A three point cross was performed using fruit flies. The three loci .pdf
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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.
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1. Gregor Mendel conducted experiments with garden peas in the 1800s to discover the basic principles of heredity and inheritance. He found that traits are passed from parents to offspring through discrete units called genes.
2. Through his experiments with monohybrid and dihybrid crosses, Mendel determined that genes exist in pairs and are inherited independently, with one allele from each parent. This explained his observed 3:1 and 9:3:3:1 phenotypic ratios.
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This document provides 9 practice problems involving Punnett squares. Each problem describes a genetic cross between organisms and asks the reader to: 1) state the genotypes of the parents involved in the cross, 2) draw and complete a Punnett square showing the possible genotypes of the offspring, and 3) state the phenotypic ratios expected among the offspring. The problems cover genetic crosses involving traits such as whisker length in seals, horn presence in mythical creatures, flower color and seed coat traits in peas, eye color in humans, ear shape in dogs, and toe curling in humans.
4 Dr D next hands you a different tray of vials of flies .pdf
4. Dr. D next hands you a different tray of vials of flies. He has better records for these flies. He
had mated a fly with no bristles to a fly with no antennae. The F1 offspring of this cross all
appeared wild type - they had both bristles and antennae. He mated one of male F1s with a
female fly that did not have bristles or antennae. The vlais of flies he has given you contain their F
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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
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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
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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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Type of mutation Cistron A or B
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2. point A
3. point B
4. point B
5. deletion A
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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
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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
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6- The gene for curled wings (C-c) is located nearly at the end of the.docx
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dominant tan allele and c representing the recessive black allele, while gene " W " designates the
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Problems:
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R--------35--------E--------17---------F
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a p).
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(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
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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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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?
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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.
6- The gene for curled wings (C-c) is located nearly at the end of the.docx
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(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?
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Dragon Genetics Hands on LabIntroductionThere are many patterns o.docx
Dragon Genetics Hands on LabIntroduction:
There are many patterns of inheritance. This lab will explore Mendelian traits where alleles are dominant or recessive. You will also have to understand genes that express incomplete dominance, codominance and are sex-linked. We will utilize this lab to explore those terms and to use our knowledge of gene combinations to create an organism.
Utilize this link to learn more about Mendelian traits, incomplete dominance and codominance.(Opens in new window) This link helps you to learn more about sex-linked genes. (Opens in new window)Part 1 Procedure: Checking for Understanding (Worth up to 15pts)
A. For each genotype, tell me if it’s homozygous or heterozygous. If it is homozygous also tell me if it is dominant or recessive.
1. BB
2. jj
3. Pp
4. RR
5. ff
B. Dimples (D) is dominant to (d) no dimples. Brown Eyes (B) are dominant to (b) blue eyes. For each genotype indicate the phenotype.
1. Dd
2. DD
3. dd
4. BB
5. bb
6. Bb
C. For each phenotype list the genotype(s). Trait: Red flower (R); White flower (r)
1. Red flower (homozygous)
2. Red flower (heterozygous)
3. White flowerPart 2: Creating the Parents (Worth up to 48 pts)
1. Flip a coin to determine the genotype for each trait and record it in the data table.
a. Heads= allele 1, Tails = allele 2 ( Example: if you flipped heads twice, your dragon will have two copies of allele 1 for her genotype)
2. Determine the phenotype resulting from the allele pair for each trait.
3. Repeat steps 1-2 for each trait and complete the Mama Dragon’s Table 1.
a. Pay attention to the notes under the trait; it means it does not fully follow Mendelian rules.
Example:
For the trait Eye Size. I flipped my coin 2 times. Flip 1= heads (E); Flip 2= heads (E). So I would put EE in the Genotype box and Two Small Eyes in the Phenotype box for that trait.
Table 1: Genotypes and Phenotypes for Mama Dragon
Trait
Allele 1
Allele 2
Genotype
Phenotype
Eye Size & Number
Two Small (E)
One large (e)
Eye Color
(incomplete Dom)
Red (R)
White (r)
Tail Shape
Curly (T)
Straight (t)
Tail Color
Purple (P)
Orange (p)
Teeth
Sharp (S)
Round (s)
Feet
Four toes (F)
Two toes (f)
Horn Color
(Codominant)
Purple (P)
White (W)
Ear Shape
Pointy (Y)
Round (y)
Claws
Long (L)
Short (l)
Skin Color
(Codominant)
Blue (B)
Yellow (Y)
Mane Texture
(incomplete Dom)
Curly (M)
Straight (m)
Dragon Hemophilia
(Sex- Linked)
Hemophilia (XH)
Normal Clotting (Xh)
Table 2: Genotype and Phenotypes for Daddy Dragon
Trait
Allele 1
Allele 2
Genotype
Phenotype
Eye Size & Number
Two Small (E)
One large (e)
Eye Color
(incomplete Dom)
Red (R)
White (r)
Tail Shape
Curly (T)
Straight (t)
Tail Color
Purple (P)
Orange (p)
Teeth
Sharp (S)
Round (s)
Feet
Four toes (F)
Two toes (f)
Horn Color
(Codominant)
Purple (P)
White (W)
Ear Shape
Pointy (Y)
Round (y)
Claws
Long (L)
Short (l)
Skin Color
(Codominant)
Blue (B)
Yellow (Y)
Mane Texture
(incomplete Dom)
Curly (M)
Straigh ...
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1.
Phenotype
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Miniature wings Brown eyes 4330
Normal wings Brown eyes 610
Miniature wings Red eyes 590
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.
