Answer the following questions in reference to Drosophila melanogast.pdf
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Answer the following questions in reference to Drosophila melanogaster (fruit fly) that has 6 chromosomes. How many combinations are possible when chromosomes assort independently into gametes (ignoring crossing over)? 6 raised to the 2nd power 2 raised to the 6th power 6 x 2 = 12 2 raised to the 3rd power 3 raised to the 2nd power How many tetrads align at the metaphase plate of meiosis I? 0 12 6 3 Solution Ans.1 No. of possible combination during independent assortment = 2n Where, 2 indicates ploidy of the somatic cells undergoing meiosis, n = number of haploid chromosomes. Given, 2n = 6 Hence, n = 3. Thus, No. of possible combination during independent assortment = 2n = 23 Correct option: 2 raised to 3rd power. Ans. 2. Number of tetrad during meiosis = Number of haploid chromosomes (n) To understand it, let’s number the three haploid chromosomes as A, B and C. A diploid cell has two copies of each chromosome- one inherited from mother and the other from father. So, diploid cell has following chromosomes- A1, A2, B1, B2, C1 and C3, where 1= maternal chromosomes and 2 = fraternal chromosomes. All the chromosomes remain independent of each other, randomly scattered in nucleus. We will proceed with one chromosome pair, for simplicity. The following events occur when a diploid cell enters meiosis- A1 and A2 are called homologues pair. During S-phase, DNA is duplicated. The resultant- I. A1A1 remain joined to each other. Are called sister chromatids. The same happens to A2A2, and rest other chromosomes. II. Homologous chromosome A1 and A2 pair together prophase I. Since each of A1 and A2 are duplicated, they are (A1A1) and (A2A2) respectively. III. Homologous pairing forms a tetrad – A1A1/A2A2 i.e. four chromosomes remaining together. Note: Both the maternal and fraternal chromosomes and their duplicated copies remain together as a tetrad. That is, the number of tetrad in half of diploid number (because A1 from mother and A2 from father remain together as one single homologous chromosome). Thus, number of treads in drosophila = n = 3 Correct option: D (3).
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Activity- Simulating meiosis 1- Cut out the chromosomes on the second.docx
Activity- Simulating meiosis 1. Cut out the chromosomes on the second to last page of this lab. One color of chromosomes is inherited from one parent, and the other is inherited from the other parent. Use a paper clip to keep identical sister chromatids together. 2. Use the circles representing cells provided on the last two pages. Meiosis I Prophase I 1. Put all four of the chromosomes that you cut out and place them in the large oval cell labeled A pole and B pole. 2. Bring the members of the homologous pairs together forming a tetrad- Why do you think these are called a tetrad? 3. Crossing-over is the exchange of genetic material between two homologous chromosomes. It creates genetic recombination during meiosis. Simulate crossingover by cutting off one yellow and blue circle from two nonsister chromatids of a single homologous pair and reattaching the circles on with tape. Why did we exchange from nonsister and not from sister chromatids? MetaphaseI Move the homologous chromosome pairs to the equator (middle of the cell) in such a way that the members of each pair can be moved towards opposite sides of the cell AnaphaseI Separate the each member of the homologous pair, moving them toward opposite poles of the cell TelophaseI Do the 2 chromosomes inherited from the father have to stay together after meiosis I? So, what are two ways that meiosis can create genetic recombination? Interkinesis Interkinesis is the periods of time between meiosis I and meiosis II. Meiosis II Prophase II Choose the chromosomes from each pole to represent those in the new parental cell undergoing meiosis II. Place them inside the top two cells on the next page. Metaphase II Move the chromosomes (made out of the sister chromatids) to the equator of this cell. How many chromosomes are at this equator? Anaphase II Pull apart the two sister chromatids Telophase II Once the sister chromatids are apart they are now each called a chromosome. Move each chromosome to its separate pole. Now these chromosomes are in their own cell in a new nucleus. Conclusions Do the new nuclei contain a haploid or diploid chromosome number compared to the original cell? Suppose you had worked with both daughter cells after meiosis I. How many cells would be present when meiosis II was complete? How many chromosomes are in the parental cell undergoing meiosis II (at the beginning of meiosis II? How many chromosomes are in the daughter cell (at the end of meiosis II)? How is this possible? Explain below. To summarize meiosis: the original parent cell has the diploid ( 2 n ) number of chromosomes, and the daughter cells at the end of meiosis II have the number of chromosomes. Meiosis is cell division in which the chromosome number does what?
.
Recommended
Activity- Simulating meiosis 1- Cut out the chromosomes on the second.docx
Activity- Simulating meiosis 1. Cut out the chromosomes on the second to last page of this lab. One color of chromosomes is inherited from one parent, and the other is inherited from the other parent. Use a paper clip to keep identical sister chromatids together. 2. Use the circles representing cells provided on the last two pages. Meiosis I Prophase I 1. Put all four of the chromosomes that you cut out and place them in the large oval cell labeled A pole and B pole. 2. Bring the members of the homologous pairs together forming a tetrad- Why do you think these are called a tetrad? 3. Crossing-over is the exchange of genetic material between two homologous chromosomes. It creates genetic recombination during meiosis. Simulate crossingover by cutting off one yellow and blue circle from two nonsister chromatids of a single homologous pair and reattaching the circles on with tape. Why did we exchange from nonsister and not from sister chromatids? MetaphaseI Move the homologous chromosome pairs to the equator (middle of the cell) in such a way that the members of each pair can be moved towards opposite sides of the cell AnaphaseI Separate the each member of the homologous pair, moving them toward opposite poles of the cell TelophaseI Do the 2 chromosomes inherited from the father have to stay together after meiosis I? So, what are two ways that meiosis can create genetic recombination? Interkinesis Interkinesis is the periods of time between meiosis I and meiosis II. Meiosis II Prophase II Choose the chromosomes from each pole to represent those in the new parental cell undergoing meiosis II. Place them inside the top two cells on the next page. Metaphase II Move the chromosomes (made out of the sister chromatids) to the equator of this cell. How many chromosomes are at this equator? Anaphase II Pull apart the two sister chromatids Telophase II Once the sister chromatids are apart they are now each called a chromosome. Move each chromosome to its separate pole. Now these chromosomes are in their own cell in a new nucleus. Conclusions Do the new nuclei contain a haploid or diploid chromosome number compared to the original cell? Suppose you had worked with both daughter cells after meiosis I. How many cells would be present when meiosis II was complete? How many chromosomes are in the parental cell undergoing meiosis II (at the beginning of meiosis II? How many chromosomes are in the daughter cell (at the end of meiosis II)? How is this possible? Explain below. To summarize meiosis: the original parent cell has the diploid ( 2 n ) number of chromosomes, and the daughter cells at the end of meiosis II have the number of chromosomes. Meiosis is cell division in which the chromosome number does what?
