6- The gene for curled wings (C-c) is located nearly at the end of the.docx
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6. The gene for curled wings (C/c) is located nearly at the end of the long arm of chromosome 2 of Drosophila, while the gene for smooth abdomen (S/s) is nearly at the end of the short arm of the same chromosome (the centromere is located about 2/3 of the distance along the chromosome). The gene for ebony body (E/e) is near the end of chromosome 3. A true-breeding wild type fly is crossed with a true-breeding fly that has curled wings, smooth abdomen and ebony body (all recessive traits). (a) Write the genotypes of the parents and the offspring (F1). (b) Sketch the arrangement of chromosomes 2 and 3 at metaphase of meiosis 1 in the gamete-producing cells of the F1. (Assume no crossing-over). (c) Sketch all possible combinations of the two chromosomes in the gametes. (Assume no crossing-over). (d) If the F1 is crossed with a triple recessive fly (known as a backcross), what would be the phenotypes of the progeny (assume no crossing-over). What would you expect their relative numbers to be? (e) If crossing-over occurred what difference would you expect in the results of the F1 backcross? .
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A three point cross was performed using fruit flies. The three loci .pdf
A three point cross was performed using fruit flies. The three loci are physically linked on the
same autosome. The d allele, in the homozygous state, results in the recessive ‘double dorsal’
phenotype. Similarly, the r allele results in the recessive ‘round ears’ phenotype and the e allele
in the recessive ‘endless eye’ phenotype. In contrast, the D, R, and E alleles encode for the
dominant wild type phenotypes at each of these loci.
True breeding (homozygous) flies were crossed to produce F1 offspring. Due to poor book-
keeping, the genotypes of the parent flies were not recorded – but one parent fly had one mutant
phenotype, while the other parent had the other two mutant phenotypes.
PhenotypeNumber of F2 Flies
EarsDorsalEyes
1) Round ears Double Dorsal Endless Eyes 230
2) Wild-type Wild-type Wild-type 260
3) Round ears Wild-type Wild-type 4,390
4) Wild-type Double Dorsal Endless Eyes 4,100
5) Round ears Double Dorsal Wild-type 15
6) Wild-type Wild-type Endless Eyes 15
7) Round ears Wild-type Endless Eyes 510
8) Wild-type Double Dorsal Wild-type 480
10,000 total
a) What is the phenotype of the F1 flies?
b) What is the genotype of the F1 flies? Your answer should indicate the phase of linkage.
c) What are the genotypes and phenotypes of the original parent flies (previously not recorded)?
Your answer should indicate the phase of linkage.
d) Calculate the map distance in cM between each of the three loci and determine the gene order.
Show all work for full credit. (6 points)
Solution
a. The phenotype of the F1 flies is wild-type heterozygous at all locus.
b. The genotype of F1 flies
r D E/ R d e
c. The genotypes of parent flies: r D E and R d e
The phenotypes of parent flies: Round ears, Wild-type, Wild-type and Wild-type, Double Dorsal,
Endless Eyes
d.
Gene order is R------D------E
The distance between R and D
510+480+15+15 / 10,000 * 100 = 10.2cM
The distance between D and E
230+260+15+15 / 10,000 * 100 = 5.2cM.
AP Biology Ch 12 gene linkage groups and chromosome maps
1. The document discusses linkage groups and chromosome mapping in genetics. It provides examples of crosses involving fruit flies that demonstrate genes located on the same chromosome tend to be inherited together, while genes on different chromosomes assort independently.
2. The document explains how observing recombinant vs parental offspring types allows geneticists to determine if genes are linked and calculate the distance between genes on a chromosome.
3. It provides practice problems demonstrating crosses to determine linkage and asks the reader questions about linkage and chromosome mapping concepts.
1.) The normal eye color of Drosophila is red, but strains in which .pdf
1.) The normal eye color of Drosophila is red, but strains in which all flies have brown eyes are
available. Similarly, wings are normally long, but there are strains with short wings. A female
from a pure line with brown eyes and short wings is crossed with a male from a normal pure line.
The F1 consists of normal females and short-winged males. An F2 is then produced by
intercrossing the F1. Both sexes of F2 flies show phenotypes as follows:
3/8 red eyes, long wings
3/8 red eyes, short wings
1/8 brown eyes, long wings
1/8 brown eyes, short wings
Deduce the inheritance of these phenotypes; use clearly defined genetic symbols of your own
invention. State the genotypes of all three generations and the genotypic proportions of the F1
and F2.
2.) When a cell of genotype A/a ; B/b ; C/c having all the genes on separate chromosome pairs
divides mitotically, what are the genotypes of the daughter cells?
3.) A presumed dihybrid in Drosophila, B/b ; F/f, is test-crossed with b/b ; f/f. (B = black body ;
b = brown body; F = forked bristles; f = unforked bristles.) The results are
black, forked 230
brown, forked 240
black, unforked 210
brown, unforked 250
Use the 2 test to determine if these results fit the results expected from testcrossing the
hypothesized dihybrid.
