Crossing of two bacteriophages with mutations such as T4 rapid lysis .pdf
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Crossing of two bacteriophages with mutations such as T4 rapid lysis mutant results in progeny that shows 50 colonies in E. coli B plate and 20 in E.coli K12 lambda. What is the cotransduction frequency? (1 pt) Frequency of recombinants =2 (wild type plaques in E.coli 1212 lambda) / Total number of plaques obtained in E. coli B 0.2 0.8 0.4 1.25 In sordaira, a cross between tan+ and tan- results in the following recombinants: TOTAL 1000 What is the distance of the rpo gene and the centromere 1/2 SDS/ Total progeny 1000.065mU0.87mU6.5mU A cross in the corn is carried out with three linked genes in triply heterozygous genotype for the allele pairs (A,a), (D,d) and (R,r). The dominant A results in red and aa in green leaves. The dominant D results in tall plants and dd are dwarfed. The dominant R results in ragged leaf margins and r have smooth margins. The triple heterozygous is crossed with homozygous recessive for all three alleles and the following phenotype is seen. Assume that interference in each interval is complete so the map distance in cM corresponds to the percent recombinations. red, tall, ragged 265 red, tall, smooth 24 red, dwarf, ragged 120 red, dwarf, smooth 90 green, tall, ragged 70 Green, tall, smooth 140 Green, dwarf, ragged 16 Green, dwarf, smooth 275 TOTAL 1000 1. What is the order of genes (1 point) 2. What is the distance between the genes ( 2 points) 3. Calculate interference ( 2 points) Choose an option below with all right answers R-A-D, 0.26mU for A to R,16mU for to A to D and I=66% A-R-D, 20mU for A to R,30mU for R to D,I=33% A-D-R, 20mU for D to R,30mU for A to D,1=23% When two bacterial strains use conjugation to mate. Bacterial strain met + his + leubio- was mixed with another strain met-hist-leu+bio+. But it was difficult to know the donor from the recipient. What strategy can help in distinguishing donor from recipient ( 1pt ) Use a donor strain which is antibiotic sensitive Use a donor strain which is antibiotic resistant Use a recipient strain that is antibiotic sensitive Use a U tube to separate donor and recipient with filer that can pass the DNA.
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Btt2153 7 population genetics_print1
Here are the solutions to the try these problems:
1. a) T allele frequency = (88/125) + (1/2)*(37/125) = 0.7
t allele frequency = 1 - 0.7 = 0.3
b) TT genotype frequency = 0.49, Tt genotype frequency = 0.42, tt genotype frequency = 0.09
2. p = 200/300 = 0.667, q = 100/300 = 0.333
3. a) Recessive allele frequency = 1/2500 = 0.0004
b) Dominant allele frequency = 1 - 0.0004 = 0.9996
c) Heterozygous
Problems2. This is a map for a diploid plantR--------35-------.pdf
Problems:
2. This is a map for a diploid plant:
R--------35--------E--------17---------F
One parental plant is R e f/r E F. It is mated with a homozygous recessive plant: r e f/r e f. Give
the proportion of the offspring assuming there is both no interference, and complete interference.
3. In a newly discovered insect, the three genes R, A, and P located on the same chromosome
(R= red body, A= spotted coloring, P=purple eyes, r=white body, a=solid coloring, p=white
eyes). A heterozygous parent(R A P/r a p) is mated with a homozygous recessive parent (r a p/ r
a p).
The following offspring are shown below:
(210) dominant for all traits
(16) red body, solid coloring, purple eyes
(70) red body, spotted coloring, white eyes
(50) red body, solid coloring, white eyes
(200) recessive for all traits
(55) white body, spotted coloring, purple eyes
(13) white body, spotted coloring, white eyes
(72) white body, solid coloring, purple eyes
Perform a three-point linkage to determine the gene order and relative distance between the
genes. How did the distance between the 2 genes affect the frequency of crossing-over events?
4. Here are four recombinant sequences that came from four Hfr matings during an interrupted
mating experiment. Determine the order and relative distance by setting A to 0/100 minutes.
