Genetics, DNA, and Molecular Biology - GRE
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Colorblindness is a recessive sex-linked disease that is caused by a defective gene on the X-chromosome. If a mother who is a carrier for the trait mates with a normal sighted male, what percentage of their daughters will have the disease?
Colorblindness is a recessive sex-linked disease that is caused by a defective gene on the X-chromosome. If a mother who is a carrier for the trait mates with a normal sighted male, what percentage of their daughters will have the disease?
Since colorblindness is a recessive disease, all copies of the X-chromosome must have the diseased allele in order for the person to be colorblind. Daughters have two copies of the X-chromosome: one from the mother and the other from the father. Males only have one copy of the X-chromosome (from the mother) and a Y-chromosome from the father.
Since we know that the father has normal vision, he does NOT carry the colorblind allele. Since the daughters for this couple can only potentially receive one colorblind allele (from the mother), all of their daughters will have normal vision. This means that there is a zero percent chance for colorblindness in their daughters.
The cross would look like this, taking Xb as the colorblind allele:
Parents: XXb x XY
Offspring: XX or XXb (normal daughters), XY (normal son), YXb (colorblind son)
The chance of a colorblind daughter will be zero, but the chance of a colorblind son will be 50%.
Since colorblindness is a recessive disease, all copies of the X-chromosome must have the diseased allele in order for the person to be colorblind. Daughters have two copies of the X-chromosome: one from the mother and the other from the father. Males only have one copy of the X-chromosome (from the mother) and a Y-chromosome from the father.
Since we know that the father has normal vision, he does NOT carry the colorblind allele. Since the daughters for this couple can only potentially receive one colorblind allele (from the mother), all of their daughters will have normal vision. This means that there is a zero percent chance for colorblindness in their daughters.
The cross would look like this, taking Xb as the colorblind allele:
Parents: XXb x XY
Offspring: XX or XXb (normal daughters), XY (normal son), YXb (colorblind son)
The chance of a colorblind daughter will be zero, but the chance of a colorblind son will be 50%.
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Hemophilia A is a blood clotting disorder that is usually inherited as an X-linked recessive trait. If a woman who does not have hemophilia A has a daughter with a man who does not have hemophilia A, what is the probability that the daughter will have hemophilia A if the maternal grandfather did have hemophilia A?
Hemophilia A is a blood clotting disorder that is usually inherited as an X-linked recessive trait. If a woman who does not have hemophilia A has a daughter with a man who does not have hemophilia A, what is the probability that the daughter will have hemophilia A if the maternal grandfather did have hemophilia A?
X-linked recessive inheritance dictates that expression of themutant phenotype will only occur if the individual is homozygous for the mutation on the X-chromosomes. Therefore, a female must have inherited two mutant X-chromosomes to have hemophilia A, while a male only requires one mutant X-chromosome to have the disorder. By virtue of the father not having hemophilia A, we know the daughter is inheriting at least one wild-type X-chromosome, and therefore there is zero chance she will be homozygous and have hemophilia A.
X-linked recessive inheritance dictates that expression of themutant phenotype will only occur if the individual is homozygous for the mutation on the X-chromosomes. Therefore, a female must have inherited two mutant X-chromosomes to have hemophilia A, while a male only requires one mutant X-chromosome to have the disorder. By virtue of the father not having hemophilia A, we know the daughter is inheriting at least one wild-type X-chromosome, and therefore there is zero chance she will be homozygous and have hemophilia A.
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Two parents are heterozygous for an allele that determines hair color for this species. These parents have offspring with the following genotypic ratios:
50% Bb, 25% BB, 25% bb
The B allele produces black hair, and the b allele produces white hair. However, the phenotypic expression of this gene's traits do not follow Mendelian patterns. If allele B exhibits incomplete dominance, which of the following is true for the offspring?
Two parents are heterozygous for an allele that determines hair color for this species. These parents have offspring with the following genotypic ratios:
50% Bb, 25% BB, 25% bb
The B allele produces black hair, and the b allele produces white hair. However, the phenotypic expression of this gene's traits do not follow Mendelian patterns. If allele B exhibits incomplete dominance, which of the following is true for the offspring?
Incomplete dominance indicates that there is no dominant allele. In these cases, the phenotype associated with inheriting one copy of each allele (the heterozygotes, Bb) is often a blending of the phenotypes associated with homozygosity of each allele. As such, a genotype of BB will result in black hair, bb will produce white hair, and Bb will result in grey hair.
