This relationship is an example of _____. See Concept 54A. mutualismB. commensalismC. predationD. parasitismE. herbivory
Answer: A (Mutualism)
Explanation:
The relationship between termites and cellulose digestive microorganisms is Mutualism. This is because Termites feed on cellulose, which they get from feeding on dry woods.
Termites themselves can not degrade cellulose but the depend on microorganisms in their guts. These microorganisms releases enzymes that solely degrade cellulose.
The symbiotic relationship between termites and cellulose digestive microorganisms is Mutualism because they both depend on each other for survival. Microorganisms would not survive outside of the termite, and the termite would not be able to degrade food if it didn't have cellulose digestive microorganisms to aid in digestion.
____________ is an entity set that contains the commonly shared characteristics of it entities or subtypes. It includes characteristics that are not common to all entities within the set and becomes the parent to one or more subtypes in the hierarchy.
Answer:
A supertype entity
Explanation:
A supertype is a form of entity which has one or even more subgroups to relate to. A subtype is an entity type sub-group of entities that is relevant to the organisation and shares similar characteristics or relations which are different from certain sub-groups.
The entity supertype is a different form of entity type which has relationship with one or even more subtypes and includes subtypes ' of specific attributes.
23. Which Response core capability involves providing life-sustaining and human services to the affected population, to include hydration, feeding, sheltering, temporary housing, evacuee support, reunification, and distribution of emergency supplies?
Answer:
The correct answer is mass care services.
Explanation:
The tendency to provide first aid, shelter, distribution of fundamental needs, feeding centers, and other related goods and services on an instantaneous basis is referred to as mass care. Such services majorly have an application when the individuals encounter with a large-scale unfortunate incident.
These may comprise natural disasters like floods, earthquakes, terrorism, hurricanes, and other occurrences, which can impart serious threats to the population.
The fruit fly Drosophila melanogaster has an allele that confers resistance to DDT and similar insecticides. Laboratory strains of D. melanogaster have been established from flies collected in the wild in the 1930s (before the widespread use of insecticides) and the 1960s (after 20 years of DDT use). Lab strains established in the 1930s have no alleles for DDT resistance. In lab strains established in the 1960s, the frequency of the DDT-resistance allele is 37%. Which statement is correct?
A. Resistance to DDT evolved in some fruit flies in order to allow them to survive
B. Alloles for DDT resistance arose by mutation during the period of DDT use because of selection for pesticide resistance
C. The evolutionary fitness associated with the heritable trait of DDT resistance changed once DDT use became widespread.
Answer:
B. Alleles for DDT resistance arose by mutation during the period of DDT use because of selection for pesticide resistance
Explanation:
This question boils down to the concept of GENE MUTATION. Genetic variations can arise from gene mutations or from genetic recombination (a normal process in which genetic material is rearranged as a cell is getting ready to divide). These variations often alter gene activity or protein function, which can introduce different traits in an organism.
The Biodata dated since (1930s) illustrated that the Laboratory strains of D. melanogaster failed to prove the resistance of the organism to DDT and similar insecticides.
But, over time when the organism are being introduced to the use of DDT, the frequency of DDT resistance was 37%.
It implies that Alleles for DDT resistance arose by mutation during the period of DDT use because of selection for pesticide resistance. If a trait is advantageous and helps the individual survive and reproduce, the genetic variation is more likely to be passed to the next generation. Over time, as generations of individuals with the trait continue to reproduce, the advantageous trait becomes increasingly common in a population, making the population different than an ancestral one.
In cats, females usually have two X chromosomes, and males usually have one X and one Y chromosome. In cats, black fur color is caused by an X-linked allele; the other allele at this locus causes orange color. The heterozygote is tortoiseshell. What kinds of offspring would you most commonly expect from the cross of an XX black female and an XY orange male? In cats, females usually have two X chromosomes, and males usually have one X and one Y chromosome. In cats, black fur color is caused by an X-linked allele; the other allele at this locus causes orange color. The heterozygote is tortoiseshell. What kinds of offspring would you most commonly expect from the cross of an XX black female and an XY orange male?
a. tortoiseshell females.
b. tortoiseshell males tortoiseshell females.
c. black males orange females.
d. orange males black females.
e. orange males orange females.
f. black males
Answer:
a and f (Tortoise shell females and black males).
Explanation:
Let black fur colour allele be AA, orange will be aa and tortoiseshell will be Aa. Since the trait is X-linked;
Black female will be [tex]X^{A} X^{A}[/tex]Orange male will be [tex]X^{a}Y[/tex] (Y chromosome is assumed to not carry any trait, hence, the male can either be black or orange, they can never be heterozygote)Crossing
[tex]X^{A} X^{A}[/tex] x [tex]X^{a}Y[/tex]
Progeny: [tex]X^{A}X^{a}[/tex], [tex]X^{A}X^{a}[/tex], [tex]X^{A}Y[/tex], [tex]X^{A}Y[/tex]
Two heterozygote/tortoiseshell fur colour femalesTwo black malesThe correct answer is a and f.
From the cross of an XX black female with an XY orange male, the most common offspring would be tortoiseshell females and black males, due to the inheritance of X-linked alleles for fur color and the phenomenon of X inactivation leading to tortoiseshell patterning in heterozygous females.
In cats, the genetics of fur color is particularly interesting due to an X-linked gene determining the coat color. Black fur is caused by one allele (which we'll call B), and orange by another allele (let's call it o). When considering the cross of an XX black female (XBXB) with an XY orange male (XoY), we have to remember that X inactivation will occur in the female offspring, leading to a potential tortoiseshell phenotype if they inherit different alleles.
