Answer:
Explanation:
The parasites living at the respiratory system or tongue worms will hide under the soft tissues of the oral cavity, beneath the tongue and even beneath the throat or esophagus. These worms remain undetected during diagnosis. These worms are not subjected to the treatment of acids which worms in the digestive system are exposed to. Thus these worm parasites in the respiratory system or tongue survive comparatively for long as compared to the worms in the digestive system.
name Streptococcus agalactiae, what morphology would you expect these cells to have?
Answer:
round and in chains
Explanation:
Streptococcus agalactiae is a gram positive bacteria. It is facultative anaerobe and forms a part of microbiota in gastro intestinal and urinary tract of healthy humans. It can cause infections in immuno compromised beings.Its genus name describes its morphology. Coccus are round spherical shaped bacteria. Strepto means that the bacteria are present in chain form. Hence this bacteria is spherical and arranged in chains.
There are several other types of bacterial morphology. For example: Staphylococcus means that the bacteria is again spherical but this time arranged in groups. Diplococcus means that the spherical bacteria is arranged in a pair. Similarly, bacillus is used to describe a rod shaped bacteria.
Two pea plants heterozygous for the characters of pod color and pod shape are crossed. Draw a Punnett square to determine the phenotypic ratios of the offspring.
Answer:
I will attach a file with the punnett square
Explanation:
You have to write every possible combination of gametes for each individual. In this case, both individuals will have the same possible combination of gametes because they both are heterozygous for both traits. Then you have to make the crossing between each gamete and you get the punnett square.
A cross between pea plants heterozygous for pod color and shape produces a 9:3:3:1 phenotypic ratio, as predicted by a dihybrid cross Punnett square. This indicates 9 offspring would be green and round, 3 would be green and wrinkled, 3 would be yellow and round, and 1 would be yellow and wrinkled.
Explanation:In this cross scenario, we're considering two characters i.e., pod color and pod shape and the pea plants are heterozygous for both traits. A heterozygous plant can be denoted as GgRr, where 'G' and 'R' are dominant alleles for green color and round shape, and 'g' and 'r' are recessive alleles for yellow color and wrinkled shape.
To examine all possible combinations of these traits in the offspring, we use a type of Punnett square called a dihybrid cross. This square is 4x4 and it helps predict the genotypes and phenotypes of the offspring. The phenotypic ratio for a dihybrid cross where both parents are heterozygous for both traits is typically 9:3:3:1. This means 9 offspring will be green and round, 3 will be green and wrinkled, 3 will be yellow and round, and 1 will be yellow and wrinkled.
However, actual results can vary from these predicted ratios because the distribution of traits among offspring is determined by chance.
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Write down the possible genotypes, as determined by the phenotype.
a. If the phenotype is that of the dominant trait (for example, purple flowers), then the genotype is either homozygous dominant or heterozygous (PP or Pp in this example).
b. If the phenotype is that of the recessive trait, the genotype must be homozygous recessive (for example, pp).
c. If the problem says "true-breeding," the genotype is homozygous.
a. If the phenotype is that of the dominant trait (for example, purple flowers), then the genotype is either homozygous dominant or heterozygous (PP or Pp in this example).
To draw the genotype here, you have two possible options: PP homozygous dominant or Pp heterozygous dominant.
b. If the phenotype is that of the recessive trait, the genotype must be homozygous recessive (for example, pp).
Yes, in this case, if you want a recessive trait to show on the phenotype your only option is to have pp alleles.
c. If the problem says "true-breeding," the genotype is homozygous.
In this case, your only option is P or p.
Either if it's dominant ( capital letter) or recessive (lower letter)
Sketch a section of a phospholipid bilayer of a membrane, and label the hydrophilic head and hydrophobic tail of one of the phospholipids.
Answer:
The hydrophobic tails are sandwiched in between because they are ‘water-hating’ leaving the hydrophilic ends (which are ‘water-loving’) to interact with the watery environment inside and outside of the cell. The hydrophobic ends are mainly made of fatty acid chains ( and have no charges) while the hydrophilic ends are made of phosphate molecules that are negatively charged.