.
I need help answering both these questionsQuestion 5 1. A dip.pdf
I need help answering both these questions
Question 5: 1. A diploid cell contains three pairs of homologous chromosomes: A1 and A2, B1
and B2, and C1 and C2. Assuming no crossing-over occurs, what possible combinations of
chromosomes will be present in: a) daughter cells following mitosis: b) metaphase I of meiosis:
Solution
1. A)In mitosis the daugther produced are genetically identical , so all the geneticall material is
being inherited from mother cells to daughter cells, So the only possibility is
A1/A2/B1/B2/C1/C2. Since all the genetic material is being transfered from the parent cell to
daughter cells
B) The Meiosis 1 is meant to split the homologus pair into different cells, which will now be n (
Haploid) , So there can be 8 possible combinations of chromosomes, So if we take an example it
could inherit A1/B1/C1 or A2/M1/S1.
Activity 4 Understanding the basics of meiosisMitosis and mei.docx
Activity 4: Understanding the basics of meiosis
Mitosis and meiosis both lead to new daughter cells, but meiosis sets up organisms for sexual
reproduction. Meiosis produces cells (sperm and egg in humans) with only one set of chromosomes so that when fertilization occurs, it results in a new cell with two sets of chromosomes (one from the egg and one from the sperm). This is how there is genetic recombination of DNA resulting in unique individuals.
Before we talk about what can go wrong in meiosis, you need to be clear about the process itself and how genetic recombination in particular occurs. To this end, complete the following set of questions.
Questions
1. In the following diagram, draw what a cell with 1 chromosome would look like in the stages of meiosis. Prophase 1 is filled in for you and includes replicated homologous chromosomes; the black replicated chromosome is from the reproducing male’s mom and the grey replicated chromosome is from dad.
2. In your diagram from question 1:
a. How many tetrads are formed? _______________
b. How many chromosomes are in the sperm? _______________
3. In what stage(s) of meiosis:
a. Are tetrads formed? ________________________________
b. Does crossing over occur? ________________________________
c. Do the chromosomes move to the poles? ________________________________
d. Do replicated chromosomes separate? ________________________________
e. Does the cytoplasm divide? ________________________________
4. What processes in meiosis result in genetically unique daughter cells? When do these processes occur? (Note: There are two main processes; discuss both).
5. Compare and contrast meiosis with mitosis to complete the following table.
Table 2. Comparison of key characteristics between meiosis and mitosis.
Characteristics
Mitosis
Meiosis
Type of organisms it occurs in
# of chromosomes in human parent cell
Number of times chromosomes replicate
Number of cell divisions
Crossing over occurs? (Y/N)
Type of daughter cells produced
Number of daughter cells produced
Daughter cells identical to parent cell? (Y/N)
Daughter cells are: 1n or 2n?
# of chromosomes in human daughter cells
Activity 3: Chromosomes in mitosis and meiosis
In the nucleus of the cell are the chromosomes that are composed of the hereditary material DNA. In every somatic (body) cell of a human there are 46 chromosomes. Each species may have a different number of chromosomes than another species.
Since each somatic cell of an organism contains the same number of chromosomes, there must be a duplication of material before the nucleus divides during mitosis. In each somatic cell, there are two sets of chromosomes; this is referred to as the 2n (diploid) number, in which n means number of chromosomes. In humans, 2n = 46 chromosomes.
In each gamete (sex) cell, there is only one set of chromosomes; this is referred to as the 1n (haploid) number. In humans, 1n = 23 chromosomes. This means the.
Cytotaxonomy plant taxonomy
This ppt contains all about the various chromosomal properties that can be used in studying plant taxonomy and classification of plants.
Given the information regarding mitosis and meiosis, clearly and con.pdf
Given the information regarding mitosis and meiosis, clearly and convincingly explain in your
own words how an error in MITOSIS could possibly lead to a gamete containing new genetic
information and thereby giving rise to a new phenotype. Use detail be comprehensive. (7 points)
Solution
The process of mitosis is divided into two phases: Interphase and M-phase.
Interphase is further divided into G1, S and G2 phase. In G1, protein synthesis occurs for S-
phase, in S phase, DNA replicates and in G2 phase, protein synthesis occurs for M-phase to start
properly. Now, during S phase, the ‘2n’ nucleus converts into 4n.
So, the cell entering M-phase is already ‘4n’.
M-phase is the actual mitotic phase, in which the cell divides into two. It contains four phases,
prophase, metaphase, anaphase and telophase. Your question is related with these four phases. In
prophase, the entire DNA converts into chromosomes; the process is known as condensation of
DNA. In metaphase, these chromosomes align on the equator. Now, if alignment at equator is
not correct, then anaphase cannot be started due to metaphase check point.
In anaphase, the aligned chromosomes move towards the opposite poles. This phase is the main
phase, related with formation of gametes. The cell is already ‘4n’. ‘2n’ chromosomes move
towards one end, and the rest of ‘2n’ move towards the other end. Now, suppose an error occurs
in mitosis, and ‘3n’ chromosomes move towards one end, and 1n goes towards the other cell;
then two gametes, ‘3n’ and n will be formed.
This ‘3n’ and ‘n’ gametes are gametes having different genetic information in comparison to
their ‘2n’ counterparts.
Interestingly, Meiosis II will result in gamete formation is mitotic division (equational division).
Haploid cells with \'2c\' amount of DNA leave Meiosis I. Now, in anaphase, suppose one
chromosome\' 2c DNA migrates to one end (due to error) and no DNA migrates to the other end;
then two gametes will be formed; one having all normal chromosomes + one double
chromosome with 2c DNA, and the other having no DNA.
Please note: normally gametes are 1n (haploid) and 1c.