4.) In mice, dwarfism is caused by an X-linked recessive allele, and pink coat is caused by an
autosomal dominant allele (coats are normally brownish). If a dwarf female from a pure line is
crossed with a pink male from a pure line, what will be the phenotypic ratios in the F1 and F2 in
each sex? (Invent and define your own gene symbols.)
Solution
Answer 2. If a cell divides mitotically (equational division), the genetic constitution of the
daughter cells remains the same as the parents. So a parent with the genotype A/a; B/b; C/c will
give rise to daughter cells with the genotype A/a; B/b; C/c.
However, if the cell was to divide meiotically (reductional division), the genetic constitution
would be different as only one half of the parental chromosomes goes to the daughter cell. So the
genotypes would be as follows:
ABC or AbC or ABc or Abc or aBC or abC or aBc or abc
Answer 3. For the given cross
BbFf X bbff
For such a test cross the resultant phenotypic ration obtained is 1 (BF):1 (Bf):1 (bF):1 (bf).
For dihybrid cross, the degree of freedom will be 4 – 1 = 3
Chi square test is as follows
BF
Bf
bF
bf
Totals
Observed phenotype
230
210
240
250
930
Expected phenotype
930x ¼= 232.5
930x ¼= 232.5
930x ¼= 232.5
930x ¼= 232.5
930
Observed – expected
-2.5
-22.5
7.5
17.5
(Obs – Exp)2
Expected
6.25/232.5= 0.0268
506.25/232.5= 2.1774
56.25/232.5= 0.2419
306.25/232.5= 1.3172
3.7633
Looking at the probability value table for degree of freedom = 3,
df
.90
.70
.50
.30
.20
.10
.05
.01
3
0.58
1.42
2.37
3.67
4.64
6.25
7.82
11.35
Hence we can deduce that there is more than 20% probability for the occurrence of this event.
Answer 4. Let XD = Normal phenotype (tall)
Xd = Dwarf phenotype
P = Pink coat
p = Br.
2- In Drosophila- a cross was made between females - all expressing th.docx
In a cross between female flies expressing three X-linked recessive traits (scute bristles, sable body, and vermillion eyes) and wild-type males, all F1 generation females were wild-type while males expressed all three mutations. Analysis of over 1000 F2 generation offspring found deviations from expected numbers of double crossovers, suggesting the genes are arranged in a specific order along the X chromosome with some interference between crossovers.
29 What is the phenotypic distribution for barredeye wild.pdf
2.9) What is the phenotypic distribution for barredeye, wild-type, and intermediate-eye in the F2
generation? Remember to also include information about the sex of the individuals. Identify the
homozygous, heterozygous, and hemizygous indivduals. Which phenotypes are observed in the
F2 that were not in the parental? 3) Dihybrid cross: The dark-bodied and rudimentary winged
genes are on two different autosomal chromosomes. We will use this cross to determine the ratios
of phenotypes in the F2 generation when there are two genes involved. 3.1) Use the following
symbols to write out the cross: Male: 8 Female: ? Dark body color: DB Wild-type body color: wt
Rudimentary wings: Rw Wild-type wings: wtRiw 3.2) What would the cross look like if we mated a
Dark-Bodied/Rudimentary winged (double mutant) male with a female that is wild type for both
traits? Place the sex of the individual (symbol) on the first set of lines and their respective
genotypes in the second set. Sex: Genotypes x 3.3) What are the gametes produced by either sex
in the parental generation? Place the symbol for sex on the first set of lines and write down the
gametes produced by that sex in the following set of lines, respectively. Sex:3.3) What are the
gametes produced by either sex in the parental generation? Place the symbol for sex on the first
set of lines and write down the gametes produced by that sex in the following set of lines,
respectively. Sex: Gametes produced: 3.4) Using the gametes that you produced from the
previous question, fill in the Punnet square and show the resulting offspring from the cross. Place
female gametes to the left of the Punnet square and male gametes above. 3.5) What phenotype
should the F1 offspring have? 3.6) What would the F1 cross look like if we mated a male F1 and a
female F1? Place the sex of the individual on the first set of lines and their respective genotypes in
the second set. Sex: Genotypes x 3.7) What are the gametes produced by either sex in the F1
generation? Place the symbol for sex on2.4) Using the gametes that you produced from the
previous question, fill in the Punnet square and show the resulting offspring from the cross. Place
female gametes to the left of the Punnet square and male gametes above. 2.5) What is the
phenotypic distribution for barredeye, wild-type, and intermediate-eye? Remember to also include
information about the sex of the individuals. Identify the homozygous, heterozygous, and
hemizygous indivduals. Why aren't there any wild-type females? 2.6) What would the F1 cross
look like if we mated a male F1 and a female F1? Place the sex (use the symbol) of the individual
on the first set of lines and their respective genotypes in the second set (the genotypes should
also refect the sex). Sex: Genotypes x 2.7) What are the gametes produced by either sex in the F1
generation? Place the genotypes (these genotypes should display the sex and the trait we are
studying) on the first set of lines and write down the gamet.
Ch 12 gene linkage groups and practice problems
1. The document discusses genetics concepts like linkage groups, chromosome mapping, parental and recombinant types. It provides examples of genetic crosses in pea plants and fruit flies to demonstrate these concepts.