A phospholipid bilayer, a key component of a cell membrane, comprises two layers of phospholipids each with a hydrophilic (water-attracting) head and a hydrophobic (water-repelling) tail. The bilayer formation allows the cell to create a barrier and maintain different environments on each side.
Explanation:The phospholipid bilayer of a membrane is composed of two layers of phospholipids. Each phospholipid has a hydrophilic head and a hydrophobic tail. The hydrophilic head is attracted to water and is located on the outer sides of the bilayer, in contact with the aqueous (water-based) environment. The hydrophobic tails, however, are repelled by water and are oriented towards the middle of the bilayer away from the water. This unique arrangement allows the phospholipid bilayer to function as a barrier between the inside and outside of a cell, or between different compartments within a cell.
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The existence of the gene established by which of the following?
a. George Beadle
b. Edward Tatum
c. Gregor Mendl
d. Wilhelm Johannsen
e. James Watson
Answer:
Wilhelm Johannsen.
Explanation:
Gene may be defined as the functional segment of DNA that are heritable in nature. Genes are responsible for the formation of protein product and observable characteristics of the organisms.
The Danish botanist, Wilhelm Johannsen coined the term gene. He explained the existence of the gene by the pure line experiments of the genetics. The term gene was established in opposite to the panegenes.
Thus, the correct answer is option (d).
Extensive irrigation in arid regions causes salts to accumulate in the soil. (When water evaporates, salts that were dissolved in the water are left behind in the soil.) Based on what you learned about water balance in plant cells, explain why increased soil salinity (saltiness) might be harmful to crops.
Answer:
Increased salinity will result in total crop failure in a field. This is because plants need to take up water from the soil for growth and development. However the salinity levels of this soil will, rather, draw water from the plants by osmosis. This is because the soil water will be hypertonic to the plant's cells cytoplasm and even seeds. In osmosis, water moves from the hypotonic solution to the hypertonic solution through a semi-permeable membrane until osmotic equilibrium is achieved. Planted seeds with therefore hardly germinate and plants will wilt and die, as a result.
Answer:
Increased salinity will result in total crop failure in a field. This is because plants need to take up water from the soil for growth and development. However the salinity levels of this soil will, rather, draw water from the plants by osmosis. This is because the soil water will be hypertonic to the plant's cells cytoplasm and even seeds. In osmosis, water moves from the hypotonic solution to the hypertonic solution through a semi-permeable membrane until osmotic equilibrium is achieved. Planted seeds with therefore hardly germinate and plants will wilt and die, as a result.
Explanation:
What types of feedback are used in the body? Find examples
Answer:
Positive and Negative feedback
Explanation:
In positive feedback, the released product in the process allows the increase in the product. For example- during the time of child birth, crevix due to more pressure allows contractions and release of oxytocin. The hormone oxytocin causes more contraction.
In negative feedback, the released product causes decline in the level. For example, the temperature regulation mechanism, high temperature cause start sweating when there is temperature drop then low temperatures causes vasoconstriction and temperature increased again.
Which of the following is not considered evidence that supports the endosymbiotic theory? Mitochondria and chloroplast chromosomes are structurally similar to bacterial chromosomes. Chloroplast DNA and mitochondria DNA is an exact copy of bacteria DNA. Mitochondria and chloroplast replicate/reproduce in a manner similar to prokaryotic cells. The ribosomes of mitochondria and chloroplast resemble those of bacteria.
The incorrect evidence is that Chloroplast DNA and mitochondria DNA are exact copies of bacteria DNA, as they are merely highly related but not identical due to gene transfer to the host nucleus.
Explanation:The piece of evidence that does not support the endosymbiotic theory is the claim that 'Chloroplast DNA and mitochondria DNA is an exact copy of bacteria DNA.' While it is true that mitochondrial DNA (mtDNA) and chloroplast DNA (cpDNA) are highly related to bacterial DNA, they are not exact copies. Over time, many genes originally present in the mtDNA and cpDNA have been transferred to the host cell's nuclear DNA, meaning these organelles' genomes are now reduced in size compared to those of their bacterial ancestors.
Mitochondria and chloroplasts share several key characteristics with bacteria that support endosymbiosis: binary fission similar to bacterial replication, the structural similarity of their DNA to bacterial chromosomes, and ribosomes that resemble those found in bacteria. These characteristics, along with the additional evidence such as similar biochemistry and plasma membranes, corroborate the endosymbiotic origin of these organelles.
