The presence of a mutualist might allow what to happen in terms of population dynamics? a) The species could surpass its carrying capacity.
b) The growth rate would decrease.
c) The competition coefficient would become negative.
d) N would go to 0.
The presence of a mutualist might allow what to happen in terms of population dynamics - a) The species could surpass its carrying capacity.
Population dynamics is the study of how and why populations change in size and structure over time.
The factors in population dynamics include - rates of reproduction, death, and migration.Mutualism is the interaction between two different species that leads to positive effects on per capita reproduction and/or survival of the interacting populations.as in this case, the association is beneficial for both the species which leads to an increase in the carrying capacity.Thus, the presence of a mutualist might allow what to happen in terms of population dynamics - a) The species could surpass its carrying capacity.
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Determine the sequence of genes along a chromosome based on the following recombination frequencies:
A-B 8 map units
A-C 28 map units
A-D 25 map units
B-C 20 map units
B-D 33 map units
The most probable sequence of genes along a chromosome based on the given recombination frequencies is A, D, C, B. This order is determined by analyzing recombination frequencies, with the closest genes having the lowest frequencies and confirming the sequence through the addition of map units between linked genes.
To determine the sequence of genes along a chromosome based on the given recombination frequencies, we can utilize the concept that genes closer together on a chromosome have lower recombination frequencies. The given recombination frequencies are between gene pairs: A-B (8 map units), A-C (28 map units), A-D (25 map units), B-C (20 map units), and B-D (33 map units).
Starting with the closest genes, A and B are 8 map units apart. Since the recombination frequency between B-C is 20 map units and A-C is 28, it suggests that gene C lies farther from A and B.
Additionally, B-D's recombination frequency is 33 map units, which is higher than A-D's 25 map units, suggesting that D is closer to A; thus, gene D lies between A and C. To confirm the sequence, we can verify the sum of recombination frequencies between three linked genes; for example:
A-B + B-D = 8 + 33 = 41 map units
A-D + D-C = 25 + 3 (calculated since A-C is 28 and A-D is 25) = 28 map units
Given this information, A-D-C is the more likely arrangement because the sum of the recombination frequencies matches the recombination frequency of A-C. Therefore, the most probable order of genes along the chromosome is A, D, C, B.
If a cell has completed the first meiotic division and is just beginning meiosis ii
Answer: The correct answer is : It has half the amount of DNA as the cell that began meiosis.
Explanation: Meiosis gives rise to four unique daughter cells, each one having half the number of chromosomes that the mother cell, this is meiosis I. During meiosis II the two cells go through four phases of division again, at this stage there the number of chromosomes in daughter cells is reduced.
The phrases or terms describe different fundamental processes of nucleic acids. Classify each phrase or term as relating to replication
Answer:
The replication of DNA occurs through the following phases - unwinding of strands, binding of RNA primers, elongation, removal of primers, DNA repair and termination of replication.
Explanation:
There are two types of nucleic acids one is RNA and the other is DNA. DNA makes a new copy by the process of DNA replication. The replication of DNA is semiconservative type.
It consists of different steps. The first is an unwinding of the two strands of DNA. This occurs by helicase enzyme which acts like a scissor. The point where this enzyme acts and strat the unwinding is called the origin of replication.
The second step is the synthesis of an RNA primer at the point of origin. It is formed from the DNA template. This RNA primer helps to synthesize new DNA strands and elongates the DNA. This process is called elongation. It forms two strands- leading and lagging strands.
The lagging strand consists of many RNA primers that should be removed.
The RNA primers are removed by DNA polymerase I and it helps in adding complementary strands of DNA. As a result, one new DNA forms from the older one. The final step is DNA repair, which has done by enzyme nuclease.
The enzyme nuclease removes the wrong nucleotides and DNA polymerase fills the space with correct complementary nucleotide.
Response to a stimulus, such that the effect of the stimulus is counteracted, is called a ________
Answer:
Negative Feedback
Explanation:
The balancing feedback also called as negative feedback. The negative feedback regulates by stimulus that decreases the function. This mechanism balances the release of substance. The examples of negative feedback are thermoregulation and blood sugar regulation. The excess glucose is decreased by insulin. The changes in the body temperature are balanced and normal temperature is maintained. Thus, Response to a stimulus, such that the effect of the stimulus is counteracted, is called a Negative Feedback.