Hfr1) A T1 H T2
8 26 33 53
Hfr2) E1 T1 E2 E3
4 12 24 54
Hfr3) T2 S T3 A
7 37 52 62
Hfr4) T3 C E1 E3
2 17 22 72
5. In a u-tube experiment, a researcher placed a kanamycin resistant strain of E. coli on one side
of a membrane impervious to the cells, and a wild-type strain on the other side. After a day, he
then plated a sample from both sides of the tube on plates containing kanamycin and observed
growth on both plates. He then repeated this experiment, but this time he added DNase to the
tube and observed the same results when he plated both cell types. What happened inside the
tube?
6. Pretend the following is a chromosome that underwent some changes. What were they? (the
dot is the centromere).
A B C D E F . G H I J A B C G . F E H I J
a. pericentric inversion, deletion
b. paracentric inversion, deletion
c. deletion, translocation
d. Robertsonian translocation,
pericntric inversion
7. Given:
Type of mutation Cistron A or B
1. point A
2. point A
3. point B
4. point B
5. deletion A
Results from 5 crosses:
1x2 (plaque) 2x3 (plaque) 3x4 (plaque) 4x5 (plaque)
1x3 (plaque) 2x4 (plaque) 3x5 (plaque)
1x4 (plaque) 2x5 (no plaque)
1x5 (plaque)
8. The following nucleotide sequence is found in a piece of DNA
5’-TTAACA-3’
3’-AATTGT-5’
if the cytosince undergoes a tautameric shift for one replication cycle draw two round of
replication and explain what type of mutation occurred (Transition or Transversion)?
Bonus:
Explain the relationship between mosaicism and x-inactivation using an example.
9. In a diploid fly, the three loci R E and D are linked as follows
R--------------22cm------------E----------16cm-------------D
One fly is available to you and its genotype is R.
Ftt1033 7 population genetics-2013
Here are the solutions to the try these problems:
1. a) T allele frequency = (88/125) + (1/2)*(37/125) = 0.72
t allele frequency = 1 - 0.72 = 0.28
b) TT genotype frequency = 0.72^2 = 0.52
Tt genotype frequency = 2*0.72*0.28 = 0.48
tt genotype frequency = 0.28^2 = 0.08
2. p = 200/300 = 0.667
q = 1 - p = 0.333
3. a) Recessive allele frequency = √(1/2500) = 0.02
b
Linkage_Crossing_over.pptx
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2) The strength of linkage depends on the physical distance between genes - closely located genes show complete linkage while genes farther apart show incomplete linkage with crossover occurring.
3) Tetrad analysis in fungi allows the direct observation of linkage and estimation of recombination frequencies between genes and centromeres.
Association mapping
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In one species of the plant coding for color- shape and amount of wate.pdf
In one species of the plant coding for color, shape and amount of water in the fruit are on the
same chromosome. The gene responsible for the color of the fruit has two alleles (R = red,
r=green), the gene responsible for the shape of the fruit also has two alleles (L =long, l=round),
and the gene responsible for the amount of water in the fruit also has two alleles (S dry, s=juicy).
The alleles represented by capital letters are dominant over the alleles represented by small
letters.
Crossing a pure line plant with red, long, dry flowers with a pure line plant with green, round,
juicy fruits results in an F1 plant with rough, long, dry fruits. The F1 individuals are crossed with
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22 green, long, juicy fruits
22 red, round, dry fruits
181 red, long, dry fruits
2 green, round, dry fruits
50 green, long, dry fruits
175 green, round, juicy fruits
44 red, round, juicy fruits
4 red, long, juicy fruits
i) Write the genotypes of the parents that were used to produce the F1 heterozygote. Explain,
Show all calculations (4 points)
a) LRS/LRS and lrs/lrs
b) LRS/lrs and LRS/lrs
c) L/L, R/R, s/s and l/l, r/r, S/S
d) L/L, R/R, s/s and l/l, r/r, s/s
e) Lrs/Lrs and lRS/lRS
ii) Which of these three genes (L, R, or S) is in the medium? Explain (2 points)
a) L is in the middle
b) R is in the middle
c) S is in the middle
d) none of the above
e) the answers a, b, and c are possible
iii) What is the distance between the genes R and L? Show all calculations. (2 points)
a) 8.8 cM
b) 10 cM
c) 20 cM
d) 30 cM
e) 35 cM
iv) What is the distance between the R and S genes? Show all calculations. (2 points)
a) 10 cM
b) 14 cM
c) 20 cM
d) 30 cM
e) 35 cM
iv) What is the distance between the L and S genes? Show all calculations. (2 points)
a) 5 cM
b) 10 cM
c) 15 cM
d) 20 cM
e) 30 cM
v) What is the coincidence coefficient? Show all calculations (2 points)
a) 0
b) 0.2
c) 0.4
d) 0.6
e) 0.8.