The incorrect answers are too limited in scope to be cases of incomplete dominance. The correct answer identifies that there will be three unique phenotypes.
Incomplete dominance indicates that there is no dominant allele. In these cases, the phenotype associated with inheriting one copy of each allele (the heterozygotes, Bb) is often a blending of the phenotypes associated with homozygosity of each allele. As such, a genotype of BB will result in black hair, bb will produce white hair, and Bb will result in grey hair.
The incorrect answers are too limited in scope to be cases of incomplete dominance. The correct answer identifies that there will be three unique phenotypes.
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Which of the following statements about autosomal-dominant disorders is false?
Which of the following statements about autosomal-dominant disorders is false?
Because the disorder is autosomal dominant, the statement "If an individual does not have the disorder, they can still pass on the mutant gene if one of their parents has the disorder" must be false.
If the indivdual in question does not have the disorder, that means they did not inherit ANY copies of the mutant gene, and therefore cannot pass it on.
Because the disorder is autosomal dominant, the statement "If an individual does not have the disorder, they can still pass on the mutant gene if one of their parents has the disorder" must be false.
If the indivdual in question does not have the disorder, that means they did not inherit ANY copies of the mutant gene, and therefore cannot pass it on.
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Which of the following inheritance patterns only requires a single copy of a mutant allele for presentation of a specific gene disorder?
Which of the following inheritance patterns only requires a single copy of a mutant allele for presentation of a specific gene disorder?
For autosomal dominant disorders, the individual only needs to inherent a single copy of a mutated allele to then show symptoms of that disorder. If it were recessive, both alleles would have to be mutant. X-linked recessive is incompletely correct for males since they only have one X-chromosome, and incorrect for females since 2 copies of the X-chromosome are needed, and thus 2 copies of the allele. Complex inheritance describes situations beyond a single gene, and Mendelian inheritance is not a specific method of inheritance. Note that Y-linked disorders are passed from father to son, and since males only have one copy of the Y-chromosome, if there is a genetic mutation on the Y-chromosome, the individual will be affected.
For autosomal dominant disorders, the individual only needs to inherent a single copy of a mutated allele to then show symptoms of that disorder. If it were recessive, both alleles would have to be mutant. X-linked recessive is incompletely correct for males since they only have one X-chromosome, and incorrect for females since 2 copies of the X-chromosome are needed, and thus 2 copies of the allele. Complex inheritance describes situations beyond a single gene, and Mendelian inheritance is not a specific method of inheritance. Note that Y-linked disorders are passed from father to son, and since males only have one copy of the Y-chromosome, if there is a genetic mutation on the Y-chromosome, the individual will be affected.
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A scientist is working with a breed of dog and has noticed that two traits, ear length and color, behave in normal dominant-recessive hierarchies. Long ears (A) are dominant to short ears (a) and black coloration (B) is dominant to yellow coloration (b). If he breeds a long eared, black dog (AaBb) with a short eared yellow dog (aabb), what would be the resulting phenotypic ratios of the offspring?
A scientist is working with a breed of dog and has noticed that two traits, ear length and color, behave in normal dominant-recessive hierarchies. Long ears (A) are dominant to short ears (a) and black coloration (B) is dominant to yellow coloration (b). If he breeds a long eared, black dog (AaBb) with a short eared yellow dog (aabb), what would be the resulting phenotypic ratios of the offspring?
This question can be solved by making a punnett square. The genotypes are given: AaBb x aabb.
The potential gametes the AaBb dog can produce are AB, Ab, aB, and ab. The aabb dog can only produce one gamete: ab.
Putting these gametes in our punnett square we can see that we end up with the following potential offspring: AaBb, Aabb, aaBb, and a_a_bb.
Each of these possible offspring carries a different phenotype. AaBb will carry both dominant traits and be black with long ears. Aabb will be yellow with long ears_._ aaBb will be black with short ears. Finally, aabb will be yellow with short ears. Each of these gametes is produced in the same ratio, making these phenotypes exist in a 1:1:1:1 probability.
This question can be solved by making a punnett square. The genotypes are given: AaBb x aabb.
The potential gametes the AaBb dog can produce are AB, Ab, aB, and ab. The aabb dog can only produce one gamete: ab.
Putting these gametes in our punnett square we can see that we end up with the following potential offspring: AaBb, Aabb, aaBb, and a_a_bb.