This is an organism that has two different alleles for the same genetic trait.
Answer:
Heterozygous
Explanation:
An organism is said to be heterozygous when he has two different alleles of the same trait. A homozygous individual have the two same alleles for a given trait. For example, height is a trait which has two alleles that are T and t.
So in heterozygous individuals, two different alleles will be present which is Tt and in homozygous individual two same allele of height will be present that can be TT or tt.
In heterozygous condition normally one allele will be dominant and another allele will be recessive and in homozygous condition normally both the allele will be dominant or recessive. Therefore heterozygous is the organism that has two different alleles for the same trait.
The organism that has two different alleles for the same genetic trait is called a heterozygous organism.
The organism that has two different alleles
In genetics, alleles are alternative forms of a gene that occupy the same position (locus) on a pair of chromosomes. When an organism has two different alleles for a particular trait, it means that each of the two copies of the gene it possesses is different.
This could result in the expression of a dominant allele over a recessive allele, or in the case of co-dominance or incomplete dominance, both alleles may contribute to the phenotype of the organism.
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If, in one population of pupfish all of the individuals have a blood pigment that is extraordinarily effective at carrying oxygen, but this trait is not seen in any of the other populations, what likely happened?
A) Because oxygen was low where these pupfish lived, a new allele for an effective blood pigment arose.
B) This population was lucky to have an individual with a random mutation for an effective blood pigment, and the frequency of this allele was increased in subsequent generations through natural selection.
C) The ancestral population probably had this type of blood pigment, but it was lost through genetic drift in the other 29 populations.
D) The other populations did not need this pigment, so they did not evolve it.
The most likely explanation for the distinct blood pigment in one pupfish population is a random mutation combined with the beneficial allele being favored and spread by natural selection.
Explanation:If, in one population of pupfish, all of the individuals have a blood pigment that is extraordinarily effective at carrying oxygen, but this trait is not seen in any of the other populations, it suggests a specific evolutionary event has occurred. Option B suggests that this population was lucky to have an individual with a random mutation for an effective blood pigment, and the frequency of this allele was increased in subsequent generations through natural selection. This is the most likely scenario because the allele for the effective blood pigment would offer a significant survival advantage in environments where oxygen was scarce, and those individuals with the mutation would be more likely to survive and reproduce, passing the beneficial allele on to future generations.
how do humans cause ecological issues in ecosystems?
Answer:
Explanation:
Human activities cause serious damage and ecological issues to the ecosystem which is the environmental deterioration and depletion of resources like air, water and good soil. These harmful activities that cause such destruction includes: pollution, extinction of wild life, the demand for food and shelter as the population increases, overfishing, agriculture all these have negative impacts on the ecosystem.
Assembly of amino acids according to the genetic information carried by mrna is called
The process of assembling amino acids as directed by the genetic information in mRNA is known as translation. It involves the conversion of nucleotide sequences in mRNA into a sequence of amino acids that form proteins. The genetic code, which is nearly universal, employs codons to specify amino acids during this process.
Explanation:Translation of mRNA in Protein Synthesis
The assembly of amino acids according to the genetic information carried by mRNA is called translation. During this process, the genetic code inscribed in the mRNA sequence is interpreted to synthesize proteins. Translation is a complex biological mechanism that involves various molecules and cellular structures, including mRNA, transfer RNA (tRNA), ribosomal RNA (rRNA), ribosomes, and numerous enzymes.
Each amino acid is specified by a three-nucleotide sequence known as a codon, and the process is facilitated by tRNA molecules that match these codons to specific amino acids. The ribosome acts as the site where this process unfolds, reading the mRNA codons and linking the corresponding amino acids to form a polypeptide chain, which ultimately folds into a functional protein. The genetic code that leads to this translation process is nearly universal among all species on Earth, governing the intricate translation from nucleotide sequences to amino acids—the building blocks of proteins.
This elucidation of protein synthesis is fundamental in understanding how genetic information dictates the structure and function of proteins within biological systems.