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Btt2153 7 population genetics_print1
Here are the solutions to the try these problems:
1. a) T allele frequency = (88/125) + (1/2)*(37/125) = 0.7
t allele frequency = 1 - 0.7 = 0.3
b) TT genotype frequency = 0.49, Tt genotype frequency = 0.42, tt genotype frequency = 0.09
2. p = 200/300 = 0.667, q = 100/300 = 0.333
3. a) Recessive allele frequency = 1/2500 = 0.0004
b) Dominant allele frequency = 1 - 0.0004 = 0.9996
c) Heterozygous
Problems2. This is a map for a diploid plantR--------35-------.pdf
Problems:
2. This is a map for a diploid plant:
R--------35--------E--------17---------F
One parental plant is R e f/r E F. It is mated with a homozygous recessive plant: r e f/r e f. Give
the proportion of the offspring assuming there is both no interference, and complete interference.
3. In a newly discovered insect, the three genes R, A, and P located on the same chromosome
(R= red body, A= spotted coloring, P=purple eyes, r=white body, a=solid coloring, p=white
eyes). A heterozygous parent(R A P/r a p) is mated with a homozygous recessive parent (r a p/ r
a p).
The following offspring are shown below:
(210) dominant for all traits
(16) red body, solid coloring, purple eyes
(70) red body, spotted coloring, white eyes
(50) red body, solid coloring, white eyes
(200) recessive for all traits
(55) white body, spotted coloring, purple eyes
(13) white body, spotted coloring, white eyes
(72) white body, solid coloring, purple eyes
Perform a three-point linkage to determine the gene order and relative distance between the
genes. How did the distance between the 2 genes affect the frequency of crossing-over events?
4. Here are four recombinant sequences that came from four Hfr matings during an interrupted
mating experiment. Determine the order and relative distance by setting A to 0/100 minutes.
Hfr1) A T1 H T2
8 26 33 53
Hfr2) E1 T1 E2 E3
4 12 24 54
Hfr3) T2 S T3 A
7 37 52 62
Hfr4) T3 C E1 E3
2 17 22 72
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growth on both plates. He then repeated this experiment, but this time he added DNase to the
tube and observed the same results when he plated both cell types. What happened inside the
tube?
6. Pretend the following is a chromosome that underwent some changes. What were they? (the
dot is the centromere).
A B C D E F . G H I J A B C G . F E H I J
a. pericentric inversion, deletion
b. paracentric inversion, deletion
c. deletion, translocation
d. Robertsonian translocation,
pericntric inversion
7. Given:
Type of mutation Cistron A or B
1. point A
2. point A
3. point B
4. point B
5. deletion A
Results from 5 crosses:
1x2 (plaque) 2x3 (plaque) 3x4 (plaque) 4x5 (plaque)
1x3 (plaque) 2x4 (plaque) 3x5 (plaque)
1x4 (plaque) 2x5 (no plaque)
1x5 (plaque)
8. The following nucleotide sequence is found in a piece of DNA
5’-TTAACA-3’
3’-AATTGT-5’
if the cytosince undergoes a tautameric shift for one replication cycle draw two round of
replication and explain what type of mutation occurred (Transition or Transversion)?
Bonus:
Explain the relationship between mosaicism and x-inactivation using an example.
9. In a diploid fly, the three loci R E and D are linked as follows
R--------------22cm------------E----------16cm-------------D
One fly is available to you and its genotype is R.
Ftt1033 7 population genetics-2013
Here are the solutions to the try these problems:
1. a) T allele frequency = (88/125) + (1/2)*(37/125) = 0.72
t allele frequency = 1 - 0.72 = 0.28
b) TT genotype frequency = 0.72^2 = 0.52
Tt genotype frequency = 2*0.72*0.28 = 0.48
tt genotype frequency = 0.28^2 = 0.08
2. p = 200/300 = 0.667
q = 1 - p = 0.333
3. a) Recessive allele frequency = √(1/2500) = 0.02
b
Linkage_Crossing_over.pptx
1) Linkage is the tendency of genes located near each other on the same chromosome to be inherited together during meiosis. Genes farther apart on a chromosome assort independently according to Mendel's law of independent assortment.