Each of these possible offspring carries a different phenotype. AaBb will carry both dominant traits and be black with long ears. Aabb will be yellow with long ears_._ aaBb will be black with short ears. Finally, aabb will be yellow with short ears. Each of these gametes is produced in the same ratio, making these phenotypes exist in a 1:1:1:1 probability.
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Which of the following choices represents information contained in a punnett square?
I. Potential genotype ratios of offspring
II. Possible gametes produced by parent generation
III. Allele frequencies of the population
Which of the following choices represents information contained in a punnett square?
I. Potential genotype ratios of offspring
II. Possible gametes produced by parent generation
III. Allele frequencies of the population
Punnett squares give information about the potential genotype ratios of offspring possible from the cross of two members of the parental generation. The letters represent alleles of various genes, but do not give any information about the allele frequencies. To get information about the allele frequencies, more information about the size and make-up of the population would be needed. The actual cross is between potential gametes produced by the parental generation. Each square shows the potential offspring from these potential gametes.
Punnett squares give information about the potential genotype ratios of offspring possible from the cross of two members of the parental generation. The letters represent alleles of various genes, but do not give any information about the allele frequencies. To get information about the allele frequencies, more information about the size and make-up of the population would be needed. The actual cross is between potential gametes produced by the parental generation. Each square shows the potential offspring from these potential gametes.
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Peas in pea plants can be either yellow or green, with yellow being the dominant color. The peas can also be smooth or wrinkled, with smooth being the dominant shape. Suppose that a pea plant that is heterozygous for both traits is self crossed.
What proportion of the next generation will have smooth, green peas?
Peas in pea plants can be either yellow or green, with yellow being the dominant color. The peas can also be smooth or wrinkled, with smooth being the dominant shape. Suppose that a pea plant that is heterozygous for both traits is self crossed.
What proportion of the next generation will have smooth, green peas?
The shortcut for this problem involves the standard phenotypic ratios for a dihybrid cross. Nine offspring will show both dominant traits. Three will show one dominant trait and the other recessive trait. Three will show the inverse phenotypes, with the opposite dominant trait and recessive trait combination. One offspring will show both recessive traits. Based on these ratios, we can see that three of the sixteen offspring will show the dominant smooth trait and the recessive green phenotype.
We can also solve by using a dihybrid punnett square. The cross described will be AaBb x AaBb, in which the A alleles signify color and the B alleles signify shape.
Consider the color of the peas. In order to have green peas, two recessive alleles must combine in the next generation. According to a punnett square where both sides are heterozygous for the trait, there is only a one in four chance of this taking place. Since smooth is the dominant shape for the peas, a punnett square where each side is heterozygous shows a three in four chance that pea plants will have this shape. By multiplying these two probabilities, we determine that three out of sixteen pea plants will have smooth, green peas.
The shortcut for this problem involves the standard phenotypic ratios for a dihybrid cross. Nine offspring will show both dominant traits. Three will show one dominant trait and the other recessive trait. Three will show the inverse phenotypes, with the opposite dominant trait and recessive trait combination. One offspring will show both recessive traits. Based on these ratios, we can see that three of the sixteen offspring will show the dominant smooth trait and the recessive green phenotype.
We can also solve by using a dihybrid punnett square. The cross described will be AaBb x AaBb, in which the A alleles signify color and the B alleles signify shape.
Consider the color of the peas. In order to have green peas, two recessive alleles must combine in the next generation. According to a punnett square where both sides are heterozygous for the trait, there is only a one in four chance of this taking place. Since smooth is the dominant shape for the peas, a punnett square where each side is heterozygous shows a three in four chance that pea plants will have this shape. By multiplying these two probabilities, we determine that three out of sixteen pea plants will have smooth, green peas.
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What is the structure in prokaryotic cells that often holds antibiotic resistance genes?
What is the structure in prokaryotic cells that often holds antibiotic resistance genes?
Antibiotic resistance genes are often found on plasmids, which are small DNA molecules which are easily transfered to other prokaryotic cells.
Antibiotic resistance genes are often found on plasmids, which are small DNA molecules which are easily transfered to other prokaryotic cells.
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In a small village, a doctor over-prescribed Ciprofloxin (an antiobiotic) to his patients. The result was that the majority of his patients had developed a new strain of bacteria that had become resistant to Ciprofloxin.
What is a possible route that may have caused these bacteria to become antibiotic resistant?
In a small village, a doctor over-prescribed Ciprofloxin (an antiobiotic) to his patients. The result was that the majority of his patients had developed a new strain of bacteria that had become resistant to Ciprofloxin.