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Throughout the book, the deterioration of Morrie's body is symbolically compared to ____________.
a. The orange cactus plant.
b. The tank of goldfish.
c. The spruce tree outside his window.
d.The pink hibiscus plant.
Answer: option D
Deterioration of Morrie's body is compared to a pink hibiscus plant.
Explanation:
Tuesdays with Morrire is a book written by an American author Mitch Albom. The book was written base on the visits the author made to his form sociology lecturer Morrire Schwartz who was suffering from Amyotrophic lateral sclerosis. The book use a metaphor to describe the deteriorating body of Morrie's using thr pink hibiscus plant. When the pink hibiscus plant is deteriorating, the plant petals start to fall and later dies. Morrie's body also began to deteriorates while he depend on oxygen to breathe, as times goes by, the deterioration continues just as petals of hibiscus plant and he eventually died.
Coexisting species of wild cats differ in the size of their canine teeth, which corresponds to differences in their preferred species of prey. This outcome is most likely the result of: Competitive exclusion Resource partitioning Ecological release Preemptive competition
Answer:
The correct answer is- Resource partitioning
Explanation:
According to the competitive exclusion principle, two species can not share the same niche because species that share the same niche have the same needs which leads to the interspecies competition.
This competition leads to resource partitioning which means the species who share the same niche evolved by natural selection to occupy different niche by dividing their resource which leads to the coexistence of two different species.
So here coexistence of wild cats differ in the size of their canine teeth is the result of resource partitioning.
American vultures used to be classified in the same family as African vultures. Which discovery caused scientists to reclassify American vultures as more closely related to storks?
Answer:
DNA evidence revealed the American vultures share more recent ancestor with the Storks
Explanation:
The hooded vultures that is mostly found in the African continent have a close resemblance with the American vultures and were traditionally classified to belong to the Falcon family.
However, it was observed that the American vultures shared a similar behavior with Stork which is not common to the vulture found in Africa, including the hooded vulture. The Stork and the American vulture exhibit the behavior of urinating on their legs when being overheated. When the urine gets evaporated, it helps them to cool their body temperature.
This shared behavior between the storks and the American vultures led scientists into using molecular analysis in analyzing the DNA of the hooded vultures found in Africa, the American vultures, and the stork.
Evidence from the DNA analysis later revealed that the American vultures and the storks share a more common DNA sequences than African vultures and American vultures do.
Micah visits his doctor complaining of a frontal headache and pressure over his cheekbones and eyes. He is congested and has a nasal discharge. Micah's voice has an odd nasal sound. What is the doctors diagnosis of Micah's condition?
a) tonsillitis
b) rhinitis
c) sinusitis
d) pleurisy
Answer:
The answer would likely be C as it seems that his nasal passage is mostly affected. Based on diagrams of sinusitis, the cheekbones and eyes bones areas are likely to be impacted by the nasal blockage.
Explanation:
Final answer:
Micah's doctor would likely diagnose his condition as sinusitis based on the presence of a frontal headache, pressure over the cheekbones, nasal congestion, and a nasal discharge, which indicates inflammation of the sinuses.
Explanation:
Symptoms and Diagnosis of Micah's Condition
Based on the symptoms described, Micah's doctor would likely diagnose his condition as sinusitis. Sinusitis is an inflammation of the sinuses, marked by headaches, pressure over the cheekbones, nasal congestion, and a nasal discharge. The presence of an odd nasal sound in Micah's voice further supports this diagnosis as it indicates that his sinuses, which are connected to the nasal passages, are affected.
Rhinitis, which is inflammation of the nasal cavity, often accompanies sinusitis but is not characterized by the same type of headache and pressure over the cheekbones. This combination of symptoms, with the addition of sinus involvement, is more aligned with rhinosinusitis. While tonsillitis and pleurisy are conditions that also affect parts of the respiratory system, their symptoms do not match Micah's presentation.
Bacterial rhinosinusitis, which is an infection and inflammation of the paranasal sinuses usually occurring after a viral infection, is commonly caused by pathogens such as Streptococcus pneumoniae, Haemophilus influenzae, and Moraxella catarrhalis.