Genetics chapter 5 part 1
This document provides an overview of genetic linkage and mapping. It discusses how Thomas Morgan established the chromosome theory of inheritance by discovering genetic linkage between genes located on the same chromosome. Genes located close together on a chromosome assort together more often than genes farther apart due to less recombination between linked genes. The document describes Morgan's early experiments in Drosophila that demonstrated genetic linkage between eye color and wing size genes. It also explains how Alfred Sturtevant created the first genetic linkage map by quantifying recombination frequencies between linked genes on the X chromosome. The concept of map units (centiMorgans) is introduced as a way to represent the physical distance between genes based on their recombination rate.
Consider the fruit fly Drosophila melanogaster, a common model organ.pdf
Consider the fruit fly Drosophila melanogaster, a common model organism used in genetics.
Drosophila, like peas, have many visible phenotypic traits that can easily be studied in the
laboratory. In this organism, gray body color is dominant over ebony body color and long wings
are dominant over vestigial (stunted)wings. The following crosses are the P1 generation. All P1
organisms are true-breeding (homozygous). Choose symbols for your genes. Remember that the
same gene must have the same letter -use the capital for a dominant allele and lowercase for a
recessive allele. Indicate the genotype and phenotype for the F1 organisms for each cross. Then
indicate the genotypic and phenotypic ratio for the F2 generations (F1 crossed with F1) or the
next generation as noted..
1- gray body long wings Xebony body vestigial wings
2- gray body vestigial wings X ebony body long wings
3- gray body long wings X gray body vestigial wings.
Genetic experiment on the offspring of drosophila melanogaster
1) The experiment aimed to determine if mating Drosophila melanogaster in a dihybrid cross would yield results similar to Mendel's 9:3:3:1 ratio by observing inheritance of eye color (red vs white) and body color (ebony vs brown).
2) The F1 generation showed more dominant traits than recessive, but not all were dominant as in Mendel's experiments. The F2 results using a chi-square test did not match the expected 9:3:3:1 ratio.
3) Errors were made in transferring flies between generations that may have impacted the results, which were ultimately deemed inconclusive. Proper fly handling and food preparation are needed to obtain clear results.
Hour 2
The document discusses Mendel's experiments on monohybrid crosses using pea plants. It explains that a monohybrid cross involves a single trait determined by one gene, such as tall vs dwarf plants. Mendel observed that the ratio of dominant to recessive phenotypes in the F2 generation is 3:1, while the ratio of genotypes is 1:2:1. His experiments supported his laws of segregation and independent assortment.
Hour 2
The document discusses Mendel's experiments on monohybrid crosses using pea plants. It explains that a monohybrid cross involves a single trait determined by one gene, such as tall vs dwarf plants. Mendel observed that the ratio of dominant to recessive phenotypes in the F2 generation is 3:1, while the ratio of genotypes is 1:2:1. His experiments supported his laws of segregation and independent assortment.
Multiple alleles.pdf
This document discusses multiple alleles, which are two or more alternative forms of a gene that occupy the same locus on homologous chromosomes. It provides examples of traits determined by multiple alleles, including coat color in rabbits (with alleles for agouti, chinchilla, himalayan, and albino), human blood groups (A, B, AB, and O alleles), eye color in fruit flies (with alleles producing shades from red to white), and feather patterns in ducks. The relationship between genotypes and resulting phenotypes is explained for each example. The document also briefly discusses self-sterility in plants as another example of multiple alleles.
10.2 dihybrid crosses
This document discusses dihybrid crosses in genetics. It explains how to calculate the expected genotypic and phenotypic ratios in the F1 and F2 generations of a dihybrid cross using a Punnett square. The ratios are expected to be 9:3:3:1. Crossing over can occur between non-sister chromatids during prophase I, resulting in an exchange of alleles and increasing genetic variation. Autosomes are differentiated from sex chromosomes. An example dihybrid cross problem is shown and solved.
Chapter 4. Genetics and Evolution.pptx
Genetics
-. Basic Principles of Mendelian Genetics and Patterns of Inheritance
-Molecular Genetics & Inheritance
-. Protein Synthesis
- Mutations
-. Manipulation of DNA
-. ABO blood groups and Rh Factors
Evolution
- Theories on the origin of life on Earth
-. Theories of Evolution
Configuration management is particularly important during the transit.pdf
Configuration management is particularly important during the transition from system
development to production. Identify four specific areas where close attention to configuration
management is crucial during this phase transition and explain why.
Solution
A \" Configuration \" consists of the functional, physical and interface characteristics of existing
or planned hardware, firmware, software or a combination thereof as set forth in technical
documentation and ultimately achieved in a product. The configuration is formally expressed in
relation to a functional, Allocated or product configuration baseline.
Configuration managment permits the orderly development of a system, subsystem or
configuration item. A good configuration managment program ensures that designs are traceable
to requirements, that change is controlled and documented, that interfaces are defined and
understood, and that there is consistency between the product and its supporting documentation.
Configuration managment provides documentation hat describes what is supported to be
produced, what is being produced, waht has been produced, and what modifications have been
made to what was produced.
A key concept that affects planning is the configuration item (CI). CI decisions will determine
what configurations will be managed. CIs are an aggregation of hardware, firmware, or computer
software, or any of their discrete portions, which satisfies an end-use function and is designated
for separate configuration management. Any item required for logistic support and designated
for separate procurement is generally identified as CI. Components can be designated CIs
because of crucial interfaces or the need to be integrated with operation with other components
within or outside of the system. An item can be designated CI if it is developed wholly or
partially with government funds, including nondevelopmental items (NDI) if additional
development of technical data is required. All CIs are directly traceable to the WBS.
CONFIGURATION MANAGEMENT STRUCTURE :-
Configuration management comprises four interrelated efforts:
Also directly associated with configuration management are data management and interface
management. Any configuration management planning effort must consider all six elements.
