Goal to help you link cell division to Mendelian genetics; show wit.pdf
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Goal: to help you link cell division to Mendelian genetics; show with diagrams why/how Mendel’s dihybrid cross produces a ratio of 9:3:3:1 among F2 progeny. The genes for Mendel’s characters seed shape and seed color, are on different chromosomes. Each gene has two alleles. Starting, as Mendel did, with “true breeding” parents, show in writing and as part of a diagram, what their genotype is, the genotype of their gametes, the genotype of their progeny (the F1), the genotypes of their progeny’s gametes, and the genotypes of the F2. Include statements about the proportion of each kind of the genotypes you identify. To minimize the number of figures you should draw, I suggest that you illustrate the following stages: Parents genotypes (as I did in class, use one color for pollen donor (male function), another for plant fertilized by the pollen, different chromosomes different sizes/shapes) Anaphase I of meiosis in parents Anaphase II of meiosis in parents Gametes produced by meiosis in parents and their proportions. F1 genotypes Anaphase I of meiosis in F1 Anaphase II of meiosis in F2 Gametes produced by meiosis in F2 and their proportions. F2 Genotypes and their proportions; do this with letters only, not pictures. Solution To start with we have homozygotic parents : First parent: Round color seed (RR) and Yellow in color (YY) = Genotype : RRYY : Gametes RY and RY Second parent : wrinkled seed (rr) and green in colour (yy) = Genotype : rryy : Gametes ry and ry first offspring F1 = round and yellow RrYy Gamtes of F1 = RY, Ry, rY, ry Now crossing two individuals from F1 to obtain F2 generation now to see the proportion of all the phenotypes: Total number of F2 offspring Round and yellow = (RRYY) or (RrYy) or (RRYy) or (RrYY) = 9 Round and green = (RRyy) or (Rryy) = 3 Wrinkled and yellow = (rrYY) or (RRYy) = 3 Wrinkled and green = (rryy) = 1 Therefore the ratio is 9:3:3:1 in the F2 progeny.1st parent (RY)1st parent (RY)2nd parent (ry)RrYyRrYy2nd parent (ry)RrYyRrYy.
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Biology 103 Laboratory Exercise – Genetic Problems .docx
Biology 103 Laboratory Exercise – Genetic Problems
Introduction
Although the science of genetics has become a highly sophisticated discipline dealing
with the interactions of hereditary factors at the molecular level, it has its roots in the
basic laws of heredity initially discovered and presented by Gregor Mendel more than
one hundred years ago. Mendel's success in discovering these laws was due largely to his
application of the simple rules of mathematical probability - the laws of chance - to his
observations concerning the inheritance of certain characteristics in the garden pea plant.
Reginald Punnett and the Punnett Square
The Punnett square is a diagram used by biologists to determine genotypic probability
within the offspring from a particular genetic cross. The Punnett square shows every
possible genotypic combination of maternal alleles with the paternal alleles for a genetic
cross. Punnett squares only give probabilities for genotypes, not phenotypes. The square
diagram was designed by the British geneticist, Reginald Punnett (1865-1967) and first
presented to the science community in 1905. Punnett’s Mendelism (1905) is considered
the first popular science book to introduce genetics to the public.
Solving Genetic Problems
R
R'
R
RR RR'
R'
RR' R'R'
Maternal alleles
A
A
a
Aa
Aa
Paternal
Alleles
a
Aa
Aa
The first step in solving a genetic problem is to establish the genetic symbols you will use
in your problem solution. Stay consistent by using these same symbols throughout the
problem solving process.
Represent dominant and recessive alleles (different forms of a gene) using traditional
genetic symbols. Dominant alleles should be represented with the capital version of an
alphabetic letter while using the lower case version to show recessiveness. For example:
B = black color, b = white color.
Each individual gene or trait is diploid (2n) in nature and therefore, must be represented
with two alleles. Continuing with the alleles mentioned previously, an individual may
have the genetic makeup BB, Bb, or bb when using those alleles.
Remember that gametes (sperm and egg) are haploid (n) and can only provide one allele
per trait. For example: B or b
An individual’s genotype contains the possible gametes that can be expected to be
produced by that individual. Much of genetics revolves around the probability of the
makeup of gametes. If the individual is homozygous, all of the gametes produced will
possess the same kind of allele. For example, an individual with the genotype BB would
be expected to produce only B gametes and individuals with genotype bb would produce
only b gametes.
If the individual is heterozygous, that is the individual’s genotype contains one dominant
allele and one recessive allele (Bb), the gametes produced will possess one or the other of
the two forms of the gene – B or b. ...
Interaction of genes complimentary, supplementary and epistasis
Interaction of genes- Complimentary, Supplementary and Epistasis
Recommended
Biology 103 Laboratory Exercise – Genetic Problems .docx
Biology 103 Laboratory Exercise – Genetic Problems
Introduction
Although the science of genetics has become a highly sophisticated discipline dealing
with the interactions of hereditary factors at the molecular level, it has its roots in the
basic laws of heredity initially discovered and presented by Gregor Mendel more than
one hundred years ago. Mendel's success in discovering these laws was due largely to his
application of the simple rules of mathematical probability - the laws of chance - to his
observations concerning the inheritance of certain characteristics in the garden pea plant.
Reginald Punnett and the Punnett Square
The Punnett square is a diagram used by biologists to determine genotypic probability
within the offspring from a particular genetic cross. The Punnett square shows every
possible genotypic combination of maternal alleles with the paternal alleles for a genetic
cross. Punnett squares only give probabilities for genotypes, not phenotypes. The square
diagram was designed by the British geneticist, Reginald Punnett (1865-1967) and first
presented to the science community in 1905. Punnett’s Mendelism (1905) is considered
the first popular science book to introduce genetics to the public.
Solving Genetic Problems
R
R'
R
RR RR'
R'
RR' R'R'
Maternal alleles
A
A
a
Aa
Aa
Paternal
Alleles
a
Aa
Aa
The first step in solving a genetic problem is to establish the genetic symbols you will use
in your problem solution. Stay consistent by using these same symbols throughout the
problem solving process.
Represent dominant and recessive alleles (different forms of a gene) using traditional
genetic symbols. Dominant alleles should be represented with the capital version of an
alphabetic letter while using the lower case version to show recessiveness. For example:
B = black color, b = white color.
Each individual gene or trait is diploid (2n) in nature and therefore, must be represented
with two alleles. Continuing with the alleles mentioned previously, an individual may
have the genetic makeup BB, Bb, or bb when using those alleles.