Which three of the following statements are consistent with the images?a. In a sea turtle's flippers, heat is transferred from (3) to (1).b. In a sea turtle, blood warms as it flows from the body (1) to the tip of the flipper (2).c. In a sea turtle's flippers, heat is transferred from (1) to (3). d. At a dolphin's testes, heat is transferred from (2) to (1).e. In a dolphin, blood cools as it flows from the aorta to the testes. f. At a dolphin's testes, heat is transferred from (1) to (2).
Answer:
C
Explanation:
HCl secretions convert pepsinogen to the active hormone pepsin. What cells in the gastric pits produce pepsinogen?
A. parietal cells
B. chief cells
C. G cells
D. paracrine cells(also known as enteroendocrine cells)
Answer:
option B. chief cells
Explanation:
The inner wall of stomach contain gastric pits. The pits contain certain cells that secretes chemicals. The cheif and parietal cells are present in gastric pits. The Hcl secreted by parietal cells. The hcl is involved in conversion of pepsinogen to pepsin. The pepsinogen is the inactive form and it become activated by hcl. The inactive zymogen (pepsinogen) is released into gastric juice. The chief and mucus cells secrete pepsinogen. The active pepsin is responsible for digesting proteins. Thus, option B is correct.
Which of the following choices would be the most logical third step in Connell's experimental procedure, permitting him to either accept or reject his hypothesis of competitive exclusion?
The experimental group:________
a. The half of each rock that contained both Balanus and Chthamalus was the experimental group.
b. The experimental group tests to see if the presence of Balanus in some manner prevents Chthamalus from growing.
Answer: B) The experimental group tests to see if the presence of Balanus in some manner prevents Chthamalus from growing.
Explanation:
The most logical third step in Connell's experimental procedure, permitting him to either accept or reject his hypothesis of competitive exclusion is the experimental group tests to see if the presence of Balanus in some manner prevents Chthamalus from growing.
Balanus is a species of crustacean, as know as sea acorns, whereas Chthamalus is a species of barnacles that can easily be found in the northern hemisphere's coasts.
To test Connell's hypothesis of competitive exclusion, the logical step would be to observe and compare the growth of Chthamalus in areas where both species are present, and where Balanus is excluded.
Explanation:In Connell's experimental procedure, the most logical third step to accept or reject his hypothesis of competitive exclusion would be to observe the growth and interaction patterns between the Balanus and Chthamalus in the experimental group. This can be done by comparing the areas where both species are present with those where Balanus is excluded. If the growth of Chthamalus is significantly more in the absence of Balanus, this would serve as an evidence of competitive exclusion by Balanus. If no significant difference is noted, then the hypothesis might need to be reconsidered or rejected.
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"Although some hearing conditions run in families, Dawn, age 61, is experiencing an age-related hearing condition. Dawn has __________." presbyopia presbycusis tinnitus sensorineural hearing loss
Answer:
The correct answer is presbycusis.
Explanation:
As we grow older and imperceptibly, human beings gradually lose their ability to hear. This process is called presbycusis.
It is a condition that occurs normally as we get older and usually affects both ears equally.
To this day, no method has been developed to prevent this.
There are many reasons why you can suffer presbycusis, although the most common is caused for the changes generated in the inner ear.
Losing an auditory part is quite uncomfortable since it affects us in all the activities we do every day.
Before exploring some of the features of the different types of eukaryotes, we should first review some of the fundamental differences between eukaryotes and prokaryotes. The two groups of organisms differ fundamentally in the structure of their individual cells. For each of the following statements, identify whether it refers to prokaryotes, eukaryotes, or both groups.
Drag each statement into the appropriate bin.
a. Prokaryotes only.
b. Eukaryotes only.
c. Both prokaryotes and eukaryotes.