Configuration Control :-
Configuration Control is the systematic proposal, justification, prioritization, evaluation,
coordination, approval or disapproval, and implementation of all approved changes in the
configuration of a system/CI after formal establishment of its baseline. In other words, it is how a
system (and its CIs) change control process is executed and managed.
Configuration Control provides management visibility, ensures all factors associated with a
proposed change are evaluated, prevents unnecessary or marginal changes, and establishes
change priorities. In DoD it consists primarily of a change process that formalizes documentation
and provides a management structure for change app.
Cholesterol is an essential component of cellular membranes. It is a(.pdf
Cholesterol is an essential component of cellular membranes. It is a(n) molecule, having both
hydrophilic and hydrophobic regions. Polar oxygen-containing Hydrophobic Amphipathic
Solution
The answer to this question is \"D\". Cholesterol is an amphipathic molecule as it contains both
hydrophillic(OH group) and hydrophobic portion. Cholesterol largely have hydrophobic ring
system but it contains the hydroxyl group which is polar in nature, makes it soluble in water..
More Related Content
Similar to A three point cross was performed using fruit flies. The three loci .pdf
Dragon Genetics Hands on LabIntroductionThere are many patterns o.docx
Dragon Genetics Hands on LabIntroduction:
There are many patterns of inheritance. This lab will explore Mendelian traits where alleles are dominant or recessive. You will also have to understand genes that express incomplete dominance, codominance and are sex-linked. We will utilize this lab to explore those terms and to use our knowledge of gene combinations to create an organism.
Utilize this link to learn more about Mendelian traits, incomplete dominance and codominance.(Opens in new window) This link helps you to learn more about sex-linked genes. (Opens in new window)Part 1 Procedure: Checking for Understanding (Worth up to 15pts)
A. For each genotype, tell me if it’s homozygous or heterozygous. If it is homozygous also tell me if it is dominant or recessive.
1. BB
2. jj
3. Pp
4. RR
5. ff
B. Dimples (D) is dominant to (d) no dimples. Brown Eyes (B) are dominant to (b) blue eyes. For each genotype indicate the phenotype.
1. Dd
2. DD
3. dd
4. BB
5. bb
6. Bb
C. For each phenotype list the genotype(s). Trait: Red flower (R); White flower (r)
1. Red flower (homozygous)
2. Red flower (heterozygous)
3. White flowerPart 2: Creating the Parents (Worth up to 48 pts)
1. Flip a coin to determine the genotype for each trait and record it in the data table.
a. Heads= allele 1, Tails = allele 2 ( Example: if you flipped heads twice, your dragon will have two copies of allele 1 for her genotype)
2. Determine the phenotype resulting from the allele pair for each trait.
3. Repeat steps 1-2 for each trait and complete the Mama Dragon’s Table 1.
a. Pay attention to the notes under the trait; it means it does not fully follow Mendelian rules.
Example:
For the trait Eye Size. I flipped my coin 2 times. Flip 1= heads (E); Flip 2= heads (E). So I would put EE in the Genotype box and Two Small Eyes in the Phenotype box for that trait.
Table 1: Genotypes and Phenotypes for Mama Dragon
Trait
Allele 1
Allele 2
Genotype
Phenotype
Eye Size & Number
Two Small (E)
One large (e)
Eye Color
(incomplete Dom)
Red (R)
White (r)
Tail Shape
Curly (T)
Straight (t)
Tail Color
Purple (P)
Orange (p)
Teeth
Sharp (S)
Round (s)
Feet
Four toes (F)
Two toes (f)
Horn Color
(Codominant)
Purple (P)
White (W)
Ear Shape
Pointy (Y)
Round (y)
Claws
Long (L)
Short (l)
Skin Color
(Codominant)
Blue (B)
Yellow (Y)
Mane Texture
(incomplete Dom)
Curly (M)
Straight (m)
Dragon Hemophilia
(Sex- Linked)
Hemophilia (XH)
Normal Clotting (Xh)
Table 2: Genotype and Phenotypes for Daddy Dragon
Trait
Allele 1
Allele 2
Genotype
Phenotype
Eye Size & Number
Two Small (E)
One large (e)
Eye Color
(incomplete Dom)
Red (R)
White (r)
Tail Shape
Curly (T)
Straight (t)
Tail Color
Purple (P)
Orange (p)
Teeth
Sharp (S)
Round (s)
Feet
Four toes (F)
Two toes (f)
Horn Color
(Codominant)
Purple (P)
White (W)
Ear Shape
Pointy (Y)
Round (y)
Claws
Long (L)
Short (l)
Skin Color
(Codominant)
Blue (B)
Yellow (Y)
Mane Texture
(incomplete Dom)
Curly (M)
Straigh ...
Fly lab experiment
1. The document summarizes the results of multiple genetic crosses performed using fruit flies. It examines crosses involving dominant alleles, recessive alleles, sex-linked recessive alleles, dihybrid crosses, and dominant lethal alleles.
2. Key results showed that eye shape lobe is dominant over wild type eye shape, while incomplete wing vein is recessive to wild type wing vein. Forked bristle is recessive and sex-linked.
3. A dihybrid cross of eye shape lobe and forked bristle showed a 9:3:3:1 ratio in the F2 generation, supporting that they are inherited independently.
4. A cross involving a dominant lethal allele, aristapedia, produced a
Drosophila Melanogaster Lab Report Essay
Drosophila melanogaster, or the fruit fly, is used to study genetics through laboratory experiments observing inheritance patterns across generations. The document describes experiments crossing fruit flies with different traits, such as eye color and wing shape. The results show that some traits, like white eyes and short wings, are inherited together, indicating genetic linkage. Other traits, like dumpy wings and brown eyes, also show linkage. The experiments support Mendel's laws of inheritance and provide insights into how traits are passed down from parents to offspring.