Remember that gametes (sperm and egg) are haploid (n) and can only provide one allele
per trait. For example: B or b
An individual’s genotype contains the possible gametes that can be expected to be
produced by that individual. Much of genetics revolves around the probability of the
makeup of gametes. If the individual is homozygous, all of the gametes produced will
possess the same kind of allele. For example, an individual with the genotype BB would
be expected to produce only B gametes and individuals with genotype bb would produce
only b gametes.
If the individual is heterozygous, that is the individual’s genotype contains one dominant
allele and one recessive allele (Bb), the gametes produced will possess one or the other of
the two forms of the gene – B or b. ...
Interaction of genes complimentary, supplementary and epistasis
Interaction of genes- Complimentary, Supplementary and Epistasis
Genetics stem-12
not so easy but understandable INTRA INTER ALLELIC INTERACTION (NON MENDELIAN GENETICS)
51 Mendelian Gene cs Lab 5 52 La.docx
51
Mendelian Gene cs
Lab 5
52
Lab 5: Mendelian Gene cs
53
Introduc on
In 1866, Gregor Mendel, an Austrian Monk, published a paper en tled “Experiments in plant hybridiza-
on”. It went largely unno ced un l 1900 when it was rediscovered and subsequently became the
basis for what we now refer to as Mendelian Gene cs.
Mendel was the Þrst to recognize:
Inherited characters are determined by speciÞc factors (now recognized these as genes).
These factors occur in pairs (genes).
When both alleles of a gene are the same they are said to be homozygous, while if they are di erent
they are said to be heterozygous. When gametes form, these factors segregate so that each gamete
contains only one allele for each gene. Remember, alleles reside on the chromosomes that are divid-
ing. These original observa ons lead to what we now refer to as The law of segrega on and the law of
independent assortment.
Figure 1: Law of Segrega on
Concepts to explore:
Gregor Mendel
Law of segrega on
Homozygous
Heterozygous
Independent assortment
Dominant vs. recessive
Incomplete dominance
Co-dominance
Genotype
Phenotype
Monohybrid cross
Dihybrid cross
Punne square
Lab 5: Mendelian Gene cs
54
The law of segrega on states that during
meiosis, homologous (paired) chromo-
somes split (Figure 1). The law of inde-
pendent assortment states that during
meiosis, each homologous chromosome
has an equal chance of ending up in ei-
ther gamete, and alleles for individual
genes segregate with the chromosomes
on which they are located (Figure 2).
Using corn as an example (Figure 2):
The large chromosome has the gene for kernel color (Y = yellow, y = blue).
The small chromosome has the gene for kernel texture (S = smooth (green); s = wrinkled (red)).
When a dominant allele is present, that characteris c is expressed, regardless of the second allele. In
this case both the Yy and YY o spring will be yellow.
A recessive allele is only expressed when both alleles are recessive. In this case only the yy combina-
on is blue. The dominant allele is always represented by capital le ers, while the recessive is repre-
sented by lower case le ers.
Genotype refers to the combina on of alleles for a par cular trait. Phenotype refers to the appear-
ance of that combina on of alleles. In our example, the genotype of the diploid cell is Yy, Ss, while the
phenotype is Yellow and Smooth.
Figure 2: Law of Independent Assortment
Figure 3: Monohybrid Cross
Punne Square F1
Lab 5: Mendelian Gene cs
55
Alleles can exhibit incomplete dominance and co-dominance. An example of incomplete dominance is
the cross of two plants, one with red ßowers and one with white, whose o spring have pink ßowers.
In the case of codominance, the same cross would result in red and white striped ßowers.
If we know the genotype of two parents we can.
Lab ReportName _______________________ Section .docx
Lab Report Name: _______________________
Section: _____________________
4
Plant Genetics
Exercise 1 - The Punnett Square
Write a hypothesis describing the phenotype of the germinating tobacco seeds. Include the approximate number of seedlings and percentages that will have a dominant trait and recessive trait out 50 seeds.
Create a Punnett square using Data Table 1, showing the possible F2 generation outcomes from a cross between a heterozygous male and a heterozygous female for the germinating tobacco seeds. The dominant allele is written as “G” for green and the recessive allele for albino/yellow color is written as “g”.
Data Table 1: Hypothetical Punnett Square
Father ( ____ ____ ) x Mother ( ____ ____)
Mother ♀
Father ♂
Gametes
Count the number of seedlings that are green versus those that are yellow and record the data in Data Table 2.
Data Table 2 – Seedling data
Seedling Color
Green
Yellow
Total
Number of Seedlings
Percentage of Total
Compare the expected outcome to actual results found in Data Table 2, write this information into Data Table 3.
Data Table 3 – Frequency calculations
Frequency
Expected Values
Actual Values
Green
Yellow
Total
Questions:
What are the predicted ratios of the phenotypes (green to yellow) in the plants?
What were the actual ratios of the phenotypes (green to yellow) displayed in the tobacco plants?
If all yellow seedlings were removed from the population, would the next generation still have a chance at displaying the yellow allele? Explain your answer.
Would a cross between a homozygous and a heterozygous parent show the same ratios?
Explain why it may be important to collect data from a larger population.
Exercise 2 – Dihybrid Crossing with Corn
Count the number of purple, yellow, smooth and wrinkled seeds on the ear of corn between the lines in Figures 4-7 and record the data in Data Table 4.
Data Table 4 – Corn kernel data
Number of Kernels
Kernel Percentage (divide count by total, then multiply by 100)
Kernel Coloration
Purple
Yellow
TOTAL
Kernel Texture
Smooth
Wrinkled
TOTAL
The ear of corn is a result of a cross between plants that were both heterozygous for color and texture. Write the genotypes of both parents of this dihybrid cross. (P = purple color; p = yellow color; S = smooth shape; s = wrinkled shape)
What are the possible gametes for both parents? (Place these gametes into the Punnett square in Data Table 5 for both the male and the female.)
Calculate the phenotypic ratios that are expected for each type of seed.Purple & smooth ______________________
Purple & winkled _____________________
Yellow & smooth _____________________
Yellow & wrinkled ____________________
Data Table 5 – Punnett square for dihybrid cross
Parent Gametes ♂
Gametes
Parent Gametes ♀
Calculate the expected percentage of:Purple, smooth kernels _________ ...
please answer all questions to the activites- The study of genetics an.docx
please answer all questions to the activites.