Answer:
Prokaryotes only:
cell wall contains peptidoglycan or pseudomurein binary fission 70S ribosomes singular circular chromosomeEukaryotes only:
membranous organelles, including mitochondria, lysosomes, endoplasmic reticulum 80S ribosomes nuclear envelope compartmentalizes the chromosomesBoth prokaryotes and eukaryotes:
plasma membrane encloses the cytoplasm has both DNA and RNA includes unicellular cellsFinal answer:
Prokaryotic cells lack a nucleus and have simpler structures, while eukaryotic cells have a defined nucleus and are more complex, with multicellular organisms being exclusively eukaryotes. Both share basic cell components like the plasma membrane and ribosomes.
Explanation:
The distinctions between prokaryotic cells and eukaryotic cells are key in understanding cell structure and function. Prokaryotic cells, which include Bacteria and Archaea, are characterized by lacking a nucleus and membrane-bound organelles, and their DNA generally exists in a single, circular chromosome within a nucleoid region. On the contrary, eukaryotic cells, which make up animals, plants, fungi, and protists, contain a well-organized, membrane-bound nucleus with multiple, rod-shaped chromosomes and other internal membrane-bound organelles.
Both prokaryotic and eukaryotic cells share some fundamental components such as the plasma membrane, which serves as a barrier to the environment, the cytoplasm, which contains organic molecules and salts, a DNA genome that stores genetic information, and ribosomes, where proteins are synthesized. However, eukaryotic cells are generally larger and more complex, with multicellular organisms exclusively falling into the eukaryote category.
What type of microscope would be best for studying the structures found inside of cells?
A Transmission Electron Microscope (TEM) is best for studying the structures found inside of cells because it offers high resolution. A Scanning Electron Microscope (SEM) is better for studying the surface of cells.
Explanation:The type of microscope best suitable for studying the structures found inside of cells is a Transmission Electron Microscope (TEM). This is because a TEM uses a high-energy electron beam to illuminate the sample, resulting in much higher resolution than light microscopes. With a resolution of less than 0.5 nanometer, the TEM allows for the visualization of structures inside the cell such as organelles, protein complexes and even individual molecules to be viewed in detail.
A Scanning Electron Microscope (SEM) can also be used, but it is best for studying the surface of cells rather than internal structures. For example, SEM can be used to study the texture and shape of the cell membrane.
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Salivary amylase enzymatically breaks down glucose in the oral cavity. true or false
Answer:
FALSE
Explanation:
Digestion begins in the oral cavity when food enters in contact with the salivary amylase, the principal enzyme in saliva. Carbohydrates in the form of starches (like potatoes, rice, or pasta) are hydrolized from polysaccharides into disaccharides, amylose and amylopectin are hydrolyzed into smaller chains of glucose (dextrins and maltose). This step in the digestion of sugars is limited due to the brief exposure time of the food to the enzyme.
Therefore we can conclude that the answer is FALSE because salivary amylase enzymatically breaks down polysaccharides into disaccharides, glucose is not broken down at this point.
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How does the simple primary and secondary structure of dna hold the information needed to code for the many features of multicellular organisms?
Answer:
Explanation:
The nitrogenous base sequence of the DNA is responsible for carrying the genetic information needed to code for proteins and many features of multicellular organisms.
How does biodiversity help populations survive in changing ecosystems
Answer:
Maintaining variability in the genetic pool.
Explanation:
In a biodiverse population, there are different alleles in the genetic pool (variability) and not just a single fixed one. In a changing ecosystem it is possible that although not all of these alleles are beneficial to the population's fitness- in the context of change-, one of them is and helps the population survive and adapt to such change.
Biodiversity then helps because if the population does not have multiple alleles, but only one, it would be possible that the only allele in the genetic pool is not appropriate to face the change in the ecosystem.
Answer:
Biodiversity boosts ecosystem productivity where each species, no matter how small, has an important role to play. For example, A larger number of plant species means a greater variety of crops. Greater species diversity ensures natural sustainability for all forms of life.
Explanation:
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Muscle cells in oxygen deprivation convert pyruvate to ______ and in this step gain ______.
Answer:
Lactate, and NAD+
Explanation:
In the process of glycolysis the product is pyruvate. When muscle don't contain enough oxygen they will go through anaerobic glycolysis they don't perform the citric acid cycle and converting the pyruvate to lactate and NAD+. NAD+ is required for the keep going glycolysis.
So, the answer is Lactate and NAD+.