2) The strength of linkage depends on the physical distance between genes - closely located genes show complete linkage while genes farther apart show incomplete linkage with crossover occurring.
3) Tetrad analysis in fungi allows the direct observation of linkage and estimation of recombination frequencies between genes and centromeres.
Association mapping
This document summarizes an association mapping study of seed oil and protein contents in upland cotton. 180 cotton accessions were genotyped using 228 SSR markers and phenotyped for oil and protein content over multiple locations and years. Population structure analysis identified two subpopulations. Association analysis identified 86 marker-trait associations between 58 SSR markers and the two traits, with 15 and 12 markers associated with oil and protein content respectively. 18 markers were significantly associated with the traits in more than one environment, with 9 markers associated with both oil and protein content simultaneously and stably across locations.
In one species of the plant coding for color- shape and amount of wate.pdf
In one species of the plant coding for color, shape and amount of water in the fruit are on the
same chromosome. The gene responsible for the color of the fruit has two alleles (R = red,
r=green), the gene responsible for the shape of the fruit also has two alleles (L =long, l=round),
and the gene responsible for the amount of water in the fruit also has two alleles (S dry, s=juicy).
The alleles represented by capital letters are dominant over the alleles represented by small
letters.
Crossing a pure line plant with red, long, dry flowers with a pure line plant with green, round,
juicy fruits results in an F1 plant with rough, long, dry fruits. The F1 individuals are crossed with
a pure line plant having green, round and juicy fruits and an F2 is obtained composed of:
22 green, long, juicy fruits
22 red, round, dry fruits
181 red, long, dry fruits
2 green, round, dry fruits
50 green, long, dry fruits
175 green, round, juicy fruits
44 red, round, juicy fruits
4 red, long, juicy fruits
i) Write the genotypes of the parents that were used to produce the F1 heterozygote. Explain,
Show all calculations (4 points)
a) LRS/LRS and lrs/lrs
b) LRS/lrs and LRS/lrs
c) L/L, R/R, s/s and l/l, r/r, S/S
d) L/L, R/R, s/s and l/l, r/r, s/s
e) Lrs/Lrs and lRS/lRS
ii) Which of these three genes (L, R, or S) is in the medium? Explain (2 points)
a) L is in the middle
b) R is in the middle
c) S is in the middle
d) none of the above
e) the answers a, b, and c are possible
iii) What is the distance between the genes R and L? Show all calculations. (2 points)
a) 8.8 cM
b) 10 cM
c) 20 cM
d) 30 cM
e) 35 cM
iv) What is the distance between the R and S genes? Show all calculations. (2 points)
a) 10 cM
b) 14 cM
c) 20 cM
d) 30 cM
e) 35 cM
iv) What is the distance between the L and S genes? Show all calculations. (2 points)
a) 5 cM
b) 10 cM
c) 15 cM
d) 20 cM
e) 30 cM
v) What is the coincidence coefficient? Show all calculations (2 points)
a) 0
b) 0.2
c) 0.4
d) 0.6
e) 0.8.
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Thomas Morgan observed that genes on the same chromosome can be linked during meiosis. He conducted test crosses on Drosophila with traits for eye color and wing type. The results showed that some offspring received combinations of traits that were not expected if the genes assorted independently, demonstrating genetic linkage. Genetic linkage occurs because homologous chromosomes occasionally exchange sections during meiosis through the process of crossover. Crossover results in new combinations of genes that can become separated from their original linked combinations on the chromosome. The frequency of crossover between genes determines how far apart they will be on the genetic map of the chromosome.
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A quick recap….
There are several terms that need to be clarified so that you can more easily follow the exercise. A gene is a
piece of DNA that directs the expression of a particular characteristic (trait). Genes are located on
chromosomes, and the location where a particular gene is found is referred to as the locus (plural: loci) of that
gene. An allele is a gene for which there is an alternative expression, which can lead to the alterative form of a
trait. For example, a diploid organism carries the allele “A” on one homologous chromosome, and the allele “A”
on the other. The genotype of this organism is then AA and it is said to be homozygous. An organism may also
carry two different alleles. For example on one chromosome it could carry the allele “A” and on the other it
could carry the allele “a”. The genotype of such an organisms is then Aa, and it is described as heterozygous for
this chromosomal locus.