The study of genetics and inheritance is concerned with understanding the biological properties that are transmitted from parent to offspring. Mendel's work resulted in the two laws of inheritance: (1) the Law of Segregation and (2) the Law of Independent Assortment. Exercise 1- Law of Segregation COVh A In Mendel's experiments in pea plants, he observed seven traits of the plants including height, pea shape, pea color and observed that these traits had two easily recognizable forms (tall versus short plants, round versus wrinkled peas, and yellow versus green peas). Each of these variant forms of the same gene (tall vs. short) we now referred to as alleles. Mendel found that when studying crosses of plants and observing one trait (plant height) using monohybrid crosses, the alleles of the gene segregate from one another when gametes are formed. The law of segregation results from the separation of homologous chromosomes and, thus, alleles during meiosis. The subsequent random fertilization of gametes results in the predicted ratios observed in monohybrid crosses. In a monohybrid cross, one pair of alleles is studied. If both parents are homozygous, with one parent homozygous dominant and the other parent homozygous recessive, the first generation will be heterozygous exhibiting the dominant phenotype. As Mendel demonstrated, 75% of the second generation (F2) offspring will have the dominant phenotype and 25% the recessive phenotype (a 3:1 ratio). This prediction can be validated using the chi-square statistical test. In corn (Zea mays), for our purposes, the inheritance of the seed (kernel) color can be considered to be a single gene, with one pair of alleles (purple vs. yellow). The purple color allele is dominant, and the yellow allele is recessive. Note: For genetic accuracy, the color-determining pigments are found in the aleurone coating of the endosperm, which can be colorless, purple, or red, and in the endosperm proper, which is yellow or white. Color production is subject to epistasis and modification by other genes. In this exercise you will collect data from the second-generation ( F 2 ) offspring of a monohybrid cross. P 1 Parental Generation Procedure: 1. Obtain one (1) ear of monohybrid corn per group. 2. Count at least 20 seeds total on the ear of corn, recording how many of the seeds are purple and how many of the seeds are yellow in Table 1 below. 3. Record your totals on the classroom whiteboard. 4. You can now test the validity of your (THE CLASS) data using the chi-square test. The chi-square is a relatively simple mathematical calculation that can determine how well the observed experimental results (the actual count) correspond to the theoretical prediction (the expected results). The formula for the chi-square calculation is: x 2 = e 1 ( o 1 e 1 ) 2 + e 2 ( o 2 e 2 ) 2 = x 2 = the symbol for the term, chi-square 0 = observed count (number) ( o 1 = the purple seeds, o 2 = ye.
Gene interaction -Complementary, Supplementary,Dominant Epistasis, Recessive...
Most of the characters of living organisms are controlled/ influenced/ governed by a collaboration of several different genes. • Numerous deviations have been recorded in which different kinds of interactions are possible between the genes.
Plant Cytogenetics: Dihybrid Inheritance
Prepared for B. Sc. III Botany students, based on the syllabus prescribed by Dr. Babasaheb Ambedkar Marathwada University, Aurangabad.
Find the indicial roots then solve the following ODE using series. Yo.pdf
Find the indicial roots then solve the following ODE using series. You must show the first 4
terms of each series. xy\" + 2y\' + xy = 0
Solution
Ans-
Let X be a set and let be a family of subsets of X. Then is called a topology on X if:[1][2]
If is a topology on X, then the pair (X, ) is called a topological space. The notation X may be
used to denote a set X endowed with the particular topology .
The members of are called open sets in X. A subset of X is said to be closed if its complement is
in (i.e., its complement is open). A subset of X may be open, closed, both (clopen set), or
neither. The empty set and X itself are always both closed and open.
Basis for a topology
Main article: Basis (topology)
A base (or basis) B for a topological space X with topology T is a collection of open sets in T
such that every open set in T can be written as a union of elements of B.[3][4] We say that the
base generates the topology T. Bases are useful because many properties of topologies can be
reduced to statements about a base that generates that topology—and because many topologies
are most easily defined in terms of a base that generates them.
Subspace and quotient
Every subset of a topological space can be given the subspace topology in which the open sets
are the intersections of the open sets of the larger space with the subset. For any indexed family
of topological spaces, the product can be given the product topology, which is generated by the
inverse images of open sets of the factors under the projection mappings. For example, in finite
products, a basis for the product topology consists of all products of open sets. For infinite
products, there is the additional requirement that in a basic open set, all but finitely many of its
projections are the entire space.
A quotient space is defined as follows: if X is a topological space and Y is a set, and if f : XY is
a surjective function, then the quotient topology on Y is the collection of subsets of Y that have
open inverse images under f. In other words, the quotient topology is the finest topology on Y for
which f is continuous. A common example of a quotient topology is when an equivalence
relation is defined on the topological space X. The map f is then the natural projection onto the
set of equivalence classes.
Examples of topological spaces
A given set may have many different topologies. If a set is given a different topology, it is
viewed as a different topological space. Any set can be given the discrete topology in which
every subset is open. The only convergent sequences or nets in this topology are those that are
eventually constant. Also, any set can be given the trivial topology (also called the indiscrete
topology), in which only the empty set and the whole space are open. Every sequence and net in
this topology converges to every point of the space. This example shows that in general
topological spaces, limits of sequences need not be unique. However, often topological spaces
must b.
Design verilog code [in ISE Project Navigator Verilog Module, Veril.pdf
Design verilog code [in ISE Project Navigator: Verilog Module, Verilog Text Fixture_TB] for:
4-bits synchronous up/down counter with an enable. This counter has three inputs Clock,
ENABLE, and UP/DOWN’ and four outputs O3 (MSB),O2,O1, and O0 (LSB). The table below
defines the circuit behavior: ENABLE UP/DOWN\' Function 0 X Stop counting 0 Count down:
15,14,...,1,0,15,14, Count up: 0,1,...,14,15,0,1
Solution
What is the verilog code for 4-bit updown counter with synchronous clear?
2 ANSWERS
[Sanket Naik]
Sanket Naik, been there, done that...!
1.7k Views
module up_dn_cnt (
input clk,
input clr, //Active high clear
input up, //Active high up count enable
input dn, //Active down up count enable
output [3:0] count
);
always @(posedge clk, posedge clr) begin
if(clr == 1) begin
count <= \'b0;
end else begin
if(up == 1) begin
count <= count + 1;
end else if(dn == 1) begin
count <= count - 1;
end
end
end
endmodule.