Muscle cells in oxygen deprivation convert pyruvate to lactic acid and gain two ATPs in the process. This occurs during strenuous exercise when cells need high amounts of energy. Accumulation of lactic acid during this process can cause muscle fatigue.
Explanation:When muscle cells are in a state of oxygen deprivation, they convert pyruvate to lactic acid. This is an anaerobic process during which high amounts of energy are necessary but cannot be supplied by oxygen to the muscles. The conversion of pyruvate into lactic acid allows the recycling of the enzyme NAD+ from NADH, a necessity for the process of glycolysis to continue, resulting in a gain of two ATPs per glucose molecule.
This process occurs largely during strenuous physical activity when the energy demand is high. However, the accumulation of lactic acid in the muscles may contribute to muscle fatigue. Glycolysis, besides not utilizing glucose very efficiently, also cannot be sustained for a very long period. It is, however, useful in providing short bursts of high-intensity output.
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What would be the function of an epithelial cell with a relatively large amount of cytoplasm? View Available Hint(s) What would be the function of an epithelial cell with a relatively large amount of cytoplasm? regeneration and rapid tissue replacement absorption and secretion resist wear and tear anchorage to the extracellular matrix
Answer:absorption and secretion
Explanation:
The main function of epithelial cells with large amount of cytoplasm is absorption and secretion. They're usually present sweat and oil glands of the skin due to its secretive function.
In the first wave of a distributed denial of service (DDoS) attack, the targets that will be the "foot soldiers" are infected with the implements that will be used to attack the ultimate victim.True / False.
Answer:
In the first wave of a distributed denial of service –ddos attack, the targets that will be the “foot soldiers” are infected with the implements that will be used to attack ultimate victim is true.
Explanation:
The lookup table is used for tracking where 'mac' (media access control) address are present on the port switch. Distributed denial of service is a type of 'denial of service attack' where multiple systems which is infected with Trojan used to target single system causing dos. Botnets are used to do ddos attack steal the data, allows the 'attacker' to access device, and connections and Spen spam. The 'command and control software' is used to control the botnet.In the cross between a female A/a;B/b;c/c;D/d;e/e and male A/a;b/b;C/c;D/d;e/e, (Assume independent assortment of all genes and complete dominance.)What proportion of the progeny will be phenotypically identical to
(1) the female parent?
(2) the male parent,
(3) either parent and
(4) neither parent?What proportion of the progeny will be genotypically identical to
(1) the female parent?
(2) the male parent,
(3) either parent, and
(4) neither parent?
Answer:
Phenotypically identical
(1) the female parent: 9/64
(2) the male parent: 9/64
(3) either parent: 9/32
(4) neither parent: 23/32
Genotypically identical
(1) the female parent: 1/16
(2) the male parent: 1/16
(3) either parent: 1/8
(4) neither parent: 7/8
Explanation:
To find out the answers, we will have to find out probability of each gene separately:
female male
A/a;B/b;c/c;D/d;e/e A/a;b/b;C/c;D/d;e/e
If we will look at only A gene combinations in both the parents the results will be as under:
Parentals : A/a x A/a
The 4 gametes in the progeny will be as under :
AA = 1/4
Aa = 1/4 Combined probability of Aa = 1/4+ 1/4 = 1/2
Aa = 1/4
aa = 1/4
If we will look at only B gene combinations in both the parents the results will be as under:
Parentals : B/b x b/b
The 4 gametes in the progeny will be as under :
Bb = 1/4 Combined probability of Bb = 1/4+ 1/4 = 1/2
Bb = 1/4
bb = 1/4 Combined probability of bb = 1/4+ 1/4 = 1/2
bb = 1/4
If we will look at only C gene combinations in both the parents the results will be as under:
Parentals : c/c x C/c
The 4 gametes in the progeny will be as under :
Cc = 1/4 Combined probability of Cc = 1/4+ 1/4 = 1/2
Cc = 1/4
cc = 1/4 Combined probability of cc = 1/4+ 1/4 = 1/2
cc = 1/4
If we will look at only D gene combinations in both the parents the results will be as under:
Parentals : D/d x D/d
The 4 gametes in the progeny will be as under :
DD = 1/4
Dd = 1/4 Combined probability of Dd = 1/4+ 1/4 = 1/2
Dd = 1/4
dd = 1/4
If we will look at only E gene combinations in both the parents the results will be as under:
Parentals : e/e x e/e
The 4 gametes in the progeny will be as under :
ee = 1/4
ee = 1/4 Combined probability of ee = 1/4+ 1/4 + 1/4 +1/4 = 1
ee = 1/4
ee = 1/4
It is given that genotype of first parent is A/a;B/b;c/c;D/d;e/e or AaBbccDdee .