What is a possible route that may have caused these bacteria to become antibiotic resistant?
Antibiotic resistant genes are commonly transferred through plasmids. Plasmids are small, circular, extra-chromosomal pieces of DNA that contain supplemental genes, which can be passed from bacteria to bacteria.
Antibiotic resistant genes are commonly transferred through plasmids. Plasmids are small, circular, extra-chromosomal pieces of DNA that contain supplemental genes, which can be passed from bacteria to bacteria.
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Which of these processes is the means by which a bacterium can directly uptake and incorporate foreign DNA from the environment into its genome?
Which of these processes is the means by which a bacterium can directly uptake and incorporate foreign DNA from the environment into its genome?
Transformation is defined as the process by which bacteria can incorporate exogenous DNA from the environment into their genome via direct uptake. Transduction and conjugation are also processes by which exogenous DNA is incorporated, but involve other methods.
Transformation is defined as the process by which bacteria can incorporate exogenous DNA from the environment into their genome via direct uptake. Transduction and conjugation are also processes by which exogenous DNA is incorporated, but involve other methods.
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Sexually transmitted diseases are a common problem among young people in the United States. One of the more common diseases is caused by the bacterium Neisseria gonorrhoeae, which leads to inflammation and purulent discharge in the male and female reproductive tracts.
The bacterium has a number of systems to evade host defenses. Upon infection, it uses pili to adhere to host epithelium. The bacterium also uses an enzyme, gonococcal sialyltransferase, to transfer a sialyic acid residue to a gonococcal surface lipooligosaccharide (LOS). A depiction of this can be seen in Figure 1. The sialyic acid residue mimics the protective capsule found on other bacterial species.
Once infection is established, Neisseria preferentially infects columnar epithelial cells in the female reproductive tract, and leads to a loss of cilia on these cells. Damage to the reproductive tract can result in pelvic inflammatory disease, which can complicate pregnancies later in the life of the woman.
A doctor uses a new antibiotic to fight off Neisseria infection. Shortly thereafter, resistance develops and the doctor finds that inoculating wild type bacteria with the pure DNA of resistant Neisseria organisms produces resistant colonies. Which of the following is the method of modification that the doctor is using?
Sexually transmitted diseases are a common problem among young people in the United States. One of the more common diseases is caused by the bacterium Neisseria gonorrhoeae, which leads to inflammation and purulent discharge in the male and female reproductive tracts.
The bacterium has a number of systems to evade host defenses. Upon infection, it uses pili to adhere to host epithelium. The bacterium also uses an enzyme, gonococcal sialyltransferase, to transfer a sialyic acid residue to a gonococcal surface lipooligosaccharide (LOS). A depiction of this can be seen in Figure 1. The sialyic acid residue mimics the protective capsule found on other bacterial species.
Once infection is established, Neisseria preferentially infects columnar epithelial cells in the female reproductive tract, and leads to a loss of cilia on these cells. Damage to the reproductive tract can result in pelvic inflammatory disease, which can complicate pregnancies later in the life of the woman.
A doctor uses a new antibiotic to fight off Neisseria infection. Shortly thereafter, resistance develops and the doctor finds that inoculating wild type bacteria with the pure DNA of resistant Neisseria organisms produces resistant colonies. Which of the following is the method of modification that the doctor is using?
Transformation is the uptake of nucleic acid by competent cells, as was described in this question. Conjugation invovles cell-to-cell DNA transfer and transduction involves the use of a viral vector.
Transformation is the uptake of nucleic acid by competent cells, as was described in this question. Conjugation invovles cell-to-cell DNA transfer and transduction involves the use of a viral vector.
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Sexually transmitted diseases are a common problem among young people in the United States. One of the more common diseases is caused by the bacterium Neisseria gonorrhoeae, which leads to inflammation and purulent discharge in the male and female reproductive tracts.
The bacterium has a number of systems to evade host defenses. Upon infection, it uses pili to adhere to host epithelium. The bacterium also uses an enzyme, gonococcal sialyltransferase, to transfer a sialyic acid residue to a gonococcal surface lipooligosaccharide (LOS). A depiction of this can be seen in Figure 1. The sialyic acid residue mimics the protective capsule found on other bacterial species.
Once infection is established, Neisseria preferentially infects columnar epithelial cells in the female reproductive tract, and leads to a loss of cilia on these cells. Damage to the reproductive tract can result in pelvic inflammatory disease, which can complicate pregnancies later in the life of the woman.