More Related Content
Similar to Goal to help you link cell division to Mendelian genetics; show wit.pdf
Genetics stem-12
not so easy but understandable INTRA INTER ALLELIC INTERACTION (NON MENDELIAN GENETICS)
51 Mendelian Gene cs Lab 5 52 La.docx
51
Mendelian Gene cs
Lab 5
52
Lab 5: Mendelian Gene cs
53
Introduc on
In 1866, Gregor Mendel, an Austrian Monk, published a paper en tled “Experiments in plant hybridiza-
on”. It went largely unno ced un l 1900 when it was rediscovered and subsequently became the
basis for what we now refer to as Mendelian Gene cs.
Mendel was the Þrst to recognize:
Inherited characters are determined by speciÞc factors (now recognized these as genes).
These factors occur in pairs (genes).
When both alleles of a gene are the same they are said to be homozygous, while if they are di erent
they are said to be heterozygous. When gametes form, these factors segregate so that each gamete
contains only one allele for each gene. Remember, alleles reside on the chromosomes that are divid-
ing. These original observa ons lead to what we now refer to as The law of segrega on and the law of
independent assortment.
Figure 1: Law of Segrega on
Concepts to explore:
Gregor Mendel
Law of segrega on
Homozygous
Heterozygous
Independent assortment
Dominant vs. recessive
Incomplete dominance
Co-dominance
Genotype
Phenotype
Monohybrid cross
Dihybrid cross
Punne square
Lab 5: Mendelian Gene cs
54
The law of segrega on states that during
meiosis, homologous (paired) chromo-
somes split (Figure 1). The law of inde-
pendent assortment states that during
meiosis, each homologous chromosome
has an equal chance of ending up in ei-
ther gamete, and alleles for individual
genes segregate with the chromosomes
on which they are located (Figure 2).
Using corn as an example (Figure 2):
The large chromosome has the gene for kernel color (Y = yellow, y = blue).
The small chromosome has the gene for kernel texture (S = smooth (green); s = wrinkled (red)).
When a dominant allele is present, that characteris c is expressed, regardless of the second allele. In
this case both the Yy and YY o spring will be yellow.
A recessive allele is only expressed when both alleles are recessive. In this case only the yy combina-
on is blue. The dominant allele is always represented by capital le ers, while the recessive is repre-
sented by lower case le ers.
Genotype refers to the combina on of alleles for a par cular trait. Phenotype refers to the appear-
ance of that combina on of alleles. In our example, the genotype of the diploid cell is Yy, Ss, while the
phenotype is Yellow and Smooth.
Figure 2: Law of Independent Assortment
Figure 3: Monohybrid Cross
Punne Square F1
Lab 5: Mendelian Gene cs
55
Alleles can exhibit incomplete dominance and co-dominance. An example of incomplete dominance is
the cross of two plants, one with red ßowers and one with white, whose o spring have pink ßowers.
In the case of codominance, the same cross would result in red and white striped ßowers.
If we know the genotype of two parents we can.
Lab ReportName _______________________ Section .docx
Lab Report Name: _______________________
Section: _____________________
4
Plant Genetics
Exercise 1 - The Punnett Square
Write a hypothesis describing the phenotype of the germinating tobacco seeds. Include the approximate number of seedlings and percentages that will have a dominant trait and recessive trait out 50 seeds.
Create a Punnett square using Data Table 1, showing the possible F2 generation outcomes from a cross between a heterozygous male and a heterozygous female for the germinating tobacco seeds. The dominant allele is written as “G” for green and the recessive allele for albino/yellow color is written as “g”.
Data Table 1: Hypothetical Punnett Square
Father ( ____ ____ ) x Mother ( ____ ____)
Mother ♀
Father ♂
Gametes
Count the number of seedlings that are green versus those that are yellow and record the data in Data Table 2.
Data Table 2 – Seedling data
Seedling Color
Green
Yellow
Total
Number of Seedlings
Percentage of Total
Compare the expected outcome to actual results found in Data Table 2, write this information into Data Table 3.
Data Table 3 – Frequency calculations
Frequency
Expected Values
Actual Values
Green
Yellow
Total
Questions:
What are the predicted ratios of the phenotypes (green to yellow) in the plants?
What were the actual ratios of the phenotypes (green to yellow) displayed in the tobacco plants?
If all yellow seedlings were removed from the population, would the next generation still have a chance at displaying the yellow allele? Explain your answer.
Would a cross between a homozygous and a heterozygous parent show the same ratios?
Explain why it may be important to collect data from a larger population.
Exercise 2 – Dihybrid Crossing with Corn
Count the number of purple, yellow, smooth and wrinkled seeds on the ear of corn between the lines in Figures 4-7 and record the data in Data Table 4.
Data Table 4 – Corn kernel data
Number of Kernels
Kernel Percentage (divide count by total, then multiply by 100)
Kernel Coloration
Purple
Yellow
TOTAL
Kernel Texture
Smooth
Wrinkled
TOTAL
The ear of corn is a result of a cross between plants that were both heterozygous for color and texture. Write the genotypes of both parents of this dihybrid cross. (P = purple color; p = yellow color; S = smooth shape; s = wrinkled shape)
What are the possible gametes for both parents? (Place these gametes into the Punnett square in Data Table 5 for both the male and the female.)
Calculate the phenotypic ratios that are expected for each type of seed.Purple & smooth ______________________
Purple & winkled _____________________
Yellow & smooth _____________________
Yellow & wrinkled ____________________
Data Table 5 – Punnett square for dihybrid cross
Parent Gametes ♂
Gametes
Parent Gametes ♀
Calculate the expected percentage of:Purple, smooth kernels _________ ...
please answer all questions to the activites- The study of genetics an.docx
please answer all questions to the activites.