2. Recombinant types result from crossovers during meiosis, where alleles switch positions on homologous chromosomes. The frequency of recombinants allows calculation of distance between genes.
3. A sample genetics problem is presented, asking the reader to analyze a dihybrid cross in pea plants and determine the chromosome arrangement and distance between genes.
Linkage and recombination of gene
- A three-point test cross was conducted in Drosophila to map the locations of three genes - black body (b), purple eyes (pr), and curved wings (c) - on the same chromosome.
- Offspring were classified into eight phenotypic classes. Reciprocal classes occurred in roughly equal numbers, indicating recombination between loci.
- Map distances were calculated between loci based on percentages of recombinants. A discrepancy in the direct vs. indirect distance between b and c was resolved by accounting for double crossovers.
- The study demonstrated how three-point test crosses can order loci and more accurately measure map distances compared to two-point crosses by detecting double crossovers.
Insect wings
This document summarizes the key aspects of insect wing anatomy and function. It describes how wings develop as outgrowths of the integument, with tracheae forming the supportive wing veins. It discusses wing venation patterns, circulation, articulation, diversity across insect orders, and variations in size, texture, coloration, resting position and more. The Comstock-Needham system for describing major longitudinal veins and vein branches is also summarized.
Recommended
A three point cross was performed using fruit flies. The three loci .pdf
A three point cross was performed using fruit flies. The three loci are physically linked on the
same autosome. The d allele, in the homozygous state, results in the recessive ‘double dorsal’
phenotype. Similarly, the r allele results in the recessive ‘round ears’ phenotype and the e allele
in the recessive ‘endless eye’ phenotype. In contrast, the D, R, and E alleles encode for the
dominant wild type phenotypes at each of these loci.
True breeding (homozygous) flies were crossed to produce F1 offspring. Due to poor book-
keeping, the genotypes of the parent flies were not recorded – but one parent fly had one mutant
phenotype, while the other parent had the other two mutant phenotypes.
PhenotypeNumber of F2 Flies
EarsDorsalEyes
1) Round ears Double Dorsal Endless Eyes 230
2) Wild-type Wild-type Wild-type 260
3) Round ears Wild-type Wild-type 4,390
4) Wild-type Double Dorsal Endless Eyes 4,100
5) Round ears Double Dorsal Wild-type 15
6) Wild-type Wild-type Endless Eyes 15
7) Round ears Wild-type Endless Eyes 510
8) Wild-type Double Dorsal Wild-type 480
10,000 total
a) What is the phenotype of the F1 flies?
b) What is the genotype of the F1 flies? Your answer should indicate the phase of linkage.
c) What are the genotypes and phenotypes of the original parent flies (previously not recorded)?
Your answer should indicate the phase of linkage.
d) Calculate the map distance in cM between each of the three loci and determine the gene order.
Show all work for full credit. (6 points)
Solution
a. The phenotype of the F1 flies is wild-type heterozygous at all locus.
b. The genotype of F1 flies
r D E/ R d e
c. The genotypes of parent flies: r D E and R d e
The phenotypes of parent flies: Round ears, Wild-type, Wild-type and Wild-type, Double Dorsal,
Endless Eyes
d.
Gene order is R------D------E
The distance between R and D
510+480+15+15 / 10,000 * 100 = 10.2cM
The distance between D and E
230+260+15+15 / 10,000 * 100 = 5.2cM.
AP Biology Ch 12 gene linkage groups and chromosome maps
1. The document discusses linkage groups and chromosome mapping in genetics. It provides examples of crosses involving fruit flies that demonstrate genes located on the same chromosome tend to be inherited together, while genes on different chromosomes assort independently.
2. The document explains how observing recombinant vs parental offspring types allows geneticists to determine if genes are linked and calculate the distance between genes on a chromosome.
3. It provides practice problems demonstrating crosses to determine linkage and asks the reader questions about linkage and chromosome mapping concepts.
1.) The normal eye color of Drosophila is red, but strains in which .pdf
1.) The normal eye color of Drosophila is red, but strains in which all flies have brown eyes are
available. Similarly, wings are normally long, but there are strains with short wings. A female
from a pure line with brown eyes and short wings is crossed with a male from a normal pure line.
The F1 consists of normal females and short-winged males. An F2 is then produced by
intercrossing the F1. Both sexes of F2 flies show phenotypes as follows:
3/8 red eyes, long wings
3/8 red eyes, short wings
1/8 brown eyes, long wings
1/8 brown eyes, short wings
Deduce the inheritance of these phenotypes; use clearly defined genetic symbols of your own
invention. State the genotypes of all three generations and the genotypic proportions of the F1
and F2.
2.) When a cell of genotype A/a ; B/b ; C/c having all the genes on separate chromosome pairs
divides mitotically, what are the genotypes of the daughter cells?
3.) A presumed dihybrid in Drosophila, B/b ; F/f, is test-crossed with b/b ; f/f. (B = black body ;
b = brown body; F = forked bristles; f = unforked bristles.) The results are
black, forked 230
brown, forked 240
black, unforked 210
brown, unforked 250
Use the 2 test to determine if these results fit the results expected from testcrossing the
hypothesized dihybrid.