7.genetics and inheritance
The document discusses genetics concepts like alleles, dominant and recessive traits, genotype and phenotype. It uses examples like pea plant height and mouse fur color to demonstrate inheritance patterns. Mendel's experiments with pea plants are cited to explain monohybrid crosses and how they were used to determine dominance relationships between alleles and predict offspring traits and genotypes.
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.
BIO Virtual fly lab (final draft)
The document summarizes Gregor Mendel's experiments with pea plants and how they laid the foundation for genetics. It then describes how the author will simulate Mendel's experiments virtually using fruit flies. The author performs a series of crosses between fruit flies with different traits and uses a chi-square analysis to determine if the results match expected Mendelian ratios. The crosses are designed to show traits behaving as dominant, recessive, recessive sex-linked, and dihybrid traits.
1. Meiosis And Dihybrid Cross
The document discusses meiosis, genetic crosses, and inheritance patterns. It begins by outlining meiosis and its role in genetic variation. It then discusses dihybrid crosses and Mendel's results, which showed independent assortment and ratios of 9:3:3:1. The document further explains linkage, recombination frequency, and genetic maps. It concludes by covering sex-linked inheritance and examples like color blindness and hemophilia.
In fruit flies, the brown gene, which controls eye color, is located .pdf
In fruit flies, the brown gene, which controls eye color, is located on an autosome. Brown
mutants (bw) have brown eyes, while wild-type flies (bw^+) have red eyes. A gene on the X
chromosome, miniature, controls wing size. Miniature wings are caused by the m allele, which is
recessive to the allele for normal wings (m^+). True-breeding miniature-winged, red-eyed
females are crossed to true-breeding normal-winged, brown-eyed males. Predict the phenotypes
(including their ratios of the F1 flies. If the F1 flies are intercrossed with each other, what
phenotypes will appear in the F2 and in what proportions?
Solution
1.
Phenotype
Normal wings Red Eyes 4418
Miniature wings Brown eyes 4330
Normal wings Brown eyes 610
Miniature wings Red eyes 590
2. The F1 was self-crossed to give rise to the F2 generation. Of these offspring, 3/4 had brown
eyes while 1/4 had Red eyes. Phenotypic ratio is 3:1.
Biology_FundamentalsOfGenetics(A).pdf
Heredity involves the transmission of genetic information from one generation to the next. Genes provide continuity between generations and control an organism's structure, function, and development. Gregor Mendel discovered the basic rules of heredity through experiments with pea plants. He found that traits segregate and assort independently during reproduction. Mendel's laws of segregation and independent assortment allow scientists to predict an offspring's traits using Punnett squares. Genetics explains inheritance of traits and introduces concepts like dominance, recessive alleles, genotypes and phenotypes.
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.
Soalan bioday
A male homozygous dominant black mouse is bred with a female homozygous recessive brown mouse. The offspring would be expected to be 100% black, as the black phenotype is dominant over brown.
1- The invasive spotted lanternfly has made it to Vermont- It feeds on.docx
1. The invasive spotted lanternfly has made it to Vermont. It feeds on apples and maple trees and deposits honeydew that attracts other pests and encourages mold growth. The Vermont Agency of Agriculture is therefore very concerned. As a scientist working for the US Department of Agriculture, you're working to better understand the genetic bases of traits in this emerging crop pest species. You've identified three important recessive traits, cold tolerance ( t ), super sweet honeydew ( h ) , and rapid development (d) that could affect their ability to colonize Vermont. Limited cold tolerance ( T ) , regular honeydew ( H ) , and normal development ( D ) are dominant. You know these traits are linked to some degree based on your observations of inheritance patterns. You want to map these traits to the chromosome and determine the genetic distance between the loci. You cross purebreeding spotted lanternflies with cold tolerance, super sweet honeydew, and rapid development with purebreeding spotted lanternflies with limited cold tolerance, regular honeydew, and normal development. You then testcross the resulting F1 progeny and observe the following 1000 F2 progeny. A) Determine the gene order and fill in the progeny genotypes in the table above using the gene annotations noted above. (4 pts) Draw a genetic map indicating the gene order and genetic distance between all 3 gene pairs in units of map distance (or cM ); show your work for full credit. Note: You can calculate distance of outer loci by summing. ( 9 pts) B) Determine if the observed frequency of double crossovers is consistent with the expected frequency by calculating the interference value (I). Interpret the results as the % fewer double recombinants than expected. (4 pts)
.
Genetics chapter 5 part 1
This document provides an overview of genetic linkage and mapping. It discusses how Thomas Morgan established the chromosome theory of inheritance by discovering genetic linkage between genes located on the same chromosome. Genes located close together on a chromosome assort together more often than genes farther apart due to less recombination between linked genes. The document describes Morgan's early experiments in Drosophila that demonstrated genetic linkage between eye color and wing size genes. It also explains how Alfred Sturtevant created the first genetic linkage map by quantifying recombination frequencies between linked genes on the X chromosome. The concept of map units (centiMorgans) is introduced as a way to represent the physical distance between genes based on their recombination rate.
Consider the fruit fly Drosophila melanogaster, a common model organ.pdf
Consider the fruit fly Drosophila melanogaster, a common model organism used in genetics.