The study of genetics and inheritance is concerned with understanding the biological properties that are transmitted from parent to offspring. Mendel's work resulted in the two laws of inheritance: (1) the Law of Segregation and (2) the Law of Independent Assortment. Exercise 1- Law of Segregation COVh A In Mendel's experiments in pea plants, he observed seven traits of the plants including height, pea shape, pea color and observed that these traits had two easily recognizable forms (tall versus short plants, round versus wrinkled peas, and yellow versus green peas). Each of these variant forms of the same gene (tall vs. short) we now referred to as alleles. Mendel found that when studying crosses of plants and observing one trait (plant height) using monohybrid crosses, the alleles of the gene segregate from one another when gametes are formed. The law of segregation results from the separation of homologous chromosomes and, thus, alleles during meiosis. The subsequent random fertilization of gametes results in the predicted ratios observed in monohybrid crosses. In a monohybrid cross, one pair of alleles is studied. If both parents are homozygous, with one parent homozygous dominant and the other parent homozygous recessive, the first generation will be heterozygous exhibiting the dominant phenotype. As Mendel demonstrated, 75% of the second generation (F2) offspring will have the dominant phenotype and 25% the recessive phenotype (a 3:1 ratio). This prediction can be validated using the chi-square statistical test. In corn (Zea mays), for our purposes, the inheritance of the seed (kernel) color can be considered to be a single gene, with one pair of alleles (purple vs. yellow). The purple color allele is dominant, and the yellow allele is recessive. Note: For genetic accuracy, the color-determining pigments are found in the aleurone coating of the endosperm, which can be colorless, purple, or red, and in the endosperm proper, which is yellow or white. Color production is subject to epistasis and modification by other genes. In this exercise you will collect data from the second-generation ( F 2 ) offspring of a monohybrid cross. P 1 Parental Generation Procedure: 1. Obtain one (1) ear of monohybrid corn per group. 2. Count at least 20 seeds total on the ear of corn, recording how many of the seeds are purple and how many of the seeds are yellow in Table 1 below. 3. Record your totals on the classroom whiteboard. 4. You can now test the validity of your (THE CLASS) data using the chi-square test. The chi-square is a relatively simple mathematical calculation that can determine how well the observed experimental results (the actual count) correspond to the theoretical prediction (the expected results). The formula for the chi-square calculation is: x 2 = e 1 ( o 1 e 1 ) 2 + e 2 ( o 2 e 2 ) 2 = x 2 = the symbol for the term, chi-square 0 = observed count (number) ( o 1 = the purple seeds, o 2 = ye.
Gene interaction -Complementary, Supplementary,Dominant Epistasis, Recessive...
Most of the characters of living organisms are controlled/ influenced/ governed by a collaboration of several different genes. • Numerous deviations have been recorded in which different kinds of interactions are possible between the genes.
Plant Cytogenetics: Dihybrid Inheritance
Prepared for B. Sc. III Botany students, based on the syllabus prescribed by Dr. Babasaheb Ambedkar Marathwada University, Aurangabad.
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Find the indicial roots then solve the following ODE using series. Yo.pdf
Find the indicial roots then solve the following ODE using series. You must show the first 4
terms of each series. xy\" + 2y\' + xy = 0
Solution
Ans-
Let X be a set and let be a family of subsets of X. Then is called a topology on X if:[1][2]
If is a topology on X, then the pair (X, ) is called a topological space. The notation X may be
used to denote a set X endowed with the particular topology .
The members of are called open sets in X. A subset of X is said to be closed if its complement is
in (i.e., its complement is open). A subset of X may be open, closed, both (clopen set), or
neither. The empty set and X itself are always both closed and open.
Basis for a topology
Main article: Basis (topology)
A base (or basis) B for a topological space X with topology T is a collection of open sets in T
such that every open set in T can be written as a union of elements of B.[3][4] We say that the
base generates the topology T. Bases are useful because many properties of topologies can be
reduced to statements about a base that generates that topology—and because many topologies
are most easily defined in terms of a base that generates them.
Subspace and quotient
Every subset of a topological space can be given the subspace topology in which the open sets
are the intersections of the open sets of the larger space with the subset. For any indexed family
of topological spaces, the product can be given the product topology, which is generated by the
inverse images of open sets of the factors under the projection mappings. For example, in finite
products, a basis for the product topology consists of all products of open sets. For infinite
products, there is the additional requirement that in a basic open set, all but finitely many of its
projections are the entire space.
A quotient space is defined as follows: if X is a topological space and Y is a set, and if f : XY is
a surjective function, then the quotient topology on Y is the collection of subsets of Y that have
open inverse images under f. In other words, the quotient topology is the finest topology on Y for
which f is continuous. A common example of a quotient topology is when an equivalence
relation is defined on the topological space X. The map f is then the natural projection onto the
set of equivalence classes.
Examples of topological spaces
A given set may have many different topologies. If a set is given a different topology, it is
viewed as a different topological space. Any set can be given the discrete topology in which
every subset is open. The only convergent sequences or nets in this topology are those that are
eventually constant. Also, any set can be given the trivial topology (also called the indiscrete
topology), in which only the empty set and the whole space are open. Every sequence and net in
this topology converges to every point of the space. This example shows that in general
topological spaces, limits of sequences need not be unique. However, often topological spaces
must b.
Design verilog code [in ISE Project Navigator Verilog Module, Veril.pdf
Design verilog code [in ISE Project Navigator: Verilog Module, Verilog Text Fixture_TB] for:
4-bits synchronous up/down counter with an enable. This counter has three inputs Clock,
ENABLE, and UP/DOWN’ and four outputs O3 (MSB),O2,O1, and O0 (LSB). The table below
defines the circuit behavior: ENABLE UP/DOWN\' Function 0 X Stop counting 0 Count down:
15,14,...,1,0,15,14, Count up: 0,1,...,14,15,0,1
Solution
What is the verilog code for 4-bit updown counter with synchronous clear?
2 ANSWERS
[Sanket Naik]
Sanket Naik, been there, done that...!
1.7k Views
module up_dn_cnt (
input clk,
input clr, //Active high clear
input up, //Active high up count enable
input dn, //Active down up count enable
output [3:0] count
);
always @(posedge clk, posedge clr) begin
if(clr == 1) begin
count <= \'b0;
end else begin
if(up == 1) begin
count <= count + 1;
end else if(dn == 1) begin
count <= count - 1;
end
end
end
endmodule.
DNA and RNA are both stands of nucleotides but RNA can carry out for .pdf
DNA and RNA are both stands of nucleotides but RNA can carry out for more functions than
DNA. How docs the structure of RNA differ from DNA? What additional activities/function can
RNA do that DNA cannot?
Solution
The structure of ribonucleic acid is single stranded in which thymine replaces uracil as
complimentary base and other adenine, guanine, cytosine are the nucleotides but
deoxyribonucleic acid is double stranded & duplex.