Suppose this gamete with double DNA (2c) fertilizes with a gamete with half DNA (1c), then the
resulting cell will have “all normal chromosomes + one chromosome with ‘3c’ amount of DNA”.
This condition if occurring with chromosome 21 is known as Down\' syndrome.
It is important to note that errors in Mitosis result in non-disjunction; which further results in
many Mendelian disorders; as well as ploidy defects.
** Hope it helps..
Linkage
Linkage refers to the presence of two different genes on the same chromosome . Two genes that occur on the same chromosome are said to be linked, and those that occur very close together are tightly linked.
The homolog of the #3 chromosome is the X-chromosome Y-chromosome .pdf
The homolog of the #3 chromosome is the X-chromosome Y-chromosome other #3
chromosome chromosome that synapses with the #3 chromosome during mitosis
Solution
Ans. Each diploid organism inherits 2 copies of a chromosome (here, chromosome number 3 or
#3) – one from its mother and another from father.
That is, a human has two copies of #3 chromosomes- inherited one copy from each parent. For
simplicity of clarification, we represent mater chromosome as #3M and fraternal chromosome as
#3F
Now, note the following events carefully –
I. When a cell prepares for meiosis, it duplicates it chromosomes. That is now there are 2 copies
of #3M (one old, one new copy) and 2 copies of #3F (one old, one new copy) chromosomes, too.
II. Both the copies of #3M remain joined together with the help of cohesion rings. Both the
chromosomes, old #3M and new #3M are called sister chromatids of each other.
III. So, after S-phase, there are two pairs of sister chromatids – one #3M pair (has two #3 sister
chromatids joined together by cohesin rings) and one #3F pair.
IV. In Leptotene stage prophase I of meiosis 1, for the first time in meiotic cell cycle, the term
“homologous chromosomes” is used. Here, the term mean “. A pair of genetically similar
chromosomes having same genes on same loci are called homologous chromosomes or
homologues”. It means chromosomes #3M and #3F are homologues chromosomes because they
share the same gene loci – IRRESPECTIVE of whether synapsis has occurred or not.
V. Note that synapsis occurs in zygotene stage, (after leptotene stage in sequence).
VI. No doubt, the homologous chromosomes undergo synapsis during meiosis, but #3M and #3F
are called homologous chromosomes even if they don’t synapse (say, for a hypothetical
condition) because they same genes on same loci.
Therefore, the term “synapsis” is not required to mention two chromosomes as homologous
chromosomes- it is the character of “sharing same gene loci” that makes two chromosomes
homologous.
Thus, correct option is- C. other #3 chromosome.
Note: Different texts may discriminate the term differently. It the best possible explanations
presented here..
Lab TemplateWeek 4 MeiosisSubmitted by your name here.docx
Lab Template
Week 4: Meiosis
Submitted by: <your name here>
As you complete the lab, record your answers in this template. Save the document as LastName_FirstName_BIO1020_W4A3, and submit it to the Dropbox. Full lab instructions and the rubric with which you will be evaluated can be found in the online classroom.
Activity
Cellular reproduction in Eukaryotes involves either mitosis or, in the case of sex cells, meiosis. Mitosis involves the reproduction of a cell into two identical daughter cells. Meiosis, however, is a reduction division where a parental diploid cell produces four haploid gametes. Upon fusion, two haploid gametes (in humans the sperm and the egg) will result in one diploid zygote. In this activity you will track chromosomes through meiosis using colored beads.
Experiment 1
Questions
1. Why is crossing over important in heredity? (10 points)
2. Provide two ways that meiosis I and meiosis II are different. (10 points)
3. (10 points)
a. In the lab, how many chromosomes were present in each cell when meiosis I started?
b. How many chromosomes were present in each daughter cell at the end of meiosis II?
4. If humans have 46 chromosomes in each of their body cells, determine how many chromosomes you would expect to find in the following (10 points):
Sperm:
Egg:
Daughter cell from mitosis:
Daughter cell from meiosis II:
5. Why is it necessary to reduce the chromosome number of gametes, but not of other cells of an organism? (10 points)
41
Meiosis
Lab 4
42
Lab 4: Meiosis
43
Introduc on
Meiosis only occurs in organisms that reproduce sexually. The process generates haploid (1n) cells
called gametes (sperm cells in males and egg cells in fe-
males), or spores in some plants, fungi, and pro sts, that
contain one complete set of chromosomes. Haploid cells
fuse together during fer liza on to form a diploid cell with
two copies of each chromosome (2n).
Genes are the units of heredity that have speciÞc loci
(loca ons) on the DNA strand and code for inheritable
traits (such as hair color). Alleles are alterna ve forms of the same gene (brown vs. blue eyes). Homol-
ogous chromosomes contain the same genes as each other but o en di erent alleles. Non-sex cells
(e.g. bone, heart, skin, liver) contain two alleles (2n), one from the sperm and the other from the egg.
Mitosis and meiosis are similar in many ways. Meiosis, however, has two rounds of division—meiosis I
and meiosis II. There is no replica on of the DNA between meiosis I and II. Thus in meiosis, the parent
cell produces four daughter cells, each with just a single set of chromosomes (1n).
Meiosis I is the reduc on division– the homologous pairs of chromosomes are separated so that each
daughter cell will receive just one set of chromosomes. During meiosis II, sister chroma ds are sepa-
rated (as in mitosis).
Concepts to explore:
Meiosis
Diploid cells
Haploid cells
...
genetic linkage and gene mapping
genetic linkage, types, crossing over, types,
gene mapping for diploid and haploid organisms, types
Are hydrophobic interactions weak or strongSolutionHydrophobi.pdf
Are hydrophobic interactions weak or strong?
Solution
Hydrophobic interactions describe the relations between water and hydrophobes (Low-water
soluble molecule). Hydrophobes are non-polar molecules and usually have a long chain of
carbon that do not interact with water molecules.
Strength of hydrophobic bonds are relatively stronger than other weak intermolecular forces (i.e.,
Vander Waals forces and hydrogen bonds). The strength of hydrophobic interactions depends on
several factors like:
1. Temperature: As the temperature increases, the strength of hydrophobic interaction also
increases also. However, at extreme temperatures, hydrophobic interactions denature.
2. Number of carbons on hydrophobes: Molecules with greatest number of carbon atoms will
have the strongest hydrophobic interactions.