Where does fertilization typically occur?
Answer:
The process of fertilization occurs in the fallopian tubes that associate the ovaries with the uterus. Fertilization takes place when a sperm cell effectively encounters with an egg cell in the fallopian tube. After the process of fertilization, the newly formed fertilized cell is known as a zygote. From here, the zygote will mediate down towards the fallopian tube and into the uterus.
The zygote than gets attached to the lining of uterus, this is known as implantation.
Fertilization occurs when a sperm and an egg combine in the uterine tubes. The sperm must encounter the egg within a few days for fertilization to happen. The egg can only survive independently for around 24 hours after ovulation.
Explanation:Fertilization occurs when a sperm and an oocyte (egg) combine and their nuclei fuse. This process takes place in the uterine tubes, which is the most typical site for sperm to encounter the egg. If the sperm does not encounter the egg immediately, it can survive in the uterine tubes for another 3-5 days. In comparison, an egg can survive independently for only approximately 24 hours following ovulation.
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Which of the following organisms arose during the Mesozoic Era? a. mammals c. insects b. birds d. salamanders
The Mesozoic era is featured by the arousal of b. birds.
Explanation:The Mesozoic Time is the age of the dinosaurs and kept going nearly 180 million a long time from around 250 to 65 million a long time back. It includes periods called the Triassic, Cretaceous, and Jurassic periods. A mass-extinction stamped the starting and conclusion of the Mesozoic Period. The occasion that caused the move from the Paleozoic time to the Mesozoic period was the most prominent termination this soil has seen.
There's noteworthy prove that birds developed inside theropod dinosaurs, particularly, that winged creatures are individuals of Maniraptora, a category which belongs to dinosaurs period.
What are the genotypic differences between F-, F+ and Hfr cells?
Answer:
The fertility factor or the F factor signifies a plasmid in some bacteria, which allows the conduction of genetic substance from a donor cell to the recipient by the process of conjugation, leading to recombination.
The genotypic difference between the F- cells, F+ cells, and the Hfr cells are that the F- cells are devoid of the F factor, the F+ cells possess autonomous F factor, that is, a segment of DNA, which can replicate autonomously in the cell. In case of Hfr, the F factor is integrated into its chromosomal DNA, thus, they carry an integrated F factor.
Describe how Mendel cross-fertilized and self-fertilized pea plants.
Pea plants have several visible differences, and since it is a common plant Mendel saw it as a perfect subject for his experiments, since pea have stamen and stigma these plants can self-pollinate, fertilizing the stigma with stamen's pollen, to prevent this, Mendel removed their stamen.
Cross-fertilization is the process in which Mendel collected pollen from a plant and transferred it to another plant with its stamen's previously cut.Self-fertilization is when the plant's pollen fertilizes itself.
I added an illustrative image, I hope you find this information useful! Good luck!
Final answer:
Mendel conducted experiments on pea plants by cross-fertilizing and self-fertilizing them to study inheritance patterns. Cross-fertilization involved manually transferring pollen between different true-breeding plants, while self-fertilization occurred naturally in subsequent generations.
Explanation:
Gregor Mendel is famed for his experiments on pea plants that formed the basis of modern genetics. To carry out his studies on inheritance, Mendel performed both cross-fertilization and self-fertilization techniques on true-breeding plants, primarily the species Pisum sativum. For cross-fertilization, Mendel manually transferred pollen from the anther (male reproductive organ) of one pea plant to the stigma (female reproductive organ) of another, ensuring that the two parent plants had different traits. Before doing this, he removed the anthers from the recipient plant to prevent self-fertilization and ensure controlled breeding.
When Mendel's cross was successful, he obtained what he termed as the P or parental generation. He would then cultivate the resulting seeds to produce the first generation of offspring, known as the F1 generation. After examining the traits in the F1 generation, Mendel permitted these plants to self-fertilize, meaning they produced seeds using their pollen, creating the F2 generation. This meticulous approach allowed Mendel to discover the inheritance patterns that would lead to his famous laws of heredity.