The genotype of an organism is the listing of the two alleles for each trait that it possesses. The phenotype of an
organism is a description of the way a trait is displayed in the structure, behavior, or physiology of the organism.
Some alleles are dominant to others and mask the presence of other alleles. The dominant condition is indicated
by uppercase letters (e.g., “A”). The alleles that are masked are called recessive alleles. The recessive condition
is indicated by lowercase letters (e.g., “a”). When both dominants are present in the genotype (AA), the organism
is said to be homozygous dominant for the trait, and the organisms will show the dominant phenotype (trait
expression A). When both recessives are present in the genotype (aa), the organism is said to be homozygous
recessive for the trait, and the organisms will show the recessive phenotype (trait expression a). In the case of
complete dominance, the dominant allele completely masks the recessive allele, and an organism with a
heterozygous genotype (Aa) will show the dominant phenotype (trait expression A).
Evolutionary Agents
Evolution is a process resulting in changes in gene frequencies (= the genetic make-up) of a population over
time. The mechanisms of evolution include selection (which can cause change over time & adaptation), and
forces that provide variation and cause change over time (but not adaptation). Factors that change gene
frequencies over time are referred to as evolutionary agents.
A powerful way to detect the presence of evolutionary agents is the use of the Hardy–Weinberg model. This
model can be applied to traits that are influenced by several loci; the simplest case is for a trait that is regulated
by one locus with two alleles.
With the Hardy–Weinberg model, the frequency of genotypes in the population can be predicted from the
probability of encounters between gametes bearing the different alleles. With alleles R .
genetic linkage and gene mapping
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The Molecular Life of Diatoms
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Collège de France
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http://events.embo.org/13-diatom/
3. You have co-infected E. coli with two strains of T4 phage. One is.pdf
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and turbid (tu). The other is wildtype for these genes. The products of this infection are plated
and classified for each phenotype, with the following results: m r tu 3467 + + + 3729 m r + 853
m + tu 162 m + + 520 + r tu 474 + r + 172 + + tu 965 a. Determine recombination frequencies
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between the genes in terms of recombination frequencies?
Solution
(1) First, try to know the parental genotype
Parent 1: + + +
Parent 2: m r tu
Now to know the distance between genes, we need to know the recombinant progeny
Distance between m and r
Recombinant progeny
Number
m + tu
162
m + +
520
+ r tu
474
+ r +
172
Total
1328
The map distance between m and r is Recombination frequency= (Number of
recombinants/Total progeny)x100=1328/10,342x100 = 12.8 m.u.
Recombinant progeny
Number
m r +
853
m + tu
162
+ r +
172
+ + tu
965
Total
2152
Using the same approach, the r–tu distance is Recombination frequency= (Number of
recombinants/Total progeny)x100 =(2152)/10,342x100 = 20.8 m.u
Recombinant progeny
Number
m r +
853
m + +
520
+ r tu
474
+ + tu
965
total
2812
Using the same approach, the m -tu distance is Recombination frequency= (Number of
recombinants/Total progeny)x100 =(2812)/10,342x100 = 27.2 m.u.
b. As the genes m and tu are the farthest apart, you can easily predict the gene order m r tu
c. m--------12.8 m.u---------r---------------------20.8 m.u--------------------------tu
Recombinant progeny
Number
m + tu
162
m + +
520
+ r tu
474
+ r +
172
Total
1328.
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2) They found the two tomato genomes show only 0.6% nucleotide divergence, indicating recent admixture, but more than 8% divergence from potato.
3) Comparative analysis revealed the Solanum lineage experienced two genome triplications, one ancient and shared with other plants and one more recent, providing material for genes controlling fruit traits like color and fleshiness to evolve.
Secuencian el genoma del tomate
Por primera vez, el genoma del tomate (Solanum lycopersicum) , ha sido descifrado completamente. Esto es en un paso importante hacia la mejora del rendimiento, la nutrición, la resistencia a las enfermedades, el sabor y el color del tomate y otros cultivos similares. La secuencia completa del genoma de la variedad Heinz 1706 así como la secuencia de un pariente silvestre, fue publicado en el último número de la revista Nature .
Los investigadores del Consorcio formado por investigadores de numerosos países informó que los tomates tienen unos 35.000 genes organizados en 12 cromosomas.