Drosophila, like peas, have many visible phenotypic traits that can easily be studied in the
laboratory. In this organism, gray body color is dominant over ebony body color and long wings
are dominant over vestigial (stunted)wings. The following crosses are the P1 generation. All P1
organisms are true-breeding (homozygous). Choose symbols for your genes. Remember that the
same gene must have the same letter -use the capital for a dominant allele and lowercase for a
recessive allele. Indicate the genotype and phenotype for the F1 organisms for each cross. Then
indicate the genotypic and phenotypic ratio for the F2 generations (F1 crossed with F1) or the
next generation as noted..
1- gray body long wings Xebony body vestigial wings
2- gray body vestigial wings X ebony body long wings
3- gray body long wings X gray body vestigial wings.
Genetic experiment on the offspring of drosophila melanogaster
1) The experiment aimed to determine if mating Drosophila melanogaster in a dihybrid cross would yield results similar to Mendel's 9:3:3:1 ratio by observing inheritance of eye color (red vs white) and body color (ebony vs brown).
2) The F1 generation showed more dominant traits than recessive, but not all were dominant as in Mendel's experiments. The F2 results using a chi-square test did not match the expected 9:3:3:1 ratio.
3) Errors were made in transferring flies between generations that may have impacted the results, which were ultimately deemed inconclusive. Proper fly handling and food preparation are needed to obtain clear results.
Hour 2
The document discusses Mendel's experiments on monohybrid crosses using pea plants. It explains that a monohybrid cross involves a single trait determined by one gene, such as tall vs dwarf plants. Mendel observed that the ratio of dominant to recessive phenotypes in the F2 generation is 3:1, while the ratio of genotypes is 1:2:1. His experiments supported his laws of segregation and independent assortment.
Hour 2
The document discusses Mendel's experiments on monohybrid crosses using pea plants. It explains that a monohybrid cross involves a single trait determined by one gene, such as tall vs dwarf plants. Mendel observed that the ratio of dominant to recessive phenotypes in the F2 generation is 3:1, while the ratio of genotypes is 1:2:1. His experiments supported his laws of segregation and independent assortment.
Multiple alleles.pdf
This document discusses multiple alleles, which are two or more alternative forms of a gene that occupy the same locus on homologous chromosomes. It provides examples of traits determined by multiple alleles, including coat color in rabbits (with alleles for agouti, chinchilla, himalayan, and albino), human blood groups (A, B, AB, and O alleles), eye color in fruit flies (with alleles producing shades from red to white), and feather patterns in ducks. The relationship between genotypes and resulting phenotypes is explained for each example. The document also briefly discusses self-sterility in plants as another example of multiple alleles.
10.2 dihybrid crosses
This document discusses dihybrid crosses in genetics. It explains how to calculate the expected genotypic and phenotypic ratios in the F1 and F2 generations of a dihybrid cross using a Punnett square. The ratios are expected to be 9:3:3:1. Crossing over can occur between non-sister chromatids during prophase I, resulting in an exchange of alleles and increasing genetic variation. Autosomes are differentiated from sex chromosomes. An example dihybrid cross problem is shown and solved.
Chapter 4. Genetics and Evolution.pptx
Genetics
-. Basic Principles of Mendelian Genetics and Patterns of Inheritance
-Molecular Genetics & Inheritance
-. Protein Synthesis
- Mutations
-. Manipulation of DNA
-. ABO blood groups and Rh Factors
Evolution
- Theories on the origin of life on Earth
-. Theories of Evolution
Similar to A three point cross was performed using fruit flies. The three loci .pdf (20)
Dragon Genetics Hands on LabIntroductionThere are many patterns o.docx
Dragon Genetics Hands on LabIntroductionThere are many patterns o.docx
In fruit flies, the brown gene, which controls eye color, is located .pdf
In fruit flies, the brown gene, which controls eye color, is located .pdf
2- In Drosophila- a cross was made between females - all expressing th.docx
2- In Drosophila- a cross was made between females - all expressing th.docx
1- The invasive spotted lanternfly has made it to Vermont- It feeds on.docx
1- The invasive spotted lanternfly has made it to Vermont- It feeds on.docx
Consider the fruit fly Drosophila melanogaster, a common model organ.pdf
Consider the fruit fly Drosophila melanogaster, a common model organ.pdf
Genetic experiment on the offspring of drosophila melanogaster
Genetic experiment on the offspring of drosophila melanogaster
More from rydeberghal13313
Configuration management is particularly important during the transit.pdf
Configuration management is particularly important during the transition from system
development to production. Identify four specific areas where close attention to configuration
management is crucial during this phase transition and explain why.
Solution
A \" Configuration \" consists of the functional, physical and interface characteristics of existing
or planned hardware, firmware, software or a combination thereof as set forth in technical
documentation and ultimately achieved in a product. The configuration is formally expressed in
relation to a functional, Allocated or product configuration baseline.
Configuration managment permits the orderly development of a system, subsystem or
configuration item. A good configuration managment program ensures that designs are traceable
to requirements, that change is controlled and documented, that interfaces are defined and
understood, and that there is consistency between the product and its supporting documentation.
Configuration managment provides documentation hat describes what is supported to be
produced, what is being produced, waht has been produced, and what modifications have been
made to what was produced.
A key concept that affects planning is the configuration item (CI). CI decisions will determine
what configurations will be managed. CIs are an aggregation of hardware, firmware, or computer
software, or any of their discrete portions, which satisfies an end-use function and is designated
for separate configuration management. Any item required for logistic support and designated
for separate procurement is generally identified as CI. Components can be designated CIs
because of crucial interfaces or the need to be integrated with operation with other components
within or outside of the system. An item can be designated CI if it is developed wholly or
partially with government funds, including nondevelopmental items (NDI) if additional
development of technical data is required. All CIs are directly traceable to the WBS.