One stage in the lysogeniclife cycle has been shown to cause
dramatic
changes in the phenotype and pathogenicity of bacteria (e.g
makingbacterium
cause botulism). What stage is this?
Answer:
In the stage of integration of the viral DNA into the host's cell, it is created the prophages. In this interaction, lysogenic conversion may occur, when a prophage induces a change in the phenotype of its host
Explanation:
The lysogenic cycle is a stage of reproduction of some viruses that infect bacteria, inserting their DNA into the bacterial cell. Together with the lytic cycle, they make up the two cycles of viral reproduction.
In the lytic cycle, the genetic material inserted synthesizes RNA that will form the capsids of the new phages and will eventually cause lysis and cell death. In the lysogenic cycle, viral DNA is integrated into the genome of the bacteria. The bacteria will live and reproduce, also replicating viral DNA in new bacteria without the virus manifesting itself. At the moment in which the DNA of the virus separates from the genome of the bacterium, the lytic cycle begins.
In the process of integrating the viral DNA into the bacterial genome, changes in the phenotype of the bacterium, generally associated with its pathogenicity, can occur, to increase the host's survival capabilities, this process is called lysogenic conversion. Even a non-pathogenic bacterium can permanently become pathogenic by lysogenic conversion.
Characteristics of Azotobacter Azotobacter is a genus of bacteria that live in soil and have the following characteristics: They are bacilli. They are gram-negative. They are obligate aerobes. They can fix nitrogen. (Unlike some other nitrogen-fixing bacteria, which associate with the roots of plants, Azotobacter species are free-living.) Select the four statements that are true for bacteria in the genus Azotobacter. They are poisoned by O_2. They appear purple after Gram staining. They have a relatively thin layer of in their cell wall. They require amino acids or other organic molecules as a source of nitrogen. They use O_2 for cellular respiration. They can carry out anaerobic respiration in an environment that lacks O_2. They are shaped like rods. They have the appearance of coils or corkscrews. They can convert atmospheric nitrogen to ammonia.
Answer:
- They have a relative thin layer in their cell wall
-They use O2 for cellular respiration
-They are shaped like rods
-They can convert atmospheric nitrogen to ammonia
Explanation:
The genus Azotobacter comprises a group of species of bacilli (rod-shaped), strict aerobics (they need oxygen to live) and gram-negative, which is to say they have a thin layer of peptidoglycan in their cell wall, that when performing a gram staining would appear pink. These bacteria can fix nitrogen, that is, convert atmospheric nitrogen (N2) into ammonia (NH3), more available to metabolize by most organisms, a process essential to life.
The effect of the Y chromosome on gonadal sex determination is mediated by a gene called ___________
Answer:
SRY
Explanation:
Sex determination in mammals including humans is genetically and hormonally controlled.
Genetically gonadal sex determination is mediated by a gene called SRY. This gene is known as the mammalian Y-chromosomal testis-determining gene. It induces sex determination in males.
Recent studies revealed that SRY plays an important role in inducing Sertoli cells differentiation. The Sertoli cells, in turn, guide testis formation.
Therefore, in males, differentiation of testis is switched on by expression of the Y-linked SRY gene.
In the history of Earth and the evolution of cells, aerobic cells appeared first and anaerobic cells appeared later.
a. True
b. False
Answer:
False
Explanation:
Scientist believe that the primitive atmosphere lacked free oxygen. Oxygen is a highly reactive molecule that would have made difficult for complex macromolecules to appear and then create life. Rocks from the precambrian period support this theory.
Anaerobic cells appeared first, then photosynthesis appeared filling the atmosphere with oxygen and the aerobic organism develop cellular respiration to obtain energy, a process more efficient than anaerobic respiration.
These are the only viruses large enough to be seen under the light microscope:
A) Papillomavirus
B) Poxvirus
C) Herpes virus
D) Adenovirus
Answer:
B) Poxvirus
Explanation:
Poxviruses, a type of DNA-viruses, typically infect animal cells. They are particular big among viruses, with sizes ranging from 200 to 400nm. Their extremely large size (considering they are viruses) allow them to be visualized under light microscope.
In prokaryotes, a search for genes in a DNA sequence involves scanning the DNA sequence for long open reading frames (that is, reading frames uninterrupted by stop codons). What problem can you see with this approach in eukaryotes?