The additive activities can RNA performing are to undergo \"translation\" to code for proteins
the following are the other functions where as DNA cannot undergo translation directly to code
for proteins
Small nuclear ribonucleic acid (snRNA): It is also known as U-RNA as it contains many uridine
contents in its complex structure. This is composed of 150 nucleotides nearly to produce
spliceosomes and to act along with snRNP. This snRNA has a predominant role in splicing.
mRNA: It is produced by the transcription of DNA in the presence of RNA polymerase enzyme
and it has amino acid coding exons to code for specific proteins as per genetic code.
rRNA (ribosomal RNA): It has 50S and 30S ribosomal subunits which have aminoacyl site and
peptidyl site to carry out translation process during protein synthesis with mRNA.
tRNA (transfer RNA): It has clover leaf structure & common to all tRNA structure. It has
anticodon site to bind to the mRNA and to displace mRNA transcript from nuclear region onto
the rRNA for protein synthesis. It is considered as an adaptor molecule mainly consist of 76 to
90 ribonucleotides in length
piRNA:
Piwi-interacting RNAs (piRNA) are the small non-coding RNA molecules with 26 to 31
nucleotides that are expressed in many of the germ lines of eukaryotic animals. piRNAs interact
with piwi proteins to form RNA-protein complexes..
Distinguish between the concepts of stereotyping, prejudice, and dis.pdf
Distinguish between the concepts of stereotyping, prejudice, and discrimination.
***PSYCHOLOGY
Solution
Stereotyping: generalised belief about a group of people in which identical characters are
assigned to virtually all members of the group, regardless actual variation among the members.
Prejudice: it is attitude or feeling, favourable or unfavorable towards a person or group of people,
prior to, or not based on, actual experience. Discrimination: negative or harmful beheaviour
towards the member or members of a group simply because of their membership in that group..
38 Cats eat only animals. Something fuzzy exists. Everything thats .pdf
38 Cats eat only animals. Something fuzzy exists. Everything that\'s fuzzy is a cat. And
everything eats some- thing. So animals exist. C(a), E(r, y), A(x),Fx)
Solution
Cats eat only animals.C(x)
Somthing fuzzy exists E(x,y)
Everything that is a fuzzy is a cat. A(x)
Everything eats something F(x)
E(x,y) implies A(x)
If F(x) is true then A(x) is also true.
Since A(x) is true, C(x) implies there exist animals..
An octapeptide was hydrolyzed and the amino acids separated. It was .pdf
An octapeptide was hydrolyzed and the amino acids separated. It was found to have the
following amino acid composition: Gly (1); Ala (2); Val (1); Lys (2); Met (1); Asp (1). After
incubation with FDNB and hydrolysis, 2,4-dinitrophenylalanine was detected. Treatment with
cyanogen bromide, which cleaves at M, resulted in two fragments: a pentapeptide containing two
K residues and a tripeptide containing V, G, and A. (The residues are not listed in any particular
order.) After treatment with trypsin, which cleaves at R and K, there were 3 fragments: two
dipeptides and MGAV. What is the sequence of the peptide? Use the one-letter abbreviations for
the amino acids, and do not use dashes. An octapeptide was hydrolyzed and the amino acids
separated. It was found to have the following amino acid composition: Gly (1), Ala (2); Val (1),
Lys (2), Met (1), Asp (1). After incubation with FDNB and hydrolysis, 2,4-dinitrophenylalanine
was detected. NO2 NO2 NH H3C-CH OO Treatment with cyanogen bromide, which cleaves at
M, resulted in two fragments: a pentapeptide containing two K residues and a tripeptide
containing V, G, and A. (The residues are not listed in any particular order.) After treatment with
trypsin, which cleaves at R and K, there were 3 fragments: two dipeptides and M What is the
sequence of the peptide? Use the one-letter abbreviations for the amino acids, and do not use
dashes.
Solution
An octapeptide was hydrolyzed and the amino acids separated. It was found to have the
following amino acid composition: Gly (1); Ala (2); Val (1); Lys (2); Met (1); Asp (1).
After incubation with FDNB and hydrolysis, 2,4-dinitrophenylalanine was detected.
1-fluoro-2,4-dinitrobenzene (FDNB) is an reagent that always react with N-terminal aminoacid
of a peptide chain and after acid hydrolysis it produces 2.4-dinitrophenyl derivative of
aminoacid. As there is production of 2,4-dinitrophenylalanine, that means alanine is present at
the N-terminal end of the octapeptide.
Treatment with cyanogen bromide, which cleaves at M, resulted in two fragments: a
pentapeptide containing two K residues and a tripeptide containing V, G, and A. (The residues
are not listed in any particular order.)
Cyanogen bromide always cleaves the peptide at the carboxylic end of methionine and after
cleavage it gives two fragments of pentapeptide and tripeptide. So it indicates that methionine is
present at 5th position of the octapeptide.
After treatment with trypsin, which cleaves at R and K, there were 3 fragments: two dipeptides
and MGAV.
Trypsin always cleaves at the carboxy-terminal of arginine and lysine. After cleavage, two
dipeptide and one tetrapeptide (MGAV) is produced. So it is cleared that, the two dipeptide
would be AK and DK.
Therefore the final aminoacid sequence of the octapeptide would be H2N-AKDKMGAV-COOH
Answer: AKDKMGAV.
c. hydrophillic heads; sticking out toward the interior and exterior.pdf
c. hydrophillic heads; sticking out toward the interior and exterior of the cell
d. hydrophobic and hydrophylic sides, facing in toward another Properties of Phospholipids The
plasma membrane is made up of a phospholipid bilayer, with the of the phospholipids Multiple
Choice hydrophobic tails: sticking out toward the interior and exterior of the cell hydrophilic
heads facing in toward one another
Solution
(C).hydrophilic heads;sticking out toward the interior and exterior of the cell.
According to the fluid-mosaic model,the hydrophobic tails are present in between the
phospholipid bilayer and the hydrophilic heads facing outward top and bottom of the layer.These
heads are made up of phosphate and glycerol whereas the tail is made up of unsatuarated and
saturated fatty acids..
Biology Lab questions1. Units of amino acid are carried to the ri.pdf
Biology Lab questions:
1. Units of amino acid are carried to the ribosomes by (A) transfer RNA (B) radioactive uracil
(C) template RNA (D) messenger RNA
Solution
The units of amino acids are carried to the ribosomes by t-rna which is charged.. According to
the codon present in the ribosomes that charged t-rna of that particular amino acid are carried to
the ribosomes..!!.
Why is it important to use several sources of power and influence, r.pdf
Why is it important to use several sources of power and influence, rather than relying on one or
two?
Solution
There are 5 various sources of influence that are as follows
1. Reward power
2. Coercive power
3. Legitimate power
4. Expert power
5. Recent power
Reward power is that in which we can provide reqare for obtaining power
Coercive power is that we can take a disciplinary action to obtain power or to influence for work.
Legitimate power is that which can be granted by your position or rights that as a MD of a
company have a power towards the employees who report him
Expert power is that which is received by your knowledge and experience of any work .