CONFIGURATION MANAGEMENT STRUCTURE :-
Configuration management comprises four interrelated efforts:
Also directly associated with configuration management are data management and interface
management. Any configuration management planning effort must consider all six elements.
Configuration Control :-
Configuration Control is the systematic proposal, justification, prioritization, evaluation,
coordination, approval or disapproval, and implementation of all approved changes in the
configuration of a system/CI after formal establishment of its baseline. In other words, it is how a
system (and its CIs) change control process is executed and managed.
Configuration Control provides management visibility, ensures all factors associated with a
proposed change are evaluated, prevents unnecessary or marginal changes, and establishes
change priorities. In DoD it consists primarily of a change process that formalizes documentation
and provides a management structure for change app.
Cholesterol is an essential component of cellular membranes. It is a(.pdf
Cholesterol is an essential component of cellular membranes. It is a(n) molecule, having both
hydrophilic and hydrophobic regions. Polar oxygen-containing Hydrophobic Amphipathic
Solution
The answer to this question is \"D\". Cholesterol is an amphipathic molecule as it contains both
hydrophillic(OH group) and hydrophobic portion. Cholesterol largely have hydrophobic ring
system but it contains the hydroxyl group which is polar in nature, makes it soluble in water..
A correlation coefficient computed for n = 18 and a 10 significance.pdf
A correlation coefficient computed for n = 18 and a 10% significance level is 0.692. Using a t-
test for significance of the correlation coefficient, what is the largest critical value?
Options:
+2.120
+2.101
+1.734
+1.746
Solution
for df = n-2 = 16, t(alpha/2) = t(0.05) = 1.746 (From t table)
+1.746.
1.Using the Internet or other sources to find the definition of CASE.pdf
1.Using the Internet or other sources to find the definition of CASE tool related to software
engineering and to find one free CASE tool for data model that support ER model.
Solution
The development and maintenance of software projects is done with help of various automated
CASE(Computer Aided Software Engineering) tools or software application programs.
Features:
1) They are used to automate SDLC activities.
2) The development of project is accelerated to produce desired result and helps to fix flaws
before moving ahead to next stage in software development.
3) CASE tools are used by analysts,software project managers and engineers to develop
software system.
4) It redces time and cost of software development and enhances quality of the systems
developed.
Types:
CASE tools available across various stages of Software Development Life Cycle are :
a) Analysis tools
b) Design tools
c) Project management tools
d) Database Management tools
e) Documentation tools
f) Configuration Management tools
g) Quality Assurance tools
Free CASE tools for data model that support ER model : Toad Data Modeler, SQL Power
Architect, StarUML and Umbrello..
A fluid of constant density rho enters a duct of width W and height h.pdf
A fluid of constant density rho enters a duct of width W and height h_1, with a parabolic
velocity profile with a maximum value of V_1, as shown below. At the exit plane, the duct has
height h_2 and the flow has a parabolic velocity profile with a maximum value of V_2. The
pressures at the entry and exit stations are p_1 and p_2, respectively, and they are uniform across
the duct. Find V_2 in terms of V_1, h_1, and h_2. Find the magnitude and direction of the
horizontal force F exerted by the fluid on the step in terms of rho, V_1, W, p_1 and p_2, h_1 and
h_2. Ignore friction. Note that at the point where the flow separates off the step, the flow
streamlines can be assumed to be parallel: this observation provides information about the
pressure on the vertical face of the step.
Solution
a)
Consider a strip of height dy1 at distance y from the centre. Cross-section of the strip = W*dy1
For bulk mean velocity at entry, we have
(h1*W)*V1_mean = Integral [(W*dy1)*V] ............from y1 = -h1/2 to y1 = h1/2
(h1*W)*V1_mean = 2*Integral [(W*dy1)*V1* (1 - (2y1/h1)2)] ............from y1 = 0 to y1 = h1/2
h1*V1_mean = 2V1* Integral [dy1* (1 - (2y1/h1)2)] ............from y1 = 0 to y1 = h1/2
h1*V1_mean = 2V1* [y1 - (4/3)y13 / h12]............from y1 = 0 to y1 = h1/2
h1*V1_mean = 2V1* [(h1/2) - (4/3)(h1/2)3 / h12]
V1_mean = (2/3)V1
Similarly, we can get V2_mean = (2/3)V2
By mass conservation, rho*(h1*W)*V1_mean = rho*(h2*W)*V2_mean
h1*(2/3)V1 = h2*(2/3)V2
V2 = V1*(h1/h2)
b)
By momentum conservation,
F = Momentum in - Momentum out
= P1*(h1*W) + [rho*(h1*W)*V1_mean]*V1_mean - P2*(h2*W) -
[rho*(h2*W)*V2_mean]*V2_mean
= rho*W*[h1*V1_mean2 - h2*V2_mean2] + (P1*h1*W - P2*h2*W)
= rho*W*[h1*(2V1/3)2 - h2*(2V2/3)2] + (P1*h1*W - P2*h2*W)
= rho*W*[h1*(2V1/3)2 - h2*(2V1*(h1/h2)/3)2] + (P1*h1*W - P2*h2*W)
= (4/9)*rho*W*h1*V1*[V1 - V1*(h1/h2)] + (P1*h1*W - P2*h2*W)
= (4/9)*rho*W*h1*V12*[1 - (h1/h2)] + (P1*h1*W - P2*h2*W).
A) Know the four basic tissues, how they differ, and what their gene.pdf
A) Know the four basic tissues, how they differ, and what their general functions are
B) Understand the difference in function between flat and cuboidal/columnar epithelial cells
C) Know the basic components of connective tissue
D) Understand how adjusting the amounts of collagen, elastin, and different kinds of ground
substances affects the function of connective tissues
Solution
A) epithelial tissue- originated from ecto, meso and endodermis.