3. The shape of hydrophobes: Aliphatic organic molecules will have stronger interactions than
aromatic organic molecules..
What is another word for account based forecast What is anothe.pdf
What is another word for account based forecast?
What is another word for account based forecast?
Solution
The other word for account based forecasting is Budgeting.
A Budget is a quantitative statement in a functional area forecasting the expected values. For
example, a cash budget estimates various cash inflows and cash outflows that can arise during
the controllable period. Thus accounting based forecast uses budgets as the means for estimating
the variables, so it is also called as budgeting..
More Related Content
Similar to Answer the following questions in reference to Drosophila melanogast.pdf
I need help answering both these questionsQuestion 5 1. A dip.pdf
I need help answering both these questions
Question 5: 1. A diploid cell contains three pairs of homologous chromosomes: A1 and A2, B1
and B2, and C1 and C2. Assuming no crossing-over occurs, what possible combinations of
chromosomes will be present in: a) daughter cells following mitosis: b) metaphase I of meiosis:
Solution
1. A)In mitosis the daugther produced are genetically identical , so all the geneticall material is
being inherited from mother cells to daughter cells, So the only possibility is
A1/A2/B1/B2/C1/C2. Since all the genetic material is being transfered from the parent cell to
daughter cells
B) The Meiosis 1 is meant to split the homologus pair into different cells, which will now be n (
Haploid) , So there can be 8 possible combinations of chromosomes, So if we take an example it
could inherit A1/B1/C1 or A2/M1/S1.
Activity 4 Understanding the basics of meiosisMitosis and mei.docx
Activity 4: Understanding the basics of meiosis
Mitosis and meiosis both lead to new daughter cells, but meiosis sets up organisms for sexual
reproduction. Meiosis produces cells (sperm and egg in humans) with only one set of chromosomes so that when fertilization occurs, it results in a new cell with two sets of chromosomes (one from the egg and one from the sperm). This is how there is genetic recombination of DNA resulting in unique individuals.
Before we talk about what can go wrong in meiosis, you need to be clear about the process itself and how genetic recombination in particular occurs. To this end, complete the following set of questions.
Questions
1. In the following diagram, draw what a cell with 1 chromosome would look like in the stages of meiosis. Prophase 1 is filled in for you and includes replicated homologous chromosomes; the black replicated chromosome is from the reproducing male’s mom and the grey replicated chromosome is from dad.
2. In your diagram from question 1:
a. How many tetrads are formed? _______________
b. How many chromosomes are in the sperm? _______________
3. In what stage(s) of meiosis:
a. Are tetrads formed? ________________________________
b. Does crossing over occur? ________________________________
c. Do the chromosomes move to the poles? ________________________________
d. Do replicated chromosomes separate? ________________________________
e. Does the cytoplasm divide? ________________________________
4. What processes in meiosis result in genetically unique daughter cells? When do these processes occur? (Note: There are two main processes; discuss both).
5. Compare and contrast meiosis with mitosis to complete the following table.
Table 2. Comparison of key characteristics between meiosis and mitosis.
Characteristics
Mitosis
Meiosis
Type of organisms it occurs in
# of chromosomes in human parent cell
Number of times chromosomes replicate
Number of cell divisions
Crossing over occurs? (Y/N)
Type of daughter cells produced
Number of daughter cells produced
Daughter cells identical to parent cell? (Y/N)
Daughter cells are: 1n or 2n?
# of chromosomes in human daughter cells
Activity 3: Chromosomes in mitosis and meiosis
In the nucleus of the cell are the chromosomes that are composed of the hereditary material DNA. In every somatic (body) cell of a human there are 46 chromosomes. Each species may have a different number of chromosomes than another species.
Since each somatic cell of an organism contains the same number of chromosomes, there must be a duplication of material before the nucleus divides during mitosis. In each somatic cell, there are two sets of chromosomes; this is referred to as the 2n (diploid) number, in which n means number of chromosomes. In humans, 2n = 46 chromosomes.
In each gamete (sex) cell, there is only one set of chromosomes; this is referred to as the 1n (haploid) number. In humans, 1n = 23 chromosomes. This means the.
Cytotaxonomy plant taxonomy
This ppt contains all about the various chromosomal properties that can be used in studying plant taxonomy and classification of plants.
Given the information regarding mitosis and meiosis, clearly and con.pdf
Given the information regarding mitosis and meiosis, clearly and convincingly explain in your
own words how an error in MITOSIS could possibly lead to a gamete containing new genetic
information and thereby giving rise to a new phenotype. Use detail be comprehensive. (7 points)
Solution
The process of mitosis is divided into two phases: Interphase and M-phase.
Interphase is further divided into G1, S and G2 phase. In G1, protein synthesis occurs for S-
phase, in S phase, DNA replicates and in G2 phase, protein synthesis occurs for M-phase to start
properly. Now, during S phase, the ‘2n’ nucleus converts into 4n.
So, the cell entering M-phase is already ‘4n’.
M-phase is the actual mitotic phase, in which the cell divides into two. It contains four phases,
prophase, metaphase, anaphase and telophase. Your question is related with these four phases. In
prophase, the entire DNA converts into chromosomes; the process is known as condensation of
DNA. In metaphase, these chromosomes align on the equator. Now, if alignment at equator is
not correct, then anaphase cannot be started due to metaphase check point.
In anaphase, the aligned chromosomes move towards the opposite poles. This phase is the main
phase, related with formation of gametes. The cell is already ‘4n’. ‘2n’ chromosomes move
towards one end, and the rest of ‘2n’ move towards the other end. Now, suppose an error occurs
in mitosis, and ‘3n’ chromosomes move towards one end, and 1n goes towards the other cell;
then two gametes, ‘3n’ and n will be formed.
This ‘3n’ and ‘n’ gametes are gametes having different genetic information in comparison to
their ‘2n’ counterparts.
Interestingly, Meiosis II will result in gamete formation is mitotic division (equational division).
Haploid cells with \'2c\' amount of DNA leave Meiosis I. Now, in anaphase, suppose one
chromosome\' 2c DNA migrates to one end (due to error) and no DNA migrates to the other end;
then two gametes will be formed; one having all normal chromosomes + one double
chromosome with 2c DNA, and the other having no DNA.