Las secuencias de estos genes y su disposición en los cromosomas se describen en el artículo de Nature. Esta información permitirá a los investigadores avanzar a un ritmo más rápido y a los criadores de plantas para producir nuevas variedades con determinadas características deseadas.
Ahora que la secuencia del genoma de esta variedad de tomate se conoce también será más fácil y mucho menos costoso conocer la secuencia de otras variantes de esta planta. Teniendo como molde la secuencia de esta variedad que costó millones de dólares permitirá conocer a muy bajo costo (10,000 dólares o menos) las secuencias de otras variedades.
Para facilitar el acceso a las secuencias de los genes del tomate y las especies afines han creado un sitio web interactivo ( www.solgenomics.net ).
La secuenciación del genoma del tomate tiene implicaciones para otras especies de plantas. Fresas, manzanas, melones, plátanos y otros muchos frutos carnosos, quienes comparten algunas características con los tomates, por lo que la información sobre los genes y las vías involucradas en la maduración del fruto potencialmente se puede aplicar a ellos, ayudando a mejorar la calidad de los alimentos, la seguridad alimentaria y reducir los costos de cultivo.
SPECIES DIVERSITY FOR HIGH SCHOOL STUDENTS
This document discusses species diversity and how to calculate an index of diversity. It defines species diversity as the number of different species and abundance of each species in a community. The index of diversity takes into account both the number of species and number of individuals to better measure diversity than just counting species. It provides a formula to calculate the index and works through an example comparing diversity in two habitats. A higher index number indicates more diverse areas.
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There are three main methods for genetic mapping and transfer of DNA in bacteria: conjugation, transformation, and transduction. Conjugation involves the transfer of genetic material between bacteria through direct contact, often mediated by plasmids or sex pili. Transformation is the uptake of extracellular DNA by competent bacteria. Transduction occurs when bacteriophages transfer bacterial DNA between cells during infection. These methods are used to construct genetic maps of bacteria and their viruses through interrupted mating experiments and analysis of recombinant frequencies.
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2015 03-13 Linkage and Crossover
Thomas Morgan observed that genes on the same chromosome can be linked during meiosis. He conducted test crosses on Drosophila with traits for eye color and wing type. The results showed that some offspring received combinations of traits that were not expected if the genes assorted independently, demonstrating genetic linkage. Genetic linkage occurs because homologous chromosomes occasionally exchange sections during meiosis through the process of crossover. Crossover results in new combinations of genes that can become separated from their original linked combinations on the chromosome. The frequency of crossover between genes determines how far apart they will be on the genetic map of the chromosome.
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Foundations of Biological Sciences I Evolutionary Agents - 1 .docx
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A quick recap….
There are several terms that need to be clarified so that you can more easily follow the exercise. A gene is a
piece of DNA that directs the expression of a particular characteristic (trait). Genes are located on
chromosomes, and the location where a particular gene is found is referred to as the locus (plural: loci) of that
gene. An allele is a gene for which there is an alternative expression, which can lead to the alterative form of a
trait. For example, a diploid organism carries the allele “A” on one homologous chromosome, and the allele “A”
on the other. The genotype of this organism is then AA and it is said to be homozygous. An organism may also
carry two different alleles. For example on one chromosome it could carry the allele “A” and on the other it
could carry the allele “a”. The genotype of such an organisms is then Aa, and it is described as heterozygous for
this chromosomal locus.
The genotype of an organism is the listing of the two alleles for each trait that it possesses. The phenotype of an
organism is a description of the way a trait is displayed in the structure, behavior, or physiology of the organism.
Some alleles are dominant to others and mask the presence of other alleles. The dominant condition is indicated
by uppercase letters (e.g., “A”). The alleles that are masked are called recessive alleles. The recessive condition
is indicated by lowercase letters (e.g., “a”). When both dominants are present in the genotype (AA), the organism
is said to be homozygous dominant for the trait, and the organisms will show the dominant phenotype (trait
expression A). When both recessives are present in the genotype (aa), the organism is said to be homozygous
recessive for the trait, and the organisms will show the recessive phenotype (trait expression a). In the case of
complete dominance, the dominant allele completely masks the recessive allele, and an organism with a
heterozygous genotype (Aa) will show the dominant phenotype (trait expression A).