Function: protection, covering , secretion and excretion.
Connective tissue- originated from mesoderm
Function : connection, support, blood production and defense mechanism.
Muscular tissue- originated from mesoderm
Function: movement and locomotion.
Nervous tissue- originated from ectoderm
Function: control and coordination.
B).Flat or squamous epithelium is involving in filtration and diffusion , columnar / cuboidal
epithelium is involving in absorption, secretion and excretion
C) basic components of connective tissue include connective tissue fibres, connective tissue cells
and connective tissue matrix.
D)collagen ,elastin and reticulin involves the maintenance of regidity, toughness or elasticity of
connective tissue. In different connective tissues different amounts of these protein play role.
A console lamp in the cabin of a spaceship appears green when the shi.pdf
A console lamp in the cabin of a spaceship appears green when the ship and observer are both at
rest. When the ship is moving at 0.90c away from Earth, passengers on board see: a) a dark
lamp (the frequency is too high to be seen) b) a dark lamp (the frequency is too low to bo seen)
c) a red lamp d) a violet lamp e) a green lamp
Solution
c)red lamp.
7. Whats the meaning of these transportation symbols IT Soluti.pdf
7. What\'s the meaning of these transportation symbols? IT
Solution
Answer ;-
A) Fragile - This symbol means Product inside packaging could be easily damaged.
B) Center of Gravity -
C) Keep Dry - This symbal means package to keep out of the water and rain. And not to store it
in damp
conditions.
D) This Way Up - The arrows point towards the top of the package..
A virus is 50 nm in size. Would you recommend using a stereomicrosco.pdf
A virus is 50 nm in size. Would you recommend using a stereomicroscope. a compound light
microscope, or an electron microscope to see it?
Solution
A virus of 50nm size will be so smaller that it becomes difficult to observed under
stereomicroscope or compound light microscope due to lower resolving powers. However, an
electron microscope has higher resolving power (0.1nm resolution) and is ideal for viral studies.
Therefore, the correct option is: Electron microscope.
A division has 12 employees, 8 males and 4 females.Supose that the.pdf
A division has 12 employees, 8 males and 4 females.
Supose that the committee should have one president which has to be a female and two ordinary
members who must be males. how many different committees are possible?
Please provide working with explaination if possible.
Solution
Selcting a female from 4 = 4 C1
And 2 Males from 8 = 8 C 2
Number of total diffrent committes will be = 4 C 1 * 8 C 2 = ( 4/1) * ( 8*7/2*1) = 112.
A broth can have more than one type of growth pattern. Explain thi.pdf
A broth can have more than one type of growth pattern. Explain this using an example.
A broth can have more than one type of growth pattern. Explain this using an example.
Solution
More than one growth pattern are observed in the broth. The pattern that can be observed as
follows:
1: Pellicle: It is the growth pattern exhibited when the surface membrane produced by organisms
floats on top of the medium.
2: Sediment: If the growing organisms tend to sink to the bottom, the growth pattern is called as
sediment.
3: Uniform fine turbidity: If the growing organism give cloudy apprearance, the pattern is called
as cloudy.
4: Flocculent: If the growing organisms that clump pieces.
For ex: Broth of Bacillus subtilis exhibit flocculent and sediment growth pattern..
2. A safety net can stand the fall of the total of 2700 pounds of obj.pdf
2. A safety net can stand the fall of the total of 2700 pounds of objects from a construct ion site.
Say 12 random objects are falling down into the act at the same time, and the weight of each of
them is (225. 625). What is the probability that the net withstands the impact?
Solution
z = (2700/12 - 225) / (25/sqr(12)) = 0
Therefore , from z table ,
P = 0.5 Answer.
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(Linux)
A) What is a fixup ?
A) What is Source tree dir. and where is it?
Solution
Answer:
Fixup is a linux command which combines two or more commits into single commit, but the
commit to be merged has its message discarded.
Source tree is a free Git client for windows , linux or mac. It provides visual interface between
user and Git. It is located generally where the kernal source is..
Cobol Error Its states that my patron-line wasnt defined as a data .pdf
Cobol Error? Its states that my patron-line wasnt defined as a data name and none of my header,
column and footers are not defined as data name could you see the error?
IDENTIFICATION DIVISION.
PROGRAM-ID. PROG-ONE.
AUTHOR. MARKIED JONES.
FD PATRON-LIST.
01 PATRON-LINE PIC X(80).
ENVIRONMENT DIVISION.
INPUT-OUTPUT SECTION.
FILE-CONTROL.
SELECT PATRON-FILE ASSIGN TO INFILE.
SELECT PATRON-LIST ASSIGN TO OUTFILE.
DATA DIVISION.
FILE SECTION.
FD PATRON-FILE.
01 PATRON-RECORD.
03 PR-NAME PIC X(18).
03 PR-ADDRESS PIC X(18).
03 PR-CITY-STATE-ZIP PIC X(24).
03 PR-TARGET-CONTR PIC 9(4).
03 PR-ACTUAL-CONTR PIC 9(4).
03 PR-CONTR-DATE.
05 PR-CONTR-MONTH PIC XX.
05 PR-CONTR-DAY PIC XX.
05 PR-CONTR-YEAR PIC XX.
03 FILLER PIC X(6).
03 FILLER PIC X.
WORKING-STORAGE SECTION.