Please note: normally gametes are 1n (haploid) and 1c.
Suppose this gamete with double DNA (2c) fertilizes with a gamete with half DNA (1c), then the
resulting cell will have “all normal chromosomes + one chromosome with ‘3c’ amount of DNA”.
This condition if occurring with chromosome 21 is known as Down\' syndrome.
It is important to note that errors in Mitosis result in non-disjunction; which further results in
many Mendelian disorders; as well as ploidy defects.
** Hope it helps..
Linkage
Linkage refers to the presence of two different genes on the same chromosome . Two genes that occur on the same chromosome are said to be linked, and those that occur very close together are tightly linked.
The homolog of the #3 chromosome is the X-chromosome Y-chromosome .pdf
The homolog of the #3 chromosome is the X-chromosome Y-chromosome other #3
chromosome chromosome that synapses with the #3 chromosome during mitosis
Solution
Ans. Each diploid organism inherits 2 copies of a chromosome (here, chromosome number 3 or
#3) – one from its mother and another from father.
That is, a human has two copies of #3 chromosomes- inherited one copy from each parent. For
simplicity of clarification, we represent mater chromosome as #3M and fraternal chromosome as
#3F
Now, note the following events carefully –
I. When a cell prepares for meiosis, it duplicates it chromosomes. That is now there are 2 copies
of #3M (one old, one new copy) and 2 copies of #3F (one old, one new copy) chromosomes, too.
II. Both the copies of #3M remain joined together with the help of cohesion rings. Both the
chromosomes, old #3M and new #3M are called sister chromatids of each other.
III. So, after S-phase, there are two pairs of sister chromatids – one #3M pair (has two #3 sister
chromatids joined together by cohesin rings) and one #3F pair.
IV. In Leptotene stage prophase I of meiosis 1, for the first time in meiotic cell cycle, the term
“homologous chromosomes” is used. Here, the term mean “. A pair of genetically similar
chromosomes having same genes on same loci are called homologous chromosomes or
homologues”. It means chromosomes #3M and #3F are homologues chromosomes because they
share the same gene loci – IRRESPECTIVE of whether synapsis has occurred or not.
V. Note that synapsis occurs in zygotene stage, (after leptotene stage in sequence).
VI. No doubt, the homologous chromosomes undergo synapsis during meiosis, but #3M and #3F
are called homologous chromosomes even if they don’t synapse (say, for a hypothetical
condition) because they same genes on same loci.
Therefore, the term “synapsis” is not required to mention two chromosomes as homologous
chromosomes- it is the character of “sharing same gene loci” that makes two chromosomes
homologous.
Thus, correct option is- C. other #3 chromosome.
Note: Different texts may discriminate the term differently. It the best possible explanations
presented here..
Lab TemplateWeek 4 MeiosisSubmitted by your name here.docx
Lab Template
Week 4: Meiosis
Submitted by: <your name here>
As you complete the lab, record your answers in this template. Save the document as LastName_FirstName_BIO1020_W4A3, and submit it to the Dropbox. Full lab instructions and the rubric with which you will be evaluated can be found in the online classroom.
Activity
Cellular reproduction in Eukaryotes involves either mitosis or, in the case of sex cells, meiosis. Mitosis involves the reproduction of a cell into two identical daughter cells. Meiosis, however, is a reduction division where a parental diploid cell produces four haploid gametes. Upon fusion, two haploid gametes (in humans the sperm and the egg) will result in one diploid zygote. In this activity you will track chromosomes through meiosis using colored beads.
Experiment 1
Questions
1. Why is crossing over important in heredity? (10 points)
2. Provide two ways that meiosis I and meiosis II are different. (10 points)
3. (10 points)
a. In the lab, how many chromosomes were present in each cell when meiosis I started?
b. How many chromosomes were present in each daughter cell at the end of meiosis II?
4. If humans have 46 chromosomes in each of their body cells, determine how many chromosomes you would expect to find in the following (10 points):
Sperm:
Egg:
Daughter cell from mitosis:
Daughter cell from meiosis II:
5. Why is it necessary to reduce the chromosome number of gametes, but not of other cells of an organism? (10 points)
41
Meiosis
Lab 4
42
Lab 4: Meiosis
43
Introduc on
Meiosis only occurs in organisms that reproduce sexually. The process generates haploid (1n) cells
called gametes (sperm cells in males and egg cells in fe-
males), or spores in some plants, fungi, and pro sts, that
contain one complete set of chromosomes. Haploid cells
fuse together during fer liza on to form a diploid cell with
two copies of each chromosome (2n).
Genes are the units of heredity that have speciÞc loci
(loca ons) on the DNA strand and code for inheritable
traits (such as hair color). Alleles are alterna ve forms of the same gene (brown vs. blue eyes). Homol-
ogous chromosomes contain the same genes as each other but o en di erent alleles. Non-sex cells
(e.g. bone, heart, skin, liver) contain two alleles (2n), one from the sperm and the other from the egg.
Mitosis and meiosis are similar in many ways. Meiosis, however, has two rounds of division—meiosis I
and meiosis II. There is no replica on of the DNA between meiosis I and II. Thus in meiosis, the parent
cell produces four daughter cells, each with just a single set of chromosomes (1n).
Meiosis I is the reduc on division– the homologous pairs of chromosomes are separated so that each
daughter cell will receive just one set of chromosomes. During meiosis II, sister chroma ds are sepa-
rated (as in mitosis).
Concepts to explore:
Meiosis
Diploid cells
Haploid cells
...
genetic linkage and gene mapping
genetic linkage, types, crossing over, types,
gene mapping for diploid and haploid organisms, types
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AP Biology Ch 12 gene linkage groups and chromosome maps
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Given the information regarding mitosis and meiosis, clearly and con.pdf
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Lab TemplateWeek 4 MeiosisSubmitted by your name here.docx
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Are hydrophobic interactions weak or strongSolutionHydrophobi.pdf
Are hydrophobic interactions weak or strong?
Solution
Hydrophobic interactions describe the relations between water and hydrophobes (Low-water
soluble molecule). Hydrophobes are non-polar molecules and usually have a long chain of
carbon that do not interact with water molecules.