4.) In mice, dwarfism is caused by an X-linked recessive allele, and pink coat is caused by an
autosomal dominant allele (coats are normally brownish). If a dwarf female from a pure line is
crossed with a pink male from a pure line, what will be the phenotypic ratios in the F1 and F2 in
each sex? (Invent and define your own gene symbols.)
Solution
Answer 2. If a cell divides mitotically (equational division), the genetic constitution of the
daughter cells remains the same as the parents. So a parent with the genotype A/a; B/b; C/c will
give rise to daughter cells with the genotype A/a; B/b; C/c.
However, if the cell was to divide meiotically (reductional division), the genetic constitution
would be different as only one half of the parental chromosomes goes to the daughter cell. So the
genotypes would be as follows:
ABC or AbC or ABc or Abc or aBC or abC or aBc or abc
Answer 3. For the given cross
BbFf X bbff
For such a test cross the resultant phenotypic ration obtained is 1 (BF):1 (Bf):1 (bF):1 (bf).
For dihybrid cross, the degree of freedom will be 4 – 1 = 3
Chi square test is as follows
BF
Bf
bF
bf
Totals
Observed phenotype
230
210
240
250
930
Expected phenotype
930x ¼= 232.5
930x ¼= 232.5
930x ¼= 232.5
930x ¼= 232.5
930
Observed – expected
-2.5
-22.5
7.5
17.5
(Obs – Exp)2
Expected
6.25/232.5= 0.0268
506.25/232.5= 2.1774
56.25/232.5= 0.2419
306.25/232.5= 1.3172
3.7633
Looking at the probability value table for degree of freedom = 3,
df
.90
.70
.50
.30
.20
.10
.05
.01
3
0.58
1.42
2.37
3.67
4.64
6.25
7.82
11.35
Hence we can deduce that there is more than 20% probability for the occurrence of this event.
Answer 4. Let XD = Normal phenotype (tall)
Xd = Dwarf phenotype
P = Pink coat
p = Br.
2- In Drosophila- a cross was made between females - all expressing th.docx
In a cross between female flies expressing three X-linked recessive traits (scute bristles, sable body, and vermillion eyes) and wild-type males, all F1 generation females were wild-type while males expressed all three mutations. Analysis of over 1000 F2 generation offspring found deviations from expected numbers of double crossovers, suggesting the genes are arranged in a specific order along the X chromosome with some interference between crossovers.
29 What is the phenotypic distribution for barredeye wild.pdf
2.9) What is the phenotypic distribution for barredeye, wild-type, and intermediate-eye in the F2
generation? Remember to also include information about the sex of the individuals. Identify the
homozygous, heterozygous, and hemizygous indivduals. Which phenotypes are observed in the
F2 that were not in the parental? 3) Dihybrid cross: The dark-bodied and rudimentary winged
genes are on two different autosomal chromosomes. We will use this cross to determine the ratios
of phenotypes in the F2 generation when there are two genes involved. 3.1) Use the following
symbols to write out the cross: Male: 8 Female: ? Dark body color: DB Wild-type body color: wt
Rudimentary wings: Rw Wild-type wings: wtRiw 3.2) What would the cross look like if we mated a
Dark-Bodied/Rudimentary winged (double mutant) male with a female that is wild type for both
traits? Place the sex of the individual (symbol) on the first set of lines and their respective
genotypes in the second set. Sex: Genotypes x 3.3) What are the gametes produced by either sex
in the parental generation? Place the symbol for sex on the first set of lines and write down the
gametes produced by that sex in the following set of lines, respectively. Sex:3.3) What are the
gametes produced by either sex in the parental generation? Place the symbol for sex on the first
set of lines and write down the gametes produced by that sex in the following set of lines,
respectively. Sex: Gametes produced: 3.4) Using the gametes that you produced from the
previous question, fill in the Punnet square and show the resulting offspring from the cross. Place
female gametes to the left of the Punnet square and male gametes above. 3.5) What phenotype
should the F1 offspring have? 3.6) What would the F1 cross look like if we mated a male F1 and a
female F1? Place the sex of the individual on the first set of lines and their respective genotypes in
the second set. Sex: Genotypes x 3.7) What are the gametes produced by either sex in the F1
generation? Place the symbol for sex on2.4) Using the gametes that you produced from the
previous question, fill in the Punnet square and show the resulting offspring from the cross. Place
female gametes to the left of the Punnet square and male gametes above. 2.5) What is the
phenotypic distribution for barredeye, wild-type, and intermediate-eye? Remember to also include
information about the sex of the individuals. Identify the homozygous, heterozygous, and
hemizygous indivduals. Why aren't there any wild-type females? 2.6) What would the F1 cross
look like if we mated a male F1 and a female F1? Place the sex (use the symbol) of the individual
on the first set of lines and their respective genotypes in the second set (the genotypes should
also refect the sex). Sex: Genotypes x 2.7) What are the gametes produced by either sex in the F1
generation? Place the genotypes (these genotypes should display the sex and the trait we are
studying) on the first set of lines and write down the gamet.