The doctor wants to study the resistance further, so he creates a bacteriophage that transmits the plasmid to other bacterial cells to produce new resistant colonies. Which of the following is the method that the doctor is now using?
Sexually transmitted diseases are a common problem among young people in the United States. One of the more common diseases is caused by the bacterium Neisseria gonorrhoeae, which leads to inflammation and purulent discharge in the male and female reproductive tracts.
The bacterium has a number of systems to evade host defenses. Upon infection, it uses pili to adhere to host epithelium. The bacterium also uses an enzyme, gonococcal sialyltransferase, to transfer a sialyic acid residue to a gonococcal surface lipooligosaccharide (LOS). A depiction of this can be seen in Figure 1. The sialyic acid residue mimics the protective capsule found on other bacterial species.
Once infection is established, Neisseria preferentially infects columnar epithelial cells in the female reproductive tract, and leads to a loss of cilia on these cells. Damage to the reproductive tract can result in pelvic inflammatory disease, which can complicate pregnancies later in the life of the woman.
The doctor wants to study the resistance further, so he creates a bacteriophage that transmits the plasmid to other bacterial cells to produce new resistant colonies. Which of the following is the method that the doctor is now using?
The use of bacteriophage viral vectors to transmit genetic information is transduction, an alternative form of genetic modification to transformation or conjugation. Transformation involves direct uptake of genetic material. Conjugation involves cell-to-cell transfer of DNA.
The use of bacteriophage viral vectors to transmit genetic information is transduction, an alternative form of genetic modification to transformation or conjugation. Transformation involves direct uptake of genetic material. Conjugation involves cell-to-cell transfer of DNA.
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Which answer choice correctly pairs the mode of bacterial genetic transfer with its definition?
Which answer choice correctly pairs the mode of bacterial genetic transfer with its definition?
The three main modes of genetic transfer for prokaryotes are transformation, transduction, and conjugation. Transformation occurs when a bacterium picks up a piece of genetic material from its external environment and incorporates it into its own genome. Transduction is genetic transfer using a bacteriophage as a vector. Conjugation is direct gene transfer via sex pili.
The three main modes of genetic transfer for prokaryotes are transformation, transduction, and conjugation. Transformation occurs when a bacterium picks up a piece of genetic material from its external environment and incorporates it into its own genome. Transduction is genetic transfer using a bacteriophage as a vector. Conjugation is direct gene transfer via sex pili.
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Which of the following is a unique characteristic of prokaryotic cells?
Which of the following is a unique characteristic of prokaryotic cells?
Prokaryotes lack a nuclear membrane, which allows translation to occur at the same time as transcription.
In eukaryotic cells the mRNA has to be exported to the cytoplasm before it can be translated. This transport requires post-transcriptional modification to protect the mRNA from degradation as it leaves the nucleus, a process unnecessary to prokaryotic cells. Both prokaryotes and eukaryotes can have cell walls and cytoplasts (cytoplasm). Prokaryotes do, in fact, generate proton gradients in order to complete cellular respiration. These gradients are created across the prokaryotic cell membrane, rather than across the mitochondrial membrane.
Prokaryotes lack a nuclear membrane, which allows translation to occur at the same time as transcription.
In eukaryotic cells the mRNA has to be exported to the cytoplasm before it can be translated. This transport requires post-transcriptional modification to protect the mRNA from degradation as it leaves the nucleus, a process unnecessary to prokaryotic cells. Both prokaryotes and eukaryotes can have cell walls and cytoplasts (cytoplasm). Prokaryotes do, in fact, generate proton gradients in order to complete cellular respiration. These gradients are created across the prokaryotic cell membrane, rather than across the mitochondrial membrane.
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Which recombinant process depends on a F factor plasmid?
Which recombinant process depends on a F factor plasmid?
Beginning the process of conjugation requires the trait encoded by the F (fertility) plasmid. Transformation is the uptake of naked DNA, transduction is the transfer of genetic material via a virus, and translocation is the movement of a ribosome during protein translation.
Beginning the process of conjugation requires the trait encoded by the F (fertility) plasmid. Transformation is the uptake of naked DNA, transduction is the transfer of genetic material via a virus, and translocation is the movement of a ribosome during protein translation.
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Frederick Griffith's experiments showed that formerly harmless IIR bacteria had become deadly in mice. This change took place when the harmless IIR bacteria were exposed to the remains of heat killed IIIS bacteria. Based on this experiment, what was the process that converted the harmless bacteria to a deadly strain?