Strength of hydrophobic bonds are relatively stronger than other weak intermolecular forces (i.e.,
Vander Waals forces and hydrogen bonds). The strength of hydrophobic interactions depends on
several factors like:
1. Temperature: As the temperature increases, the strength of hydrophobic interaction also
increases also. However, at extreme temperatures, hydrophobic interactions denature.
2. Number of carbons on hydrophobes: Molecules with greatest number of carbon atoms will
have the strongest hydrophobic interactions.
3. The shape of hydrophobes: Aliphatic organic molecules will have stronger interactions than
aromatic organic molecules..
What is another word for account based forecast What is anothe.pdf
What is another word for account based forecast?
What is another word for account based forecast?
Solution
The other word for account based forecasting is Budgeting.
A Budget is a quantitative statement in a functional area forecasting the expected values. For
example, a cash budget estimates various cash inflows and cash outflows that can arise during
the controllable period. Thus accounting based forecast uses budgets as the means for estimating
the variables, so it is also called as budgeting..
What function can be used to return multiple numeric data such as in.pdf
What function can be used to return multiple numeric data such as integer and/or float entered at
the keyboard to the program?
A) Input
B)Eval
C)Output
D)return
Solution
Answer: Input
Input function can be used to return multiple numeric data such as integer and/or float entered at
the keyboard to the program. Input function will get the inputs from keyboard and seent it to
program for execution..
There are five feasible solutions to a three-objective optimization p.pdf
There are five feasible solutions to a three-objective optimization problem (solns A - E shown
below with their objective function values). Indicate whether each solution is dominated or non-
dominated. If a solution is dominated, indicate which other solution(s) dominate(s) it.
Solution
soln Dom or non-dom Dominated by
A Dominated E
B Non-dominated
C Non-dominated
D Dominated B
E Non-dominated.
An air-filled parallel-plate capacitor has capacitance Co. If two ide.pdf
An air-filled parallel-plate capacitor has capacitance Co. If two identically sized dielectric slabs
of dielectric constants K_1 and K_2 are inserted as shown in the following figure, what is the
new capacitance? (Hnt: treat this as two capacitor in combinations)
Solution
it would act as capacitors in parallel.
C= K0A/d.
Non C1= A/(d/2)*K1=2C/k1
Simlarly C2= 2C/K2
Since it can be treated as capacitor is parallel so Cfinal= 2C/K1 + 2c/k2 = 2C(K1+K2/K1*K2).
The incandescent lightbulb was actually invented by Humphry Davy in 1.pdf
The incandescent lightbulb was actually invented by Humphry Davy in 1802. It was not
commercializable in that form. Joseph Swan was successful in making a commercializable unit
in the late 1870s. He may have been granted the first light bulb patent (British patent 4933,
issued in 1880). Thomas Edison was also developing electric lighting at that time. He applied for
US patents as early as 1878 So why is Edison credited as the inventor? Partly because he was the
first to design and manufacture a robust lighting system having a bulb with an effective
incandescent material, higher vacuum inside the bulb, and high resistance so power distribution
from a central source was possible. Perhaps more importantly was the development ofsupply
sources and materials for distribution networks. A modem incandescent lightbulb is shown
below. Overall efficiency, defined as light power out divided by electrical power in, is about
2.5% for a 100 W unit. 1 1. outline of glass bulb 2. Low pressure inert gas (argon, nitrogen,
krypton, xenon) 3. Tun ten filament 5) 4. Contact wire (goes out of stem) 5. Contact wire (goes
into stem) 6. Support wires (one end embedded in stem; conduct no current) 7. Stem (glass
mount) 8. Contact wire (goes out of Identify five different systems that could be used for an
energy analysis (some combination Draw arrows for each systems\' energy flow process(es)-W
Qooeduction, Qconvection, radiation.
Solution
To conduction and convection of heat must have the medium.
in vaccum only radiation is there.
given problem the heat coming out from the tungsten filament(3).
the work supplied by the electrical current through the wires.
first humphry davy invented incandescent bulb having vaccum inside so there is no possibilty of
heat transfer through conduction or convection.
next edison filled with the low pressure inert gases. beacuse of that from hot Tungsten filamet to
inert gases heat transfer occurs through convection.
next inertgasses to glass or stem heated by the convection (fluid and solid interaction)
again glass to sourroundings is convection(solid and fluid interaction)
radiation is always there beacuse it possible with medium and with out medium
contact wires are heated up through the conduction..
The fossil record provides little information about ancient mosses. D.pdf
The fossil record provides little information about ancient mosses. Do you think nonvascular
plants could have ever been as large as trees or shrubs? Provide evidence based on your
knowledge of nonvascular plant anatomy.
Solution
2). No, the nonvascular plants could have never been as large as trees or shrubs because they
cannot provide the water and nutrients to the plant parts that are away from the source of water.
The nonvascular plants cannot control their water supply internally, it is solely depends upon
their external environment that provides water to the plant. So, they most likely grown over a
lake or rocks to get the water and propagated vegetative (clonal moss)..
The DNA sequence below is a shortened version of the sequence on the .pdf
The DNA sequence below is a shortened version of the sequence on the previous page. This
DNA sequence corresponds to the mRNA sequence only; two introns are not included. The
numbers indicate the position of nucleotides within the sequence and indicates intervening
nucleotides that are not shown. The gene is 626 nucleotides long. Recall that chromosomes are
very long DNA molecules with millions of nucleotides, so this sequence is preceded and
followed by seemingly endless sequences of nucleotides. IDNA is double-stranded. Write the
sequence of the first 10 nucleotides of the complementary strand of DNA.
Solution
ans. DNA is a double stranded which consist of two strands and one strand is complemetary to
the other.DNA contains complemetary base pairs in which adenine is linke d to tymine by two
hydrogen bonds guanine to cytosine by three hydrogen bonds
The seque of complementary strand of the DNA molecule is
5\' TGTAAACGAA.
the conducting tissue found in vascular plants that functions in tra.pdf
the conducting tissue found in vascular plants that functions in transportation of materials
throughout the plant, but does not lend structure support to the stem is
A. xylem
B. phloem
C.sporopollenin
D.endosperm
e. LYCOPHYLL
Solution
Answer : B. Phloem
Phloem is responsible for the transporting sugars,proteins, and other organic molecules in plants.
xylem is responsible for the transport of water and water soluble nutrients,taken up by the roots
and also primary component of wood. Vascular plants are able to higher than other plants due to
the rigidity of xylem cells, which support the plant..