Ch 12 gene linkage groups and practice problems
1. The document discusses genetics concepts like linkage groups, chromosome mapping, parental and recombinant types. It provides examples of genetic crosses in pea plants and fruit flies to demonstrate these concepts.
2. Recombinant types result from crossovers during meiosis, where alleles switch positions on homologous chromosomes. The frequency of recombinants allows calculation of distance between genes.
3. A sample genetics problem is presented, asking the reader to analyze a dihybrid cross in pea plants and determine the chromosome arrangement and distance between genes.
Linkage and recombination of gene
- A three-point test cross was conducted in Drosophila to map the locations of three genes - black body (b), purple eyes (pr), and curved wings (c) - on the same chromosome.
- Offspring were classified into eight phenotypic classes. Reciprocal classes occurred in roughly equal numbers, indicating recombination between loci.
- Map distances were calculated between loci based on percentages of recombinants. A discrepancy in the direct vs. indirect distance between b and c was resolved by accounting for double crossovers.
- The study demonstrated how three-point test crosses can order loci and more accurately measure map distances compared to two-point crosses by detecting double crossovers.
Insect wings
This document summarizes the key aspects of insect wing anatomy and function. It describes how wings develop as outgrowths of the integument, with tracheae forming the supportive wing veins. It discusses wing venation patterns, circulation, articulation, diversity across insect orders, and variations in size, texture, coloration, resting position and more. The Comstock-Needham system for describing major longitudinal veins and vein branches is also summarized.
Drosophila Melanogaster Genome And its developmental process
The document summarizes key aspects of the Drosophila genome and life cycle. It notes that Drosophila has advantages for genetic studies like a short life cycle and small genome. Its genome contains around 13,600 genes located on four chromosomes. The life cycle involves an egg, larva, pupa and adult stages. Segmentation and homeotic genes play important roles in development by dividing the body into segments and specifying segment identities. Maternal effect, gap, pair-rule and segment polarity genes control segmentation in a hierarchical manner. Homeotic complexes like bithorax determine body part identities in each segment.
FLIES
1) This experiment uses statistical analysis of genetic crosses in fruit flies to determine the chromosomal location of two genes - ebony bodies and scarlet eyes.
2) Reciprocal crosses were set up between ebony and scarlet flies, and the offspring from these crosses were self-crossed to generate an F2 generation.
3) It was hypothesized that the two mutations are located on the same autosomal chromosome based on their traits being inherited together in the F2 generation.
Gene structure
Drosophila melanogaster, commonly known as the fruit fly, is a widely used model organism in biological research. It has four pairs of chromosomes and a fully sequenced genome of around 139.5 million base pairs containing over 15,000 genes. Many human disease genes have homologs in fruit flies, making Drosophila useful for studying mechanisms underlying neurodegenerative diseases, aging, immunity, and cancer. Hox genes are important transcription factors that determine body plan along the head-to-tail axis. Drosophila has two Hox gene complexes containing eight Hox genes that control segment identity and structure formation. Loss or misexpression of Hox genes can lead to homeotic transformations where one body part develops in the place of another
linkage and crossing over
This document discusses linkage and crossing over of genes. It explains that genes located on the same chromosome are linked, and the closer they are, the stronger the linkage. Crossing over occurs during meiosis and leads to recombination of genes from homologous chromosomes. Linkage maps can be constructed by observing the frequency of crossing over between linked gene loci. These maps show the linear order and genetic distances between genes on chromosomes.
31 Use the following symbols to write out the cross Male.pdf
3.1) Use the following symbols to write out the cross:
Male:
Female:
Dark body color: DB
Wild-type body color: whe
Rudimentary wings: Rw
Wild-type wings: waw
3) Dihybrid cross: The dark-bodied and rudimentary winged genes are on two different autosomal
chromosomes. We will use this cross to determine the ratios of phenotypes in the F2 generation
when there are two genes involved. 3.1) Use the following symbols to write out the cross: Male: 2
Female: Dark body color: DB Wild-type body color: wtDB Rudimentary wings: Rw Wild-type wings:
wtRww 3.2) What would the cross look like if we mated a Dark-Bodied/Rudimentary winged
(double mutant) male with a female that is wild type for both traits? Place the sex of the individual
(symbol) on the first set of lines and their respective genotypes in the second set. Sex: Genotypes
3.3) What are the gametes produced by either sex in the parental generation? Place the symbol
for sex on the first set of lines and write down the gametes produced by that sex in the following
set of lines, respectively. Sex: Gametes produced:3.4) Using the gametes that you produced from
the previous question, fill in the Punnet square and show the resulting offspring from the cross.