Referent power is that which is related to your reference when you know someone and feel that
you know him means you connect with him
All the five sources of power and influence are important because with the help of all the five
sources we can do the best of himself nothing has been left. And we know every person has
different different thoughts and views. Not all the person preferred same thing . And we don\'t
know the person clearly so if we use all the five sources it helps us to increase or obtaining good
and satisfactory results .
If we use only one or two and rely on it . May be it happens that person don\'t like these two
means these two sources are not good as per that person for obtaining good response. May be it
happens that we wants more position as compared to monetary benefits so if we can use all the
five sources then It makes more chance of better influencing..
What is special about the colugos teethLower incisors are comb-li.pdf
What is special about the colugos teeth?
Lower incisors are comb-like
Only two upper incisors
Upper canines are reduced
Only one lower molar
Solution
Lower incisor are comb like.
Upper incisors are 2 in each quadrant since dental formula for colugos is 2.1.2.3 for upper jaw.
Colugos has special feature that their incisor are comb shape,which has almost 20 tines on it.
These comb in incisors are used for grooming. As is seen in stepsirrihines they hav similar
function and appearance..
write simple program in c++ in which source reads a file of integers.pdf
write simple program in c++ in which source reads a file of integers and outputs the average
Solution
Average.cpp
#include
#include
using namespace std;
int main()
{
ifstream inputFile;
inputFile.open(\"numbers.txt\");
int n;
int sum =0;
int count = 0;
double average = 0;
if (inputFile.is_open()) {
while (!inputFile.eof()) {
inputFile >> n;
sum = sum + n;
count++;
}
}
average = sum/(double)count;
cout<<\"The average is \"<.
Write an example of a unit test for a Class methods using JUnitfr.pdf
Write an example of a unit test for a Class methods using JUnitframework
Write an example of a unit test for a Class methods using JUnitframework
Solution
Example of test case java using JUnitframework:
import org.junit.Test;
import static org.junit.Assert.assertEquals;
public class TestJunit {
String message = \"Hello World\";
MessageUtil messageUtil = new MessageUtil(message);
@Test
public void testPrintMessage() {
assertEquals(message,messageUtil.printMessage());
}
}
Test suit Examples:
import org.junit.runner.RunWith;
import org.junit.runners.Suite;
//JUnit Suite Test
@RunWith(Suite.class)
@Suite.SuiteClasses({
TestJunit1.class ,TestJunit2.class
})
public class JunitTestSuite {
}
import org.junit.Test;
import org.junit.Ignore;
import static org.junit.Assert.assertEquals;
public class TestJunit1 {
String message = \"Robert\";
MessageUtil messageUtil = new MessageUtil(message);
@Test
public void testPrintMessage() {
System.out.println(\"Inside testPrintMessage()\");
assertEquals(message, messageUtil.printMessage());
}
}
import org.junit.Test;
import org.junit.Ignore;
import static org.junit.Assert.assertEquals;
public class TestJunit2 {
String message = \"Robert\";
MessageUtil messageUtil = new MessageUtil(message);
@Test
public void testSalutationMessage() {
System.out.println(\"Inside testSalutationMessage()\");
message = \"Hi!\" + \"Robert\";
assertEquals(message,messageUtil.salutationMessage());
}
}.
whats in plasmaSolutionAnswerBlood is the fluid connecti.pdf
what\'s in plasma?
Solution
Answer:
Blood is the fluid connective tissue possesses plasma (45%) and remaining blood cells such as
monocytes, macrophages, clotting factors, platelets, and nutrients. This plasma is composed of
protein compartment in which various serum albumins, globulins along with clotting factors are
existing. Plasma protein components such as “beta globulin” and “glycoproteins” have extensive
saturable binding affinity.
What is a reason we want to write in a concise manner 1. To elimina.pdf
What is a reason we want to write in a concise manner? 1. To eliminate redundant and repetitive
words 2. To allow better use of bullets 3. To avoid jargon or industry language 4. To indicate the
main point of the message
Solution
4. To indicate main point of message
We write in a concise manner to brief eth content clearly using few words.
What are the advantages of Credit Suisse’s new supplier managemente.pdf
What are the advantages of Credit Suisse’s new supplier management/evaluation system?
Please write in your own words, and atlest 2 paragraphs
Solution
Historically, the IT industry has been mostly a reactive unit. Enterprise IT departments were
essentially order-takers that were quick to adopt the latest programs. Stephen Hilton, managing
director and CIO of technology infrastructure for Credit Suisse, will have none of that. Known
for his stick-to-itiveness and indifference to flavor-of-the-week technology, Hilton has shown
leadership in building a unified culture and developed a reputation for his steady pace of
improvement.
As most Industry leaders know, since 2007 capital markets revenues have decreased rapidly,
resulting in budget cuts across the board. At the same time, most financial institutions had built
up significant fixed costs during the prior decade when revenues were high and budgets were
bloated. \"People were surprised that IT was not as variable as they thought it would be,\" says
Hilton, whose firm has a market capitalization of $42 billion and $1.3 trillion in assets under
management.
At Credit Suisse, only 30% to 40% of IT cost was allocated to staffing, reports Hilton. \"A lot of
the budget covers other IT assets, obligations, liabilities and licensing costs we built up over the
years that just are not variable in a short period of time.\" The only way you can really drive the
cost base down is through business process reengineering, he says, and that\'s exactly what his
team has been working toward.
\"It was quite a learning process for people to understand that. And if I\'m honest, that
realization, a very good realization, has actually changed the business attitude,\" Hilton says.
\"They\'re having different conversations with us now.\" Business leaders are aware that asking
for \"please do it quickly and just get it in\"-style requests aren\'t acceptable anymore. \"They
have to be more patient and more strategic in their decisions.\"
Challenge Silliness
While the old way of approaching technology may have been fitting given the short lifespan of
IT and massive innovation cycles, things began to change in the mid-2000s. For Credit Suisse
the focus shifted to optimizing the company\'s massive, complex infrastructure and improving
end user productivity. As CIO of technology infrastructure, Hilton prepared to make tough
decisions to meet company goals.
\"My view is that at the scale we now run at, process efficiency is key. I am a technologist, but
my master\'s degree is in mechanical engineering and making manufacturing decisions. So I
learned back in the late \'80s about industrialized processes and how to optimize factories. Those
skill sets are what I\'ve applied in the last five years.\" He adds, \"I very much encourage my
teams to be challenging partners. I\'ve spent five years really empowering them, saying, \'Look,
you\'re the experts. Challenge inefficiencies whenever you see them.\'\".