Here are the possible combinations of alleles the offspring might inherit:XBXo - Female heterozygous for the X-linked gene, resulting in a tortoiseshell pattern.XBY - Male with a black coat color since the only X chromosome carries the black color allele.From this cross, you would most commonly expect tortoisell females inheriting one black allele from the mother and one orange allele from the father, resulting in the distinctive mottled coat of tortoiseshell. Male offspring from this cross would most commonly be black, carrying the black allele from their mother and the Y chromosome from their father.
A balanced polymorphism (maintenance of multiple genotypes) exists through disruptive selection in seedcracker finches from Cameroon in which small- and large-billed birds specialize in cracking soft and hard seeds, respectively. If long-term climatic change resulted in all seeds becoming hard, which type of selection would then begin to operate on bill size in the finch population?
A. directional selection
B. disruptive selection would continue to operate
C. stablilizing selection
D. no selection will operate on bill size following the climate change
Answer:
The correct answer is option A. "directional selection".
Explanation:
Directional selection is a type of natural selection at which traits with extreme characteristics are selected over traits with medium characteristics. In this case, a long-term climatic change resulted in all seeds that are the fountain of food for finch becoming hard. This will result in a directional selection on bill size in the finch population, where large-billed birds will be selected over small and medium-billed birds.
A single celled organism can be classified as a prokaryote based on the absence of
Answer:
Nucleus
Explanation:
Prokaryotes are the singled cell organisms that do not have a true nucleus which means like eukaryotes they lack a membrane-bound nucleus. So the genetic material of prokaryote are present in the cytoplasm of their cell.
Apart from the nucleus prokaryotes also lack membrane-bound organelle like mitochondria or chloroplast which are present in eukaryotes and helps in ATP formation for the cell. So prokaryotes use the plasma membrane of cell for production of ATP.
Therefore an organism can be classified as a prokaryote if it lacks a nucleus and other membrane-bound organelles.
Single-celled organisms, or prokaryotes, are classified as such based on the absence of a nucleus and internal membrane-bound structures. They have a simpler structure than eukaryotic cells, with DNA located in a region called the nucleoid. The main difference between prokaryotic and eukaryotic cells is the absence of a complex nucleus in prokaryotes.
Explanation:
A single-celled organism can be classified as a prokaryote based on the absence of a nucleus or any other membrane-bound organelle. Prokaryotic cells, which are predominantly found in bacteria and archaea domains, are much simpler than their counterparts, eukaryotic cells, which are found in animals, plants, fungi, and protists. Central to a prokaryotic cell is a region called a nucleoid, where the DNA is located. In contrast, eukaryotic cells contain a nucleated center where the genetic material or DNA is stored.
Another distinguishing characteristic of prokaryotes is the shape of their DNA — it forms a single, circular chromosome. On the other hand, eukaryotic DNA forms a long string within a complex nuclear membrane that contains multiple, rod-shaped chromosomes.
Overall, the defining feature of prokaryotes is the absence of a nucleus and other internal membrane-bound structures which are instead present in eukaryotic cells.
Learn more about Prokaryote here:https://brainly.com/question/33439033
#SPJ11
Which observation is inconsistent with Haeckel's idea that "ontogeny recapitulates phylogeny"? Group of answer choices The backbone All tetrapod embryos display pharyngeal clefts, a notochord, segmentation, and paddlelike limb buds. The pharyngeal clefts and branchial arches of embryonic mammals and reptiles never acquire the form seen in adult fish. In reptile embryos, two bones develop into the articular bones of the hinge of the jaw, while these same bones become the hammer and anvil of the inner ear in marsupials. Snakes and legless lizards develop "leg buds" as embryos, only to have them reabsorbed prior to hatching.
Answer: c
Explanation:Which observation is inconsistent with Haeckel's idea that "ontogeny recapitulates phylogeny"? Group of answer choices The backbone All tetrapod embryos display pharyngeal clefts, a notochord, segmentation, and paddlelike limb buds. The pharyngeal clefts and branchial arches of embryonic mammals and reptiles never acquire the form seen in adult fish. In reptile embryos, two bones develop into the articular bones of the hinge of the jaw, while these same bones become the hammer and anvil of the inner ear in marsupials. Snakes and legless lizards develop "leg buds" as embryos, only to have them reabsorbed prior to hatching.
The _________________, which binds to a core promoter, consists of general transcription factors and RNA polymerase.
Answer:
The correct answer is basal transcription apparatus.
Explanation:
The basal transcription apparatus, which binds to a core promoter, consists of general transcription factors and RNA polymerase. This apparatus is made of proteins and synthesizes mRNA.
All of the following describe the molecular/cellular changes that occur in cones in response to light, except one. Choose the exception. Group of answer choices a) Hyperpolarization of membrane potential b) Reduction in the release of the neurotransmitter glutamate from cone synaptic terminals c) Activation of transducin d) Inactivation of phosphodiesterase.
Answer:
The exception is d) Inactivation of phosphodiesterase
Explanation:
In the presence of light:
a) The membrane potential of the photoreceptors is hyperpolarized, causing a reduction in the amount of neurotransmitter released by the photoreceptor terminal to the postsynaptic neurons
b) Sodium ions accumulate outside the plasma membrane and the receptor potential takes a form
of hyperpolarization 6. This change in membrane potential leads to the closure of calcium channels. The end result is a decrease in the secretion of the neurotransmitter glutamate by the photoreceptors
c) Visual transduction or phototransduction is the process by which a light photon generates a nervous response in the photoreceptors
d) As a consequence of phosphodiesterase activation, degradation of a molecule called cGMP (guanosine cyclic monophosphate) is stimulated