Also, it is pertinent to mention here that AA and Aa genotype will produce same kind of phenotype. In the progeny, we can calculate the probability of AA & Aa will be 1/4 + 1/2 = 3/4
Similarly with respect to gene B, BB and Bb will produce same kind of phenotype but BB genotype will not get produced so we will only find out probability of Bb alone which is 1/2
Similarly the allelic combinations of gene C which will be similar to first parent will be 1/2
The combinations of gene D which will be similar to first parent will be a combination of DD & Dd which is 1/4 + 1/2 = 3/4
For gene e the combinations which will produce same phenotype as of first parent will be 1 because all the combinations are ee.
CALCULATIONS FOR PROGENY WHICH ARE SIMILAR TO PARENTS PHENOTYPICALLY.
(1) So, the combined probability of resemblance of phenotype of progeny with first parent which is female = 3/4 x 1/2 x 1/2 x 3/4 x 1 = 9/64.
(2) The genotype of male parent is A/a;b/b;C/c;D/d;e/e or AabbCcDdee.
So, in a similar way we can find out the combined probability of resemblance of phenotype of progeny with second parent = 3/4 x 1/2 x 1/2 x 3/4 x 1 = 9/64
(3) The progeny which are similar to either parent will be 9/64 + 9/64 = 18/64 = 9/32.
(4) The progeny which will have phenotype which does not match any parent will be 1 - 9/32 = 32 - 9 /32 = 23/32.
CALCULATIONS FOR PROGENY WHICH ARE SIMILAR TO PARENTS GENOTYPICALLY.
When we will look for progeny which are genotypically similar to parents, we will look for allelic combinations which are exactly similar to parents. While finding genotypes, homozygous dominant and heterozygous will not be same. For example, AA and Aa will not produce same genotype.
(1) The probability of progeny which will be genotypically identical to female parent (AaBbccDdee) is 1/2 x 1/2 x 1/2 x 1/2 x 1 = 1/16.
(2) The probability of progeny which will be genotypically identical to male parent (AabbCcDdee) is 1/2 x 1/2 x 1/2 x 1/2 x 1 = 1/16.
(3) The progeny which will be genotypically similar to either parent will have probability = 1/16 + 1/16 = 1/8
(4) The progeny which will have genotype which does not match any parent will be = 1 - 1/8 = 7/8
Final answer:
The expected proportion of dominant phenotype offspring from a tetrahybrid cross in all loci (A-D) is 81/256. To be phenotypically identical to either parent, offspring must have the dominant phenotype, regardless of homozygosity or heterozygosity. The genotypic likeness to a parent is difficult to calculate without a Punnett square, especially in tetrahybrid crosses.
Explanation:
To answer the question regarding the inheritance patterns from a cross between a female A/a;B/b;c/c;D/d;e/e and a male A/a;b/b;C/c;D/d;e/e, we must consider each gene locus independently due to the assumption of complete dominance and independent assortment.
Phenotypic Proportions
For phenotypic proportions, we are interested in the offspring that exhibit the dominant phenotype for each of the four loci. Using the sum rule and product rule, the probability for each dominant allele being present (either homozygous dominant or heterozygous) is 3/4. So, the expected proportion of offspring with the dominant phenotype at all four loci (A-D) is 81/256 (3/4 x 3/4 x 3/4 x 3/4).
Genotypic Proportions
The genotypic ratio for a cross of Aa x Aa individuals is 1:2:1, which results in a phenotypic ratio of 3:1 for the dominant trait. This calculation can be applied to each locus separately. To be phenotypically identical to either parent, they must express the dominant trait regardless of whether they are homozygous or heterozygous for it.