Final answer:
Scanning for long open reading frames to find genes works in prokaryotes due to their less complex genome organization and lower noncoding DNA content. This method is not as effective in eukaryotes, which have more noncoding DNA, individual promoters for each gene, and complex chromatin structures that hinder straightforward gene identification.
Explanation:
In prokaryotes, scanning a DNA sequence for long open reading frames (ORFs) is a logical method for gene identification because prokaryotic genomes have less noncoding DNA and often organize genes encoding proteins of related functions into operons. This means that genes are grouped together, have a single promoter, and can be transcribed into polycistronic mRNA, making it easier to identify coding sequences.
However, using this approach in eukaryotes imposes challenges. Eukaryotic DNA contains a much higher percentage of noncoding DNA, which means that scanning for long ORFs would yield many false positives, as noncoding introns and other noncoding elements could be mistakenly identified as genes. Additionally, eukaryotic genes are typically monocistronic, each gene having its own promoter, and exist in a complex chromatin structure, which makes simple scanning insufficient.
Another complexity in eukaryotic genomes is that promoters can be located within genes or far away from them, both upstream or downstream. This means that the regulatory sequences are not always neatly placed before genes, as seen in prokaryotes. Also, eukaryotic DNA is wrapped around histone proteins, forming nucleosomes, which complicate direct access to the DNA sequence for transcription and therefore, gene identification.
Describe the relationship between linkage groups and chromosomes.
Answer:
The relationship between linkage groups and chromosomes is that the fist ones are all of the genes on a single chromosome.
Explanation:
Linkage groups are a group of genes present at different loci on the same chromosome. So a chromosome constitutes one linkage group that is inherited as a group during cell division. The haploid number of chromosomes shows the maximum number of linkage groups.
A linkage is the tendency that genes have to remain together during inheritance in the original combination.
Answer:
When genes are located on different chromosomes they segregate independently, but when they are located on the same chromosome, there is no segregation and they go together to the same gamete. This process is called gene binding.
Explanation:
When genes are located on different chromosomes they segregate independently, but when they are located on the same chromosome, there is no segregation and they go together to the same gamete. This process is called gene binding.
In the process of independent segregation, an AaBb individual produces 4 types of gametes, at a rate of 25% each. When a case of gene binding occurs, the individual AaBb produces only AB and ab gametes, at a rate of 50% each.
The link between the genes may be incomplete, because during meiosis prophase 1, when homologous chromosomes are paired, exchanges of parts occur between sister chromatids in a process called crossing-over or permutation. These exchanges result in the formation of recombinant gametes, which are chromosomes with new allele combinations.
If there was no recombination in these genes, the proportion of gametes formed by a double heterozygote would be 50% AB and 50% ab. When recombination occurs, a small proportion of recombinants are observed in the progeny.
Which of these are true regarding eukaryotes? (choose all that apply)
a. Chromosomes each have one origin of replication
b. During replication there is both a leading strand and a lagging strand
c. Each replication bubble has two replication forks
d. Replication is stopped by the Ter proteins.
Answer:
b. During replication there is both a leading strand and a lagging strand .
c. Each replication bubble has two replication forks.
Explanation:
Eukaryotic chromosomes have multiple origins of replication to replicate the long chromosomes at a higher rate.
The two DNA strands have opposite polarity, that is, 5' end of the one DNA strand is present opposite to the 3' end of the other DNA strand. DNA replication occurs only in 5' to 3' direction and the direction of the movement of the replication fork is also 5' to 3' direction.
To allow the DNA replication in 5' to 3' direction on both strands, one strand is replicated discontinuously in the direction opposite to the movement of the replication fork.
The discontinuously replicated strand is lagging strand while the other one is the leading strand.
DNA replication in eukaryotes occurs bidirectionally as two replication forks are formed at each replication bubble, one at each end of the replication bubble.
The presence of multiple origins of replication and the bidirectional process allows the replication of large eukaryotic DNA at a considerable fast speed.
Final answer:
In eukaryotic cells, DNA replication involves multiple origins of replication which allow for the efficient copying of their typically linear chromosomes. b) Both leading and lagging strands are synthesized, with c) replication proceeding with two forks per bubble.