Evolutionary Agents
Evolution is a process resulting in changes in gene frequencies (= the genetic make-up) of a population over
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frequencies over time are referred to as evolutionary agents.
A powerful way to detect the presence of evolutionary agents is the use of the Hardy–Weinberg model. This
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by one locus with two alleles.
With the Hardy–Weinberg model, the frequency of genotypes in the population can be predicted from the
probability of encounters between gametes bearing the different alleles. With alleles R .
genetic linkage and gene mapping
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The Molecular Life of Diatoms
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Collège de France
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http://events.embo.org/13-diatom/
3. You have co-infected E. coli with two strains of T4 phage. One is.pdf
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Solution
(1) First, try to know the parental genotype
Parent 1: + + +
Parent 2: m r tu
Now to know the distance between genes, we need to know the recombinant progeny
Distance between m and r
Recombinant progeny
Number
m + tu
162
m + +
520
+ r tu
474
+ r +
172
Total
1328
The map distance between m and r is Recombination frequency= (Number of
recombinants/Total progeny)x100=1328/10,342x100 = 12.8 m.u.
Recombinant progeny
Number
m r +
853
m + tu
162
+ r +
172
+ + tu
965
Total
2152
Using the same approach, the r–tu distance is Recombination frequency= (Number of
recombinants/Total progeny)x100 =(2152)/10,342x100 = 20.8 m.u
Recombinant progeny
Number
m r +
853
m + +
520
+ r tu
474
+ + tu
965
total
2812
Using the same approach, the m -tu distance is Recombination frequency= (Number of
recombinants/Total progeny)x100 =(2812)/10,342x100 = 27.2 m.u.
b. As the genes m and tu are the farthest apart, you can easily predict the gene order m r tu
c. m--------12.8 m.u---------r---------------------20.8 m.u--------------------------tu
Recombinant progeny
Number
m + tu
162
m + +
520
+ r tu
474
+ r +
172
Total
1328.
Final horizontal gene transfer
This document discusses various mechanisms of horizontal gene transfer in bacteria, including conjugation, transformation, and transduction. It provides details on the genes and proteins involved in these processes, such as the tra genes for conjugation and the com genes for competence in transformation. Examples are given for using these mechanisms to map genes and determine genetic distances. Order of events in phage lifecycles and the difference between lytic and lysogenic cycles are also covered.
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International Journal of Engineering and Science Invention (IJESI) is an international journal intended for professionals and researchers in all fields of computer science and electronics. IJESI publishes research articles and reviews within the whole field Engineering Science and Technology, new teaching methods, assessment, validation and the impact of new technologies and it will continue to provide information on the latest trends and developments in this ever-expanding subject. The publications of papers are selected through double peer reviewed to ensure originality, relevance, and readability. The articles published in our journal can be accessed online.
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Secuencian el genoma del tomate
Por primera vez, el genoma del tomate (Solanum lycopersicum) , ha sido descifrado completamente. Esto es en un paso importante hacia la mejora del rendimiento, la nutrición, la resistencia a las enfermedades, el sabor y el color del tomate y otros cultivos similares. La secuencia completa del genoma de la variedad Heinz 1706 así como la secuencia de un pariente silvestre, fue publicado en el último número de la revista Nature .
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SPECIES DIVERSITY FOR HIGH SCHOOL STUDENTS
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There are three main methods for genetic mapping and transfer of DNA in bacteria: conjugation, transformation, and transduction. Conjugation involves the transfer of genetic material between bacteria through direct contact, often mediated by plasmids or sex pili. Transformation is the uptake of extracellular DNA by competent bacteria. Transduction occurs when bacteriophages transfer bacterial DNA between cells during infection. These methods are used to construct genetic maps of bacteria and their viruses through interrupted mating experiments and analysis of recombinant frequencies.
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unos ingresos de unos 150 millones de euros en 2012, un 29 % ms que el ao anterior.
Historia
En 1907, Eugene Schueller, un joven qumico francs, en colaboracin con La Cagoule, desarroll
una frmula de tinte para el cabello llamada Aurale. Schueller formul y fabric sus propios
productos, que luego vendi a los peluqueros parisinos. En 1919, Schueller registr su empresa
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Los principios rectores de la empresa fueron la investigacin y la innovacin en el campo de la
belleza. En 1920, L'Oral emple a tres qumicos. En 1950, los equipos d.
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