Place female gametes to the left of the Punnet square and male gametes above. 3.5) What
phenotype should the F1 offspring have? 3.6) What would the F1 cross look like if we mated a
male F1 and a female F1? Place the sex of the individual on the first set of lines and their
respective genotypes in the second set. Sex: Genotypes x 3.7) What are the gametes produced
by either sex in the F1 generation? Place the symbol for sex on the first set of lines and write down
the gametes produced by that sex in the following set of lines, respectively. Sex: Gametes
produced:3.7) What are the gametes produced by either sex in the F1 generation? Place the
symbol for sex on the first set of lines and write down the gametes produced by that sex in the
following set of lines, respectively. Sex: Gametes produced: 3.8) Using the gametes that you
produced from the previous question, fill in the Punnet square and show the resulting offspring
from the cross. Place female gametes to the left of the Punnet square and male gametes above.
3.9) How many different phenotypes would you expect to observe in the F2 generation? List them.
What is the proportion of phenotypes of the F2 generation? Associate the proportion to the
appropriate phenotype..
Gutell 095.imb.2005.14.625
Gillespie J.J., McKenna C.H., Yoder M.J., Gutell R.R., Johnston J.S., Kathirithamby J., and Cognato A.I. (2005).
Assessing the Odd Secondary Structural Properties of Nuclear Small Subunit Ribosomal RNA Sequences (18S) of the Twisted-Wing Parasites (Insecta: Strepsiptera).
Insect Molecular Biology, 14(6):625-643.
Bot msc part ii sem 4_2014-15_paper 6_mayura d. joshi
GENECTIC REGULATION OF
DEVELOPMENT
IN DROSOPHILA.
BY,
MAYURA D. JOSHI
K. J. SOMAIYA COLLEGE OF
SCIENCE AND COMMERCE.
MSC--PART II--BOTANY
PAPER--VI
HIERARCHY OF GENE
ACTIVATION.
MATERNAL GENES.
SEGMENTATION GENES OF EMBRYO.
HOMEOTIC GENES OF EMBRYO.
HOMEOTIC SELECTOR GENES.
MUTATIONS IN HOMEOTIC GENES.
OTHER GENES OF THE EMBRYO.
SEX DETERMINATION IN DROSOPHILA.
THE HOMEOTIC SELECTOR
GENES OF EMBRYO.
Once the segmentation pattern has been determined, a major class of genes called the homeotic (structure-determining) selector genes are activated.
This genes specify the identity of each segment including the body part that will develop at metamorphosis.
Means this genes are master regulatory genes which specify the types of appendages and other structures that each segment will form.
Mutations in homeotic genes produce flies with structures in incorrect places.
Homeotic genes control the expression of genes responsible for specific anatomical structures.
E.g. Where antenna should form, where legs should appear etc.
Mutations in Homeotic genes.
The homeotic genes have been defined by mutations that affect the development of the fly.
That is, homeotic mutations alter the identify of particular segments, transforming them into copies of other segments.
The principal pioneer of genetic studies of homeotic mutations is Edward Lewis.
More recent molecular analysis has been done in many laboratories, including those of Thomas Kaufman, Walter Gehring, William McGinnis, Matthew Scott, and Welcome Bender.
Lewis’s pioneering studies were on a cluster of homeotic genes called the bithorax complex (BX-C).
BX-C determines the posterior identity of the fly, namely, thoracic segment T3 and abdominal segments AI-A8.
BX-C contains three genes called Ultrabithorax (Ubx), Abdominal-A (Abd-A), and Abdominal-B (Abd-B).
Each of these genes constitutes of one protein-coding transcription unit.
Mutations in these homeotic genes often are lethal and the fly typically does not survive.
Some nonlethal mutant alleles have been characterized.
These nonlethal mutant alleles allow an adult fly to develop.
The figure given here, shows the abnormal adult structures that can result from bithorax mutations.
A diagram showing the segments of a normal adult fly in the Figure (a); note that the wings are located on segment Thorax 2 (T2).
The pair of halteres (rudimentary wings used as balancers in flight) are on segment T3.
A photograph of a normal adult fly clearly showing the wings and halteres is presented in Figure (b).
Figure (c); shows one type of developmental abnormality that can result from nonlethal homeotic mutations in BX-C.
Bio2 4th period study guide
The document summarizes key concepts in biology including:
- Mendel's experiments with dominant and recessive traits in pea plants and how they are passed from parents to offspring.
- Punnett squares are used to predict offspring traits from genetic crosses using probabilities.
- Meiosis results in haploid cells while linked genes are on the same chromosome.
- DNA contains nucleotides that make up genes while RNA is involved in protein synthesis through transcription and translation.
- Mutations can occur in genes or chromosomes and can be identified through karyotypes and pedigree charts.
- Autosomes and sex chromosomes determine phenotypes from genotypes.
- Examples of inherited diseases and how their traits are transmitted.
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.
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Drosophila Melanogaster Genome And its developmental process
The document summarizes key aspects of the Drosophila genome and life cycle. It notes that Drosophila has advantages for genetic studies like a short life cycle and small genome. Its genome contains around 13,600 genes located on four chromosomes. The life cycle involves an egg, larva, pupa and adult stages. Segmentation and homeotic genes play important roles in development by dividing the body into segments and specifying segment identities. Maternal effect, gap, pair-rule and segment polarity genes control segmentation in a hierarchical manner. Homeotic complexes like bithorax determine body part identities in each segment.