Explanation:
The question pertains to the characteristics of eukaryotic DNA replication. In eukaryotic chromosomes, which are typically linear, you would find multiple origins of replication. DNA replication in eukaryotes involves b) both a leading strand and lagging strand. Each new double helix at the replication fork moves bi-directionally, creating c) two replication forks per replication bubble. Contrary to an option given in the problem, replication in eukaryotes is not stopped by Ter proteins; this is a mechanism found in bacterial DNA replication.
Each chromosome in a eukaryotic cell has many thousands of origins of replication, initiating replication at multiple points to ensure the DNA can be copied efficiently within the necessary timeframe. The replication bubbles formed by the origins coalesce as replication forks meet, resulting in the duplication of DNA. The replication process includes enzymes like DNA polymerases, helicases, and ligases, and is efficient thanks to the orchestrated function of these proteins.
The net primary productivity is usually more than 80% of gross primary productivity.
a. True
b. False
Answer:
b. False
Explanation:
As GPP stands for the gross primary productivity, it's the total amount of energy produced in the process and the NPP is the net primary productivity. NPP is the difference between the GPP and the energy used by the producer for the process of respiration. GPP is a continuous and ongoing process that consumes the energy as compared to the NPP that store energy and not being used. Thus, GPP can affect NPP but NPP does not affect GPP. For example, GPP is completely being dependent on the chlorophyll content of the plant, but NPP is independent of that content of chlorophyll.A portion of one DNA strand of the human gene responsible for cystic fibrosis is 5' ......ATAGCAGAGCACCATTCTG.....3' Write the sequence of the corresponding region of the other DNA strand of this gene, noting the polarity. What do the dots before and after the given sequence represent?
Answer:
DNA strands consists of four nitrogenous bases, pentose sugar and a phosphate group. The DNA strands are complimentary with each other as thymine binds with adenine with two hydrogen bonds and guanine binds with cytosine with three hydrogen bonds.
The given DNA strand for cystic fibrosis is 5'......ATAGCAGAGCACCATTCTG.....3'. Its complimentary DNA strand in 3'......TATCGTCTCGTGGTAAGAC......5'. The DNA strands of 5'.....CAGAATGGTGCTCTGCTAT........3′. The dots before and after the given sequence represent that the given strand is a shirt stretch of the DNA made of large number of nucleotides.
Final answer:
The sequence of the corresponding region on the other DNA strand for the given sequence is 3' ......TATCGTCTCGTGGTAAGAC.....5'. The dots indicate unspecified flanking regions.
Explanation:
For the given DNA sequence 5' ......ATAGCAGAGCACCATTCTG.....3', the sequence of the corresponding region on the other DNA strand, noting its polarity, is 3' ......TATCGTCTCGTGGTAAGAC.....5'. The dots before and after the given sequence represent the flanking regions of DNA that are not specified within this sequence. In DNA, base pairing occurs between adenine (A) and thymine (T), and between guanine (G) and cytosine (C), in antiparallel orientation, which means that the sequences on the two strands are reversed and complementary to each other.
In sheep, white vs yellow body fat is determined by a single pair of autosomal alleles. In a large, randomly mating flock, 23 percent of the sheep have white fat and 77 percent have yellow fat. You may conclude from this data that:
a. The allele for yellow is dominant to white
b. The allele for white is dominant to yellow
c. The alleles for yellow and white are codominant
d. The alleles for white and yellow are in a Hardy-Weinberg equilibrium
e. None of the above are valid conclusions
Answer:
Option A , The allele for yellow is dominant to white
Explanation:
It is given that 23 percent of the sheep have white fat and 77 percent have yellow fat.
Thus, it is clear that sheeps with yellow fat is a dominant characteristics.
Usually, the frequency of homozygous recessive allele in a given population is represented by [tex]q^2[/tex]
and here [tex]q^2 = 0.23[/tex]
Frequency of allele for white fat [tex]= q = \sqrt{0.23} \\q= 0.479[/tex]
While frequency of homozygous dominant allele and heterozygous dominant allele is represented by [tex]p^2+2pq\\[/tex]
And here [tex]p^2+2pq = 0.77[/tex]
Allele for yellow fat [tex]= 1-q\\= 1-0.479\\= 0.521[/tex]
Hence, option A is correct.