The PRIMARY structure of a protein isformed by alpha helices and .pdf
The PRIMARY structure of a protein is:
formed by alpha helices and beta sheets
formed by multiple polypeptide chains binding to one another
formed by dehydration reactions that join amino acids together
all of the above are true
The PRIMARY structure of a protein is:
Solution
option C.
The primrary structure of a protein is formed by dehydration reactions that join amino acids
together.
dehydration reaction removes H2O between two amino acids to form peptide bond. this is the
native primary structure that consists of a linear chain of amino acids.
alpha and beta structures are secondary structures.
tertiary structure involves the interaction between different polypeptide chains..
The figure below shows the chromosomes in a cell for some species (.pdf
The figure below shows the chromosomes in a cell for some species (\"pretend\" that
chromosomes are always visible for this exercise & ignore any nuclear membrane). 1. How
many chromosomes do you see? ___________ Is this haploid or diploid? What do the letters
represent? ___________ Why does the letter \"g\" appear only as lower case but the letter \"F\"
appears as both upper- AND lowercase (F & f)
Solution
1.The number of chromosomes is equal to the number of centromeres.Here in this case ,it is 6
chromosomes.
2.This is haploid as only single chromosomes are seen.
3.The letters represent the alleles of gene.
4.The letter \'g \'appears in the lower case because it is present only in recessive form while \' F\'
appears as both upper and lower case as it is present in both dominant and recessive forms..
The addButton_Click procedure should perform the following three tas.pdf
The addButton_Click procedure should perform the following three tasks:
1.add the integer entered in the numTextBox to an accumulator
2. display the integer on a separate line in the ListTextBox
3. display the accumulator\'s value in the sumLabel.
How can I accomplish this using a TextBox instead of an InputBox(with prompts)?
Solution
import java.awt.*;
import java.awt.event.*;
import java.applet.*;
/*
*/
public class Cal extends Applet
implements ActionListener
{
String msg=\" \";
int v1,v2,result;
TextField t1;
Button b[]=new Button[10];
Button add,sub,mul,div,clear,mod,EQ;
char OP;
public void init()
{
Color k=new Color(120,89,90);
setBackground(k);
t1=new TextField(10);
GridLayout gl=new GridLayout(4,5);
setLayout(gl);
for(int i=0;i<10;i++)
{
b[i]=new Button(\"\"+i);
}
add=new Button(\"add\");
sub=new Button(\"sub\");
mul=new Button(\"mul\");
div=new Button(\"div\");
mod=new Button(\"mod\");
clear=new Button(\"clear\");
EQ=new Button(\"EQ\");
t1.addActionListener(this);
add(t1);
for(int i=0;i<10;i++)
{
add(b[i]);
}
add(add);
add(sub);
add(mul);
add(div);
add(mod);
add(clear);
add(EQ);
for(int i=0;i<10;i++)
{
b[i].addActionListener(this);
}
add.addActionListener(this);
sub.addActionListener(this);
mul.addActionListener(this);
div.addActionListener(this);
mod.addActionListener(this);
clear.addActionListener(this);
EQ.addActionListener(this);
}
public void actionPerformed(ActionEvent ae)
{
String str=ae.getActionCommand();
char ch=str.charAt(0);
if ( Character.isDigit(ch))
t1.setText(t1.getText()+str);
else
if(str.equals(\"add\"))
{
v1=Integer.parseInt(t1.getText());
OP=\'+\';
t1.setText(\"\");
}
else if(str.equals(\"sub\"))
{
v1=Integer.parseInt(t1.getText());
OP=\'-\';
t1.setText(\"\");
}
else if(str.equals(\"mul\"))
{
v1=Integer.parseInt(t1.getText());
OP=\'*\';
t1.setText(\"\");
}
else if(str.equals(\"div\"))
{
v1=Integer.parseInt(t1.getText());
OP=\'/\';
t1.setText(\"\");
}
else if(str.equals(\"mod\"))
{
v1=Integer.parseInt(t1.getText());
OP=\'%\';
t1.setText(\"\");
}
if(str.equals(\"EQ\"))
{
v2=Integer.parseInt(t1.getText());
if(OP==\'+\')
result=v1+v2;
else if(OP==\'-\')
result=v1-v2;
else if(OP==\'*\')
result=v1*v2;
else if(OP==\'/\')
result=v1/v2;
else if(OP==\'%\')
result=v1%v2;
t1.setText(\"\"+result);
}
if(str.equals(\"clear\"))
{
t1.setText(\"\");
}
}
}.
A mistake occurs during lambda phage excision from the E. coli genom.pdf
A mistake occurs during lambda phage excision from the E. coli genome: a portion of the E.
colichromosome is excised with the phage genome, leaving behind a portion of the lambda
genome that includes the int gene. This defective genome is packaged and infects an E. coli
strain that is recA- (it lacks RecA). What do you think happens to this genetic material?
1) injected DNA is degraded
2) integrates into new host genome
3) The transferred portion of E. coli chromosome recombines with the new host genome
4) All of these are possible
Solution
The answer is Option no 1 that is injected DNA is degraded.
The defective genome does not carry int gene. So, the lambda phge cannot insert its genetic
material into host genome.
Homologus recombination and DNA reapir does not possible because the host genome is recA-
Strain.(rec A protein plays a major role in homologus recombination)..
Assume a System Engineering class has 27 students, of which one thir.pdf
Assume a System Engineering class has 27 students, of which one third are female and two
thirds are male. There were a total of 17 students (4 female and 13 male) that got the correct
answer to the first problem in their 8th homework assignment. What is the probability that a
male student got the correct answer?
Solution
total correct 17 srudents
4 female and 13 male
probability that male student correct answer = 13/27 = 0.481.
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Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Event Link:-
https://meetups.mulesoft.com/events/details/mulesoft-mysore-presents-exploring-gemini-ai-and-integration-with-mulesoft/
Agenda
● Java 17 Upgrade Overview
● Why and by when do customers need to upgrade to Java 17?
● Is there any immediate impact to upgrading to Mule Runtime 4.6 and beyond?
● Which MuleSoft products are in scope?
For Upcoming Meetups Join Mysore Meetup Group - https://meetups.mulesoft.com/mysore/
YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
Speaker:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Organizers:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Giridhar Meka - https://www.linkedin.com/in/giridharmeka
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Shyam Raj Prasad-
https://www.linkedin.com/in/shyam-raj-prasad/
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