Show that f(x)=57^x+1 and f^-1(x)=7x-75 are inverses Choose the ap.pdf
Show that f(x)=5/7^x+1 and f^-1(x)=7x-7/5 are inverses Choose the appropriate equality that
will show that f(x) and f^-1(x) are inverses (f.f^-1)(x)= (f^-1.f)(x)=1 (f.f^-1)(x)= (f^-1.f)(x)=x
(f.f^-1)(x)= (f^-1.f)(x)=1 (f.f^-1)(x)= (f^-1.f)(x)=x
Solution
Replace x by f^-1(x) in f(x)
=> 5/7(7x-7)/5+1
=> (7x - 7)/7 + 1 = > x - 1 + 1 => x
=> f(f^-1(x)) = x.
Read the short article Horizontal Gene Transfer in E Coli and answer.pdf
Read the short article Horizontal Gene Transfer in E Coli and answer the following
questions.(https://www.sciencedaily.com/releases/2015/05/150519105900.htm) Remember to
use your own words; plagiarism of any kind will not be tolerated (see syllabus for further
information on what constitutes plagiarism).
(1) What is horizontal gene transfer (HGT)?
(2) In the posted article, how many gene transfer elements were found for E. coli O104:H4?
(3) What is the source of these gene transfer elements?
(4) Should we be concerned about HGT in prokaryotes? If so, why?
Solution
(1) Horizontal gene transfer (HGT) is the lateral gene movement, wherein the movement of
genetic material takes place between unicellular or multicellular organisms across species. This
gene transfer is different from the vertical gene transfer in which the transmission of DNA
occurs from parent to offspring. Often genetic material swaping occurs between bacteria of same
as well as different species, through this means by aquiring one of the processes; transformation,
transduction or conjugation. During HGT, plasmids, transposons or integrons act as vectors that
transfer genes.
(2) In the given article, the team from Madurai Kamaraj University in Madurai, Tamil Nadu,
identified 38 gene transfer elements for E. coli O104:H4.
(3) The research team identified 38 gene transfer prophage elements for E.coli O104:H4. They
showed that the these elements act as genetic weapons protecting the bacteria from antibiotics.
The source of these elements were viruses (the bacteriophages), that infect bacteria, and in the
process bacteria aquired these elements from viruses.
(4) In prokaryotes, horizontal gene transfer is a major concern. Aquisition of new genes through
HGT is ability of Bacteria and Archaea to adapt to new changing environments. Druing HGT,
genes that are responsible for antibiotic resistance in one species of bacteria can be transferred to
another species of bacteria, and is the primary reason for the spread of antibiotic resistance.
Besides antibiotic resistance, HGT also gives bacteria an ability to degrade novel compounds
(such as different insecticides and pesticides). Thus the bacterial strains after HGT are much
more dangerous as they are multi drug resistant and more virulent. These transfer elements when
transferred to a pathogenic bacteia, can become a serious health hazard as they are resistant to
antibiotics available with us and thus cannot be trated easily..
Please select the best answer and click submit. The owners of fou.pdf
Please select the best answer and click \"submit.\" The owners of four companies competing for
a contract are shown in the table below. If a report is released that advocates company A. which
of the people having funded the report should result in the most skepticism?
Solution
Option d.
Debby Smith would result in the most skepticism because his company is in the second place in
the list..
One function of the spliceosome is to (Select all that apply.)A. .pdf
One function of the spliceosome is to: (Select all that apply.)
A. recognize the beginning and end of the introns.
B. recognize the 5\' end of the primary transcript.
C recognize the beginning and end of the exons.
D recognize the 3\' end of the primary transcript.
Solution
The correct answer is :A. recognize the beginning and end of the introns.
Reason: Spliceosome consists of snRNAs and snRNPs which together make a complex.This
complex binds to the beginning and end of the introns and forms a loop which is later excised
and the ends of exons are sealed..
Match the following organelles with their key functions in the cell .pdf
Match the following organelles with their key functions in the cell nucleus rough endoplasmic
reticulum Golgi apparatus lysosomes mitochondria chloroplasts peroxisomes cytosol carbon
fixation site of protein synthesis ready for cellular distribution recycling of intracellular
materials modification and packing of proteins for secreation site of protein synthesis that
contains the majority of metabolic pathways containment of toxic reactions the site of the core
genome oxidative phosphorylation
Solution
1: Nucleus: The site of core genome;g
2: Rough ER: site of protein synthesis ready for cellular distribution; b
3: Golgi apparatus: Modification and packing of proteins for secretion; d
4: Lysosome: recylcing of intracellular material; c
5: Mitochondria: Oxidative phosphorylation;h
6: Chloroplast: Carbon fixation; a
7: Peroxisomes: Containment of toxic reaction; f
8: Cytosol: Site of protein synthesis that contains majority of metabolic pathways; e.
please choose the correct answerIn the ABO blood-type phenotype, w.pdf
please choose the correct answer
In the ABO blood-type phenotype, which gene is the most epistatic?
a.) A
b.) B
c.) O
d.)I
e.) H
Solution
Answer is e) H- Epistasis is a product of gene- gene interaction. when one genes expression is
controlled by other gene, it is called epistasis. H antigen is the precursor of A and B antigens.
mo Press Submit until all questions are colored green and you have do.pdf
mo Press Submit until all questions are colored green and you have double c Question 1 Convert
the following degree measure to radians: 210 14 21 20 12 7 d) f ONone of the above.
Solution
We know that,
Radians = Degree*(pi/180)
Given Degree measures = 210o
Using the formunla
Radian Measures = 210(pi/180) = 7(pi/6) Radians
Hence (d) is the correct answer..
matching chose the stage of the light reactions which the statement.pdf
matching: chose the stage of the light reactions which the statement occurs in.
1. _Produces ATP.
2. _Passes energized electrons down an electron transport chain to create ATP.
3. _Produces NADPH.
4. _Splits water to get electrons to replace energized electrons.
5. _Replaces energized electrons from electrons that have completed the 1st electron transport
chain.
a. water-splitting photosystem.
b. NADPH-Producing photosystem.