FLIES
1) This experiment uses statistical analysis of genetic crosses in fruit flies to determine the chromosomal location of two genes - ebony bodies and scarlet eyes.
2) Reciprocal crosses were set up between ebony and scarlet flies, and the offspring from these crosses were self-crossed to generate an F2 generation.
3) It was hypothesized that the two mutations are located on the same autosomal chromosome based on their traits being inherited together in the F2 generation.
Gene structure
Drosophila melanogaster, commonly known as the fruit fly, is a widely used model organism in biological research. It has four pairs of chromosomes and a fully sequenced genome of around 139.5 million base pairs containing over 15,000 genes. Many human disease genes have homologs in fruit flies, making Drosophila useful for studying mechanisms underlying neurodegenerative diseases, aging, immunity, and cancer. Hox genes are important transcription factors that determine body plan along the head-to-tail axis. Drosophila has two Hox gene complexes containing eight Hox genes that control segment identity and structure formation. Loss or misexpression of Hox genes can lead to homeotic transformations where one body part develops in the place of another
linkage and crossing over
This document discusses linkage and crossing over of genes. It explains that genes located on the same chromosome are linked, and the closer they are, the stronger the linkage. Crossing over occurs during meiosis and leads to recombination of genes from homologous chromosomes. Linkage maps can be constructed by observing the frequency of crossing over between linked gene loci. These maps show the linear order and genetic distances between genes on chromosomes.
31 Use the following symbols to write out the cross Male.pdf
3.1) Use the following symbols to write out the cross:
Male:
Female:
Dark body color: DB
Wild-type body color: whe
Rudimentary wings: Rw
Wild-type wings: waw
3) Dihybrid cross: The dark-bodied and rudimentary winged genes are on two different autosomal
chromosomes. We will use this cross to determine the ratios of phenotypes in the F2 generation
when there are two genes involved. 3.1) Use the following symbols to write out the cross: Male: 2
Female: Dark body color: DB Wild-type body color: wtDB Rudimentary wings: Rw Wild-type wings:
wtRww 3.2) What would the cross look like if we mated a Dark-Bodied/Rudimentary winged
(double mutant) male with a female that is wild type for both traits? Place the sex of the individual
(symbol) on the first set of lines and their respective genotypes in the second set. Sex: Genotypes
3.3) What are the gametes produced by either sex in the parental generation? Place the symbol
for sex on the first set of lines and write down the gametes produced by that sex in the following
set of lines, respectively. Sex: Gametes produced:3.4) Using the gametes that you produced from
the previous question, fill in the Punnet square and show the resulting offspring from the cross.
Place female gametes to the left of the Punnet square and male gametes above. 3.5) What
phenotype should the F1 offspring have? 3.6) What would the F1 cross look like if we mated a
male F1 and a female F1? Place the sex of the individual on the first set of lines and their
respective genotypes in the second set. Sex: Genotypes x 3.7) What are the gametes produced
by either sex in the F1 generation? Place the symbol for sex on the first set of lines and write down
the gametes produced by that sex in the following set of lines, respectively. Sex: Gametes
produced:3.7) What are the gametes produced by either sex in the F1 generation? Place the
symbol for sex on the first set of lines and write down the gametes produced by that sex in the
following set of lines, respectively. Sex: Gametes produced: 3.8) Using the gametes that you
produced from the previous question, fill in the Punnet square and show the resulting offspring
from the cross. Place female gametes to the left of the Punnet square and male gametes above.
3.9) How many different phenotypes would you expect to observe in the F2 generation? List them.
What is the proportion of phenotypes of the F2 generation? Associate the proportion to the
appropriate phenotype..
Gutell 095.imb.2005.14.625
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Bot msc part ii sem 4_2014-15_paper 6_mayura d. joshi
GENECTIC REGULATION OF
DEVELOPMENT
IN DROSOPHILA.
BY,
MAYURA D. JOSHI
K. J. SOMAIYA COLLEGE OF
SCIENCE AND COMMERCE.
MSC--PART II--BOTANY
PAPER--VI
HIERARCHY OF GENE
ACTIVATION.
MATERNAL GENES.
SEGMENTATION GENES OF EMBRYO.
HOMEOTIC GENES OF EMBRYO.
HOMEOTIC SELECTOR GENES.
MUTATIONS IN HOMEOTIC GENES.
OTHER GENES OF THE EMBRYO.
SEX DETERMINATION IN DROSOPHILA.
THE HOMEOTIC SELECTOR
GENES OF EMBRYO.
Once the segmentation pattern has been determined, a major class of genes called the homeotic (structure-determining) selector genes are activated.
This genes specify the identity of each segment including the body part that will develop at metamorphosis.
Means this genes are master regulatory genes which specify the types of appendages and other structures that each segment will form.
Mutations in homeotic genes produce flies with structures in incorrect places.
Homeotic genes control the expression of genes responsible for specific anatomical structures.
E.g. Where antenna should form, where legs should appear etc.
Mutations in Homeotic genes.
The homeotic genes have been defined by mutations that affect the development of the fly.