01 PATRON-LINE-1.
03 FILLER PIC X.
03 PL-NAME PIC X(18) .
03 FILLER PIC X.
03 PL-ADDRESS PIC X(18).
03 FILLER PIC XX.
03 PL-CITY-STATE-ZIP PIC X(24).
01 WS-HEADER-LINE-1.
03 FILLER PIC X(18) VALUE SPACES.
03 FILLER PIC X(7) VALUE \"INDIANA\".
03 FILLER PIC X VALUE SPACE.
03 FILLER PIC X(10) VALUE \"UNIVERSITY\".
03 FILLER PIC X VALUE SPACE.
03 FILLER PIC X(2) VALUE \"OF\".
03 FILLER PIC X VALUE SPACE.
03 FILLER PIC X(12) VALUE \"PENNSYLVANIA\".
01 WS-HEADER-LINE-1-UL.
03 FILLER PIC X(18) VALUE SPACES.
03 FILLER PIC X(7) VALUE \"-\".
03 FILLER PIC X VALUE SPACE.
03 FILLER PIC X(10) VALUE \"-\".
03 FILLER PIC X VALUE SPACE.
03 FILLER PIC X VALUE \"-\".
01 HEADER-LINE-2.
03 FILLER PIC X(20) VALUE SPACES.
03 FILLER PIC X(8) VALUE \"COMPUTER\".
03 FILLER PIC X VALUE SPACE.
03 FILLER PIC X(7) VALUE \"SCIENCE\".
03 FILLER PIC X VALUE SPACE.
03 FILLER PIC X(10) VALUE \"DEPARTMENT\".
01 HEADER-LINE-2-UL.
03 FILLER PIC X(20) VALUE SPACES.
03 FILLER PIC X(8) VALUE ALL \"-\".
03 FILLER PIC X VALUE SPACE.
03 FILLER PIC X(7) VALUE ALL \"=\".
03 FILLER PIC X VALUE SPACE.
03 FILLER PIC X(10) VALUE ALL \"-\".
01 HEADER-LINE-3.
03 FILLER PIC X(22) VALUE SPACES.
03 FILLER PIC X(7) VALUE \"LISTING\".
03 FILLER PIC X VALUE SPACE.
03 FILLER PIC X(2) VALUE \"OF\".
03 FILLER PIC X VALUE SPACE.
03 FILLER PIC X(7) VALUE \"PATRONS\".
01 HEADER-LINE-3-UL.
03 FILLER PIC X(22) VALUE SPACES.
03 FILLER PIC X(7) VALUE \"-\".
03 FILLER PIC X VALUE SPACE.
03 FILLER PIC X(6) VALUE \"-\".
03 FILLER PIC X(15) VALUE SPACES.
03 FILLER PIC X(6) VALUE \"-\".
01 WS-COLUMN-HEADER-2.
03 FILLER PIC X(4) VALUE SPACES.
03 FILLER PIC X(4) VALUE \"NAME\".
03 FILLER PIC X(8) VALUE SPACES.
03 FILLER PIC X(7) VALUE \"ADDRESS\".
03 FILLER PIC X(12) VALUE SPACES.
03 FILLER PIC X(14) VALUE \"CITY-STATE-ZIP\".
01 WS-COLUMN-HEADER-2UL.
03 FILLER PIC X(4) VALUE SPACES.
03 FILLER PIC X(4) VALUE \"-\".
03 FILLER PIC X(8) VALUE SPACES.
03 FILLER PIC X(7) VALUE \"-\".
03 FILLER PIC X(12) VALUE SPACES.
03 FILLER PIC X(14) VALUE \"-\".
01 WS-FOOTER-LINE-1.
03 FILLER PIC X(24) VALUE
\"TOTAL # OF PATRONS:\".
03 WS-NUM-PATRONS PIC Z9.
01 WS-FOOTER-LINE-2.
03 FILLER PIC X(32) VALUE
\"PROGRAMMED BY A COSC 220 STUDENT\".
01 WS-EOF-SWITCH PIC X(80).
PROCEDURE.
Companies and people often buy and sell stocks. Often they buy the sa.pdf
Companies and people often buy and sell stocks. Often they buy the same stock for different
prices at different times. Say one owns 1000 shares a certain stock (such as Checkpoint) one may
have bought the stock in amounts of 100 shares over 10 different times with 10 different prices.
We will analyze two different methods of accounting, FIFO and LIFO accounting used for
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purchase of that item. Say we sell 250 shares of a stock, according to this method the dollar cost
average purchase price is determined by averaging the prices on the last 250 shares bought. In
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stack class in java.util. Make sure you use the stack and queue abstract data types (ADT) defined
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The stock Ticker symbol (a string). The Stock Name (a string). The number of shares of a stock
(an int). The purchase price for those shares (decimal). You can assume that the first element of
the structure is the security bought first, the second was bought second, etc. Your program
should have the user able to enter information about various stocks, the amount of shares, and the
price. The user can then enter a query about a certain stock and the cost according to the LIFO
and FIFO accounting methods for a certain number of shares. The following could be your
menu: Press 1 to enter a new stock Press 2 to find the LIFO and FIFO dollar cost average for the
number of shares sold. If 1 is pressed, the user needs to enter the stock symbol, Stock name, the
number of shares bought, and the price per share when purchased. If 2 is pressed, the user needs
to enter the stock symbol being queried and the number of shares you wish to sell. Once that
information is entered the dollar cost averaged price per share for both the LIFO method and the
FIFO method should be displayed. Implement the program described above using the Java
programming language to meet the overall requirements described above. Use the blow data.
Ticker: XOM Name: Exon Mobile Purchases in the order below. 1. 100 shares at 92.65 2. 50
shares at 97.23 3. 200 shares at 84.50 4. 100 shares at 86.58 Ticker: D Name: Dominion
Resources Purchases in the order below. 1. 350 shares at 64.98 2. 20.
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Chapter 2
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Chapter 3
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- 1. 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.
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