Solution
During water splitting photosystem ,the electrons in water are removed and passed to the
chlorophyll. The hydrogen ions extracted from the water molecule, and after two water
molecules are splitted,the oxygen atoms takes electron and form molecular oxygen (O 2 ).
In photosynthesis, water splitting donates electrons to the electron transport chain in
photosystem II.
So it splits water to get electrons to replace energized electrons.
In the photosystem II (PSII) reaction center, energy from sunlight is used to extract electrons
from water. The electrons travel via the chloroplast electron transport chain to photosystem I
(PSI), which reduces NADP+ to NADPH.
So it produces NADPH.
Let E be the set of all even integers, and let O be the set of all o.pdf
Let E be the set of all even integers, and let O be the set of all odd integers. Explain why Z E O.
A. Every integer is both even and odd.
B. Some integers are both even and odd.
C. Even integers and odd integers are integers.
D. Every integer is either even or odd.
E. No integer is both even and odd.
Solution
ans : D
take any integer z, z is either odd or even so, for all z , Z is a subset of EUO.
JavaIm not sure how to implement this code can someone help me .pdf
Java
I\'m not sure how to implement this code can someone help me / guide me? thank you
(Updated with PrefixMap interface included at the bottom of the code).
import java.util.List;
import java.util.Map;
public class Storage implements PrefixMap {
private Map keys;
private Map> prefixes;
public Storage() {
// TODO Initialise your maps
}
/*
* How many keys are stored in the map
*/
@Override
public int size() {
// TODO Implement this, then remove this comment
return 0;
}
@Override
public boolean isEmpty() {
// TODO Implement this, then remove this comment
return false;
}
/*
* Get the value corresponding to the key (or null if the key is not found)
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key is null, throw IllegalArgumentException
*/
@Override
public String get(String key) {
// TODO Implement this, then remove this comment
return null;
}
/*
* Insert the value into the data structure, using the given key. If the key
* already existed, replace and return the old value (otherwise return null)
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key or value is null, throw IllegalArgumentException
*/
@Override
public String put(String key, String value) {
// TODO Implement this, then remove this comment
return null;
}
/*
* Remove the value corresponding to the given key from the data structure,
* if it exists. Return the old value, or null if no value was found.
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key is null, throw IllegalArgumentException
*/
@Override
public String remove(String key) {
// TODO Implement this, then remove this comment
return null;
}
/*
* return the number of keys which start with the given prefix if the prefix
* contains any character other than A,C,G,T, throw MalformedKeyException
* if the prefix is null, throw IllegalArgumentException
*/
@Override
public int countKeysMatchingPrefix(String prefix) {
// TODO Implement this, then remove this comment
return 0;
}
/*
* return the collection of keys which start with the given prefix if the
* prefix contains any character other than A,C,G,T, throw MalformedKeyException
* if the prefix is null, throw IllegalArgumentException
*/
@Override
public List getKeysMatchingPrefix(String prefix) {
// TODO Implement this, then remove this comment
return null;
}
/*
* Return the number of unique prefixes
* e.g. if the tree stores keys GAT, GATTC, GATTACA, this method will return 8
* because the possible prefixes are G, GA, GAT, GATT, GATTC, GATTA, GATTAC,
GATTACA
*/
@Override
public int countPrefixes() {
// TODO Implement this, then remove this comment
return 0;
}
/*
* Return the sum of the lengths of all keys
* e.g. if the tree stores keys GAT, GATTC, GATTACA, this method will return 15
*/
@Override
public int sumKeyLengths() {
// TODO Implement this, then remove this comment
return 0;
}
}
import java.util.List;
public interface PrefixMap {
public boole.
Indexes prevent the database engine from having to scan every record .pdf
Indexes prevent the database engine from having to scan every record in the database. True
False
Solution
Answer: True
Indexes prevent the database engine from having to scan every record in the database.
Indexes provides the improvement of query performance by providing the database engine with
fast way of locating a particular record based on the a column or columns provided in where
condition..
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please choose the correct answerIn the ABO blood-type phenotype, w.pdf
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TESDA TM1 REVIEWER FOR NATIONAL ASSESSMENT WRITTEN AND ORAL QUESTIONS WITH A...
TESDA TM1 REVIEWER FOR NATIONAL ASSESSMENT WRITTEN AND ORAL QUESTIONS WITH A...
Answer the following questions in reference to Drosophila melanogast.pdf
- 1. Answer the following questions in reference to Drosophila melanogaster (fruit fly) that has 6 chromosomes. How many combinations are possible when chromosomes assort independently into gametes (ignoring crossing over)? 6 raised to the 2nd power 2 raised to the 6th power 6 x 2 = 12 2 raised to the 3rd power 3 raised to the 2nd power How many tetrads align at the metaphase plate of meiosis I? 0 12 6 3 Solution Ans.1 No. of possible combination during independent assortment = 2n Where, 2 indicates ploidy of the somatic cells undergoing meiosis, n = number of haploid chromosomes. Given, 2n = 6 Hence, n = 3. Thus, No. of possible combination during independent assortment = 2n = 23 Correct option: 2 raised to 3rd power. Ans. 2. Number of tetrad during meiosis = Number of haploid chromosomes (n) To understand it, let’s number the three haploid chromosomes as A, B and C. A diploid cell has two copies of each chromosome- one inherited from mother and the other from father. So, diploid cell has following chromosomes- A1, A2, B1, B2, C1 and C3, where 1= maternal chromosomes and 2 = fraternal chromosomes. All the chromosomes remain independent of each other, randomly scattered in nucleus. We will proceed with one chromosome pair, for simplicity. The following events occur when a diploid cell enters meiosis- A1 and A2 are called homologues pair. During S-phase, DNA is duplicated. The resultant-
- 2. I. A1A1 remain joined to each other. Are called sister chromatids. The same happens to A2A2, and rest other chromosomes. II. Homologous chromosome A1 and A2 pair together prophase I. Since each of A1 and A2 are duplicated, they are (A1A1) and (A2A2) respectively. III. Homologous pairing forms a tetrad – A1A1/A2A2 i.e. four chromosomes remaining together. Note: Both the maternal and fraternal chromosomes and their duplicated copies remain together as a tetrad. That is, the number of tetrad in half of diploid number (because A1 from mother and A2 from father remain together as one single homologous chromosome). Thus, number of treads in drosophila = n = 3 Correct option: D (3)