Frederick Griffith's experiments showed that formerly harmless IIR bacteria had become deadly in mice. This change took place when the harmless IIR bacteria were exposed to the remains of heat killed IIIS bacteria. Based on this experiment, what was the process that converted the harmless bacteria to a deadly strain?
The IIR bacteria became deadly when exposed to the remains of the IIIS bacteria. This means that the IIR bacteria managed to receive genetic material from the environment and incorporate it into their genome. This is an example of transformation, a process that results in genetic recombination. In this case, the recombination made the formerly harmless bacteria deadly in mice.
Transduction is the process by which new genetic information is introduced to a bacterium via a vector, such as a bacteriophage. Conjugation is the transfer of genetic material between bacteria via a sex pilus. Binary fission is not a means of recombination; rather, the parent cell divides to produce two identical copies of itself.
The IIR bacteria became deadly when exposed to the remains of the IIIS bacteria. This means that the IIR bacteria managed to receive genetic material from the environment and incorporate it into their genome. This is an example of transformation, a process that results in genetic recombination. In this case, the recombination made the formerly harmless bacteria deadly in mice.
Transduction is the process by which new genetic information is introduced to a bacterium via a vector, such as a bacteriophage. Conjugation is the transfer of genetic material between bacteria via a sex pilus. Binary fission is not a means of recombination; rather, the parent cell divides to produce two identical copies of itself.
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What type of prokaryotic genetic recombination requires the presence of plasmids?
What type of prokaryotic genetic recombination requires the presence of plasmids?
Conjugation is a type of genetic recombination that requires one bacterium to have the F-plasmid in order to create a sex pilus. This sex pilus will connect with another bacterium and allow DNA to pass between the bacteria.
Transduction is the transfer of genetic information to a bacterium via a vector, such as a bacteriophage. Transformation occurs when a bacterial cell receives genetic material from its surrounding environment. Binary fission does not involve recombination, and is the term for bacterial cell division that results in two identical offspring from a single parental cell.
Conjugation is a type of genetic recombination that requires one bacterium to have the F-plasmid in order to create a sex pilus. This sex pilus will connect with another bacterium and allow DNA to pass between the bacteria.
Transduction is the transfer of genetic information to a bacterium via a vector, such as a bacteriophage. Transformation occurs when a bacterial cell receives genetic material from its surrounding environment. Binary fission does not involve recombination, and is the term for bacterial cell division that results in two identical offspring from a single parental cell.
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Which of the following characteristics make plasmid DNA useful for researchers?
Which of the following characteristics make plasmid DNA useful for researchers?
Plasmids are phenomenally useful for a number of reasons. They are capable of autonomous replication inside a suitable host (such as simple prokaryotes). Because they are non-genomic DNA (not located in the nucleus), plasmids are easily isolated and separated from the host's DNA. It is also relatively easy to insert cloned DNA into plasmid vectors. This allows for expression of recombinant proteins in prokaryotic cells.
Plasmids are phenomenally useful for a number of reasons. They are capable of autonomous replication inside a suitable host (such as simple prokaryotes). Because they are non-genomic DNA (not located in the nucleus), plasmids are easily isolated and separated from the host's DNA. It is also relatively easy to insert cloned DNA into plasmid vectors. This allows for expression of recombinant proteins in prokaryotic cells.
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Which of the following is true of both bacterial conjugation and meiosis?
Which of the following is true of both bacterial conjugation and meiosis?
The only similarity between conjugation and meiosis is that both processes are types of sexual reproduction. Remember that sexual reproduction is characterized by the presence of genetic recombination (the ability to exchange genetic material between two DNA molecules). The result of both processes are daughter cells that are genetically unique from the parent cells.
In meiosis, genetic recombination occurs during crossing over in prophase I. In conjugation, genetic recombination occurs when the DNA from the donor bacterial cell is incorporated into the recipient bacterial cell. Only meiosis produces four daughter cells; conjugation produces only two.
The only similarity between conjugation and meiosis is that both processes are types of sexual reproduction. Remember that sexual reproduction is characterized by the presence of genetic recombination (the ability to exchange genetic material between two DNA molecules). The result of both processes are daughter cells that are genetically unique from the parent cells.
In meiosis, genetic recombination occurs during crossing over in prophase I. In conjugation, genetic recombination occurs when the DNA from the donor bacterial cell is incorporated into the recipient bacterial cell. Only meiosis produces four daughter cells; conjugation produces only two.
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