If the a and b loci are 20 m.u. apart in humans and an A B/ a b woman mates with an a b/ a b man, what is the probability that their first child will be A b/ a b?
Answer:
The probability of having a Ab/ab child is 10%
Explanation:
The genes A/a and B/b are linked and 20 m.u. apart.
The parental cross is:
♀ AB/ab X ♂ ab/abGametes:
The man only produces 1 type of gametes, so the probability of him producing an ab gamete is 1.
The woman produces 4: two parental (AB and ab) and two recombinant (Ab, aB).
Man: ab
Woman: AB, ab, Ab, aB
The formula to relate genetic distance with recombination frequency is:
Genetic Distance (m.u.)= Recombination Frequency X 100.
Replacing the data in the formula, we have:
20 m.u. / 100 = Recombination Frequency
0.2 = Recombination Frequency
Because the Recombination Frequency is 0.2, the woman will generate recombinant gametes 20% of times, and parental gametes the other 80%. Each recombinant gamete will appear in 10% of the cases, and each parental gamete will appear in 40% of the cases.
The probailities for each possible genotype of the progeny resulting from that cross will be:
Parental: AB/ab 40%
Parental: ab/ab 40%
Recombinant: Ab/ab 10%
Recombinant: aB/ab 10%
The probability that their first child will be A b/ a b is 20%.
Explanation:The probability that their first child will be A b/ a b can be determined using the principles of Mendelian genetics and the concept of random segregation and fertilization. Since the A and B loci are 20 m.u. apart, the chance of recombination occurring between the two loci is 20%. Therefore, the probability of their first child having the genotype A b/ a b is also 20%.
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You want to test pentapeptides (short peptides with only five amino acids) for their ability to bind to and inhibit a particular receptor. To do this, you set out to synthesize all possible pentapeptides and test each individually. Assuming you’ll use just the 20 common amino acids, how many different pentapeptides will you have to test for receptor binding
Answer:
The correct answer is 3200000.
Explanation:
If there is only one amino acid in the peptide, the probable number of combinations would be 20. If there is only two amino acid in the peptide, the probable number of combinations will be 20^2. Similarly, for three amino acids in the peptide it would be 20^3 and for four it would be 20^4.
Thus, in the case of pentapeptide, that is, five amino acids in a peptide, the probable number of combinations will be 20^5, which comes out to be 3200000. Thus, the total number of different pentapeptides will be 3200000 that needs to be tested for receptor binding.
The number of different pentapeptides that will need to be tested for receptor binding is 15,504.
Explanation:In order to calculate the number of different pentapeptides that will need to be tested, we can use the formula for combinations. Since there are 20 common amino acids to choose from and we are testing for pentapeptides (peptides with five amino acids), the formula for combinations is 20C5.
Using this formula, we can calculate the number of different pentapeptides:
20C5 = 20! / (5! * (20-5)!) = 20! / (5! * 15!) = (20 * 19 * 18 * 17 * 16) / (5 * 4 * 3 * 2 * 1) = 15,504.
Therefore, there will be 15,504 different pentapeptides that will need to be tested for receptor binding.
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Factors that should be considered in evaluating a biofuel include (choose any/all that apply, points deducted for incorrect answers):
a. should not compete with food production
b. net energy balance
c. energy balance ratio
d. total energy output
e. economic viability
f. should be carbon negative or neutral
g. should be carbon positive
Answer:
The following are the factors, which should be taken into account while evaluating a biofuel include:
a. Net energy balance
b. Should not compete with the production of food
c. Economic viability
d. Energy balance ratio
e. Should be neutral or carbon negative
The first generation biofuel is generated from sugarcane, which produces ethanol, and this eventually results in the elevation of grain prices. There was a controversy regarding the generation of raw material for its application as food or fuel.
Thus, it should not be competition with the production of food. The bio-fuels are generated with an objective to provide environment of zero carbon emission, that is, should be neutral or carbon negative, and at the same time it should be of low cost in production.
Some bases are attached to sugars whereas other bases are attached to phosphates.
a. True
b. False
Answer:
The correct answer will be option-false.