That is, homeotic mutations alter the identify of particular segments, transforming them into copies of other segments.
The principal pioneer of genetic studies of homeotic mutations is Edward Lewis.
More recent molecular analysis has been done in many laboratories, including those of Thomas Kaufman, Walter Gehring, William McGinnis, Matthew Scott, and Welcome Bender.
Lewis’s pioneering studies were on a cluster of homeotic genes called the bithorax complex (BX-C).
BX-C determines the posterior identity of the fly, namely, thoracic segment T3 and abdominal segments AI-A8.
BX-C contains three genes called Ultrabithorax (Ubx), Abdominal-A (Abd-A), and Abdominal-B (Abd-B).
Each of these genes constitutes of one protein-coding transcription unit.
Mutations in these homeotic genes often are lethal and the fly typically does not survive.
Some nonlethal mutant alleles have been characterized.
These nonlethal mutant alleles allow an adult fly to develop.
The figure given here, shows the abnormal adult structures that can result from bithorax mutations.
A diagram showing the segments of a normal adult fly in the Figure (a); note that the wings are located on segment Thorax 2 (T2).
The pair of halteres (rudimentary wings used as balancers in flight) are on segment T3.
A photograph of a normal adult fly clearly showing the wings and halteres is presented in Figure (b).
Figure (c); shows one type of developmental abnormality that can result from nonlethal homeotic mutations in BX-C.
Bio2 4th period study guide
The document summarizes key concepts in biology including:
- Mendel's experiments with dominant and recessive traits in pea plants and how they are passed from parents to offspring.
- Punnett squares are used to predict offspring traits from genetic crosses using probabilities.
- Meiosis results in haploid cells while linked genes are on the same chromosome.
- DNA contains nucleotides that make up genes while RNA is involved in protein synthesis through transcription and translation.
- Mutations can occur in genes or chromosomes and can be identified through karyotypes and pedigree charts.
- Autosomes and sex chromosomes determine phenotypes from genotypes.
- Examples of inherited diseases and how their traits are transmitted.
Similar to 6- The gene for curled wings (C-c) is located nearly at the end of the.docx (8)
Drosophila Melanogaster Genome And its developmental process
Drosophila Melanogaster Genome And its developmental process
31 Use the following symbols to write out the cross Male.pdf
31 Use the following symbols to write out the cross Male.pdf
Bot msc part ii sem 4_2014-15_paper 6_mayura d. joshi
Bot msc part ii sem 4_2014-15_paper 6_mayura d. joshi
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Denis is a dynamic and results-driven Chief Information Officer (CIO) with a distinguished career spanning information systems analysis and technical project management. With a proven track record of spearheading the design and delivery of cutting-edge Information Management solutions, he has consistently elevated business operations, streamlined reporting functions, and maximized process efficiency.
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Date: May 29, 2024
Tags: Information Security, ISO/IEC 27001, ISO/IEC 42001, Artificial Intelligence, GDPR
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Syllabus
Chapter-1
Introduction to objective, scope and outcome the subject
Chapter 2
Introduction: Scope and Specialization of Civil Engineering, Role of civil Engineer in Society, Impact of infrastructural development on economy of country.
Chapter 3
Surveying: Object Principles & Types of Surveying; Site Plans, Plans & Maps; Scales & Unit of different Measurements.
Linear Measurements: Instruments used. Linear Measurement by Tape, Ranging out Survey Lines and overcoming Obstructions; Measurements on sloping ground; Tape corrections, conventional symbols. Angular Measurements: Instruments used; Introduction to Compass Surveying, Bearings and Longitude & Latitude of a Line, Introduction to total station.
Levelling: Instrument used Object of levelling, Methods of levelling in brief, and Contour maps.
Chapter 4
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Chapter 5
Transportation: Introduction to Transportation Engineering; Traffic and Road Safety: Types and Characteristics of Various Modes of Transportation; Various Road Traffic Signs, Causes of Accidents and Road Safety Measures.
Chapter 6
Environmental Engineering: Environmental Pollution, Environmental Acts and Regulations, Functional Concepts of Ecology, Basics of Species, Biodiversity, Ecosystem, Hydrological Cycle; Chemical Cycles: Carbon, Nitrogen & Phosphorus; Energy Flow in Ecosystems.
Water Pollution: Water Quality standards, Introduction to Treatment & Disposal of Waste Water. Reuse and Saving of Water, Rain Water Harvesting. Solid Waste Management: Classification of Solid Waste, Collection, Transportation and Disposal of Solid. Recycling of Solid Waste: Energy Recovery, Sanitary Landfill, On-Site Sanitation. Air & Noise Pollution: Primary and Secondary air pollutants, Harmful effects of Air Pollution, Control of Air Pollution. . Noise Pollution Harmful Effects of noise pollution, control of noise pollution, Global warming & Climate Change, Ozone depletion, Greenhouse effect
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1. Palancharmy, Basic Civil Engineering, McGraw Hill publishers.
2. Satheesh Gopi, Basic Civil Engineering, Pearson Publishers.
3. Ketki Rangwala Dalal, Essentials of Civil Engineering, Charotar Publishing House.
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