Phenotypically identical to the female parent: The proportion is 81/256 (dominant phenotype at all loci).Phenotypically identical to the male parent: The proportion is also 81/256 since the dominant phenotype is the same for both parents.Phenotypically identical to either parent: Since both options above are the same, the proportion remains 81/256.Phenotypically identical to neither parent: This case is not possible since all dominant phenotypes will match the parents.Although beavers are relatively rare, they have considerable influence over species interactions in communities in which they are present because of the dams that they construct. The beaver is thus an example of a(n) _______.
Answer:
ecosystem engineer
Explanation:
The beaver is one of the largest rodents in the world and its fame comes from its ability to modify the environments through the cutting down of trees and the construction of dam and burrows. Reason why they are called by nature engineers or ecosystem engineers.
In general, the construction of these dams is due to different reasons, the main one being the lack of a suitable habitat, because when the environment does not meet the requirements that fit their needs, for example, when water levels do not they are enough, they build these buildings. Also, dams provide shelter for possible attacks by their predators.
In North America, where the species is native, the construction of these dams brings benefits to nature, since they help with the restoration of wetlands and in turn, generate a place suitable for the flora and fauna characteristic of this habitat. Similarly, they function as flood controllers, as they help keep water levels low. However, in the southern hemisphere, where the species is not native, the same does not happen. Since the beaver arrived, the Patagonian forest (in Argentina) was never the same again and today it is considered as a plague for not having a predator.
A primigravida is admitted to the birthing unit in early labor. A pelvic examination reveals that her cervix is 100% effaced and dilated 3 cm. The fetal head is at +1 station. In which area of the client's pelvis is the fetal occiput?
The fetal occiput is in the Ischial spines.
Explanation:During labor and delivery, the baby passes through the “pelvic bones to reach the vaginal opening”. The pelvis is located between the hip bones and is wide/flat in females. The pelvis has the uterus, cervix and vagina. The muscles in the uterus push the baby down. The baby’s head presses the cervix releasing oxytoxin. Then it dilates and allows the baby to pass through fetal station.
Fetal station is the fetus/baby is in the pelvis. Occiput is the lower part of the head/skull. The presenting part of the baby passes through the birth canal. Most of the time it would be “baby's head, shoulder, the buttocks, or the feet”. Ischial spines are “bone points” on the “mother's pelvis”. It is the “narrowest part of the pelvis”.
0 station: This is the position when baby's head is at the Ischial spines. The baby is "engaged" when largest part of the head enters the pelvis. If the presenting part lies above the Ischial spines, the station is reported as a negative number from -1 to -5.
What is the genotype of the parent with orange eyes and white skin? (Note: orange eyes are recessive.)B=black eyes G=green skinb=orange eyes g=white skina. BbGgb. bbGGc. bbGgd. bbgge BBGG
Answer:
D
Explanation:
Two genes are involved in this case: one coding for eye color and the other for skin color.
According to the question, the allele for Black eyes (B) is dominant over the allele for orange eyes (b) while on the other gene, the allele for green skin (G) is dominant over the allele for white skin (g).
This means both alleles for orange eyes (b) and white skin (g) are recessive i.e masked by their counterpart allele
Since they are recessive, they will only be phenotypically expressed if they are in their homozygous state. i.e same type of allele.
In this case, the parent has orange eyes and white skin i.e.both recessive traits
Therefore, the genotype will be bbgg.
The genotype of the parent with orange eyes and white skin is bbgg.
Explanation:The genotype of the parent with orange eyes and white skin is bbgg. In genetics, lowercase letters represent recessive alleles while uppercase letters represent dominant alleles. Since orange eyes are recessive, the parent must have the genotype bb for eye color. Similarly, since white skin is recessive, the parent must have the genotype gg for skin color.
Learn more about Genotype here:https://brainly.com/question/29156144
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A pheasant breeder starts with two birds in the P generation, one of which is AA and the other is aa. If he takes two of the birds from the F1 generation and breeds them together, what can he expect in his F2 offspring?