Explanation:
DNA is made up of repeating monomer units called nucleotides which form both the strand of the DNA. Each nucleotide is composed of a five-carbon sugar called deoxyribose, a phosphate group and four different types of nitrogenous bases.
The nitrogenous bases are of two types: purines with four nitrogen like adenine and guanine and pyrimidines with two nitrogen like thymine and cytosine.
According to the Chargaff rule, purine binds pyrimidine always in the DNA where adenine binds thymine and cytosine binds guanine.
Thus, false is the correct answer.
The first strong evidence for colinearity between a gene and its polypeptide product resulted from studies by:
a. Wilhelm Johannsen
b. Gregor Mendel
c. Charles Yanofsky
d. James Watson
e. Francis Crick
Answer:
The correct answer is option c, that is, Charles Yanofsky.
Explanation:
An American geneticist, Charles Yanofsky, worked as a faculty at Stanford University took part in the development of one gene-one enzyme hypothesis, and found attenuation, that is, a riboswitch mechanism.
In the mechanism, the messenger RNA modifies its shape in response to a small molecule, and therefore, changes its binding capability for the regulatory region of an operon and a gene. His studies suggested that genes and proteins are co-linear. If the section of the intron in the gene gets mutated, then the protein generated by the protein also gets mutated.
What are the three functions of the DNA molecule?
A. Reproduce, correct mistakes, produce new DNA
B. Replicate, correct mistakes, produce proteins
C. Replicate, mutate, produce proteins
D. Reproduce offspring, mutate, produce proteins
E. Replicate, mutate, build cells
Answer: Option C- Replicate, Mutate, Produce proteins.
Explanation:
DNA can be defined as Deoxyribonucleic Acid, which is known as the hereditary material in humans as well as in almost all other organisms. Synthesis of DNA proceeds from Deoxyribonucleotides with the help of enzymes called DNA polymerase.
The essential property of DNA is Replication, which means it can make copies of itself to pass on the genetic information and thus it is known as the hereditary material.
Sometimes during replication DNA can undergo certain errors or damage which leads to Mutation. Mutations may or may not lead to abnormal biological processes.
Another function of DNA is Transcription, the process through which synthesis of proteins is done. In this DNA makes mRNA in order to make protein.
Thus, the most suitable answer is "Option C".
The three functions of the DNA molecule are to reproduce offsprings, mutate, produce proteins
In order words, DNA carries genetic instructions and informations in all living organisms.
It contains instructions needed for an organism's reproduction, development and survival
What is DNA?DNA also known as deoxyribonucleic acid and is made up of chemical building blocks called nucleotides.
Structurally, every DNA is made up of three major components; which are as follows:
A phosphate groupA sugar group And one of four types of nitrogen basesLearn more about DNA:
https://brainly.com/question/16829508
What are the products of the light reactions that are subsequently used by the Calvin cycle?
A) oxygen and carbon dioxide
B) carbon dioxide and RuBP
C) water and carbon
D) electrons and photons
E) ATP and NADPH
Answer:
E) ATP and NADPH
Explanation:
The light-dependent reactions of photosynthesis include splitting of water in presence of sunlight followed by transfer of released electrons from PSII to PSI via intermediate carriers. The electrons finally reduce the NADP+ into NADPH.
As the electrons move from carriers between PSII and PSI, the protons are pumped into thylakoid lumen and an electrochemical gradient is generated. movement of protons back to the stroma via proton channel drives ATP synthesis.
The NADPH and ATP formed during light reactions are channeled into the Calvin cycle. Here, NADPH serves in redox reactions while ATP serves as an energy source to fix the atmospheric CO2 into glucose.
The products of the light reactions that are used in the Calvin cycle are ATP and NADPH, which provide the energy and electrons needed for the synthesis of glucose.
Explanation:The products of the light reactions that are used by the Calvin cycle are ATP (adenosine triphosphate) and NADPH (nicotinamide adenine dinucleotide phosphate). During the light reactions, light energy is converted into chemical energy in the form of these two molecules. ATP provides the energy needed for the Calvin cycle to synthesize glucose, while NADPH provides the necessary electrons. Hence, the answer is: E) ATP and NADPH.
Learn more about Calvin Cycle here:https://brainly.com/question/34149547
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