Report by Josh Jackson





Section 11-1 Control of Gene Expression
1. Cells use ______________________ to build hundreds of different________________each with a unique ____________________________.
2. Are all proteins used by a cell at any one time? If not, how do cells control this?
3. Define gene expression.
4. When are proteins produced?
5. What is the genome?
6. What are the 2 steps of gene expression?
7. What 2 scientists determined how genes are expressed in prokaryotes?
8. What gene & in what organism did Jacob & Monod make their discoveries about gene expression?
9. Name the 3 regulatory elements on the DNA of the E. coli bacterium and tell the function of each.
10. What is an operon & what 3 things is it made up of?
11. What name did Jacob & Monod give their gene & why?
12. If lactose is not present, what attaches to the operator?
13. Define repressor protein and give its function.
14. Define repression.
15. What occurs if lactose is present in the E. coli in lactose metabolism?
16. What is an inducer?
17. What is an inducer for E. coli in lactose metabolism?
18. How does the genome of eukaryotes compare with that of prokaryotes?
19. Are operons found in eukaryotes?
20. Each eukaryotic cell contains a ___________________ set of genes, but only some genes
are ______________________ at a given time.
21. What controls much of the gene expression in eukaryotes?
22. What is euchromatin?
23. Some sections of chromatin always remain coiled preventing what process?
24. Name & define the 2 kinds of segments found behind the promoter in eukaryotes.
25. Where do the processes of transcription & translation take place in prokaryotes?
26. Where do these processes take place in eukaryotes?
27. Are introns and/or exons transcribed?
28. What is pre-mRNA and how is mRNA formed from this?
Section 11-2 Gene Expression and Development
29. Multicellular, sexually reproducing organisms begin life as a _____________________with all cells containing the same _______________________.
30. Genes may be turned ______________ and _____________as various ___________________ are needed by the cells.
31. What is cell differentiation?
32. Define morphogenesis.
33. What genes determine what anatomical structures an organism will develop during morphogenesis?
34. What is a tumor and what are the 2 main types?
35. Define benign tumor.
36. Are benign tumors dangerous? Explain.
37. What treatment do doctors use with benign tumors?
38. Define malignant tumor.
39. Malignant tumors are commonly known as ____________________________.
40. What is metastasis & what happens to the body when this occurs?
41. How are malignant tumors categorized?
42. Name & describe 4 types of malignant tumors.
43. Lung cancer & breast cancer are what type of tumors?
44. When do normal cells stop dividing? Do cancer cells respond the same way? Explain.
45. What trait of cancer cells facilitates the spread of cancer cells in the body?
46. What is a carcinogen & give 5 examples?
47. What causes most lung cancer?
48. What is the effect of mutagens on cells?
49. What are oncogenes?
50. Certain ____________________ can cause cancer in plants & animals.
| Mendelian Genetics |
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| 1862 | 1868 | 1880 |
Genetic Terminology:
Blending Concept of Inheritance:
Gregor Mendel:
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Why peas, Pisum sativum?

GARDEN PEA
Mendel’s Experiments:
a. Seed shape — Round (R) or Wrinkled (r)
b. Seed Color —- Yellow (Y) or Green (y)
c. Pod Shape — Smooth (S) or wrinkled (s)
d. Pod Color — Green (G) or Yellow (g)
e. Seed Coat Color — Gray (G) or White (g)
f. Flower position — Axial (A) or Terminal (a)
g. Plant Height — Tall (T) or Short (t)
h. Flower color — Purple (P) or white (p)


Trait – plant height |
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Alleles – T tall, t short |
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P1 cross TT x tt |
genotype — Tt | |||
| t | t | phenotype — Tall | ||
| T | Tt | Tt | genotypic ratio –all alike | |
| T | Tt | Tt | phenotypic ratio- all alike | |
Trait – plant height |
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Alleles – T tall, t short |
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F1 cross Tt x Tt |
genotype — TT, Tt, tt | |||
| T | t | phenotype — Tall & short | ||
| T | TT | Tt | genotypic ratio —1:2:1 | |
| t | Tt | tt | phenotypic ratio- 3:1 | |
| Trait – Plant Height | |||||||
| Alleles – T tall, t short | |||||||
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F2 cross TT x Tt |
F2 cross tt x Tt |
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| T | t | T | t | ||||
| T | TT | Tt | t | Tt | tt | ||
| T | TT | Tt | t | Tt | tt | ||
| genotype – TT, Tt | genotype – tt, Tt | ||||||
| phenotype – Tall | phenotype – Tall & short | ||||||
| genotypic ratio – 1:1 | genotypic ratio – 1:1 | ||||||
| phenotypic ratio – all alike | phenotypic ratio – 1:1 | ||||||
Problems: Work the P1, F1, and both F2 crosses for all of the other pea plant traits & be sure to include genotypes, phenotypes, genotypic & phenotypic ratios.
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Traits: Seed Shape & Seed Color |
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Alleles: R round Y yellow |
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P1 Cross:
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| ry | Genotype: | RrYy | ||
| RY | RrYy |
Phenotype: | Round yellow seed | |
| Genotypic ratio: | All alike | |||
| Phenotypic ratio: | All Alike | |||
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Traits: Seed Shape & Seed Color |
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Alleles: R round Y yellow |
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| F1 Cross: RrYy x RrYy | ||||
| RY | Ry | rY | ry | |
| RY | RRYY |
RRYy |
RrYY |
RrYy |
| Ry | RRYy |
RRyy |
RrYy |
Rryy |
| rY | RrYY |
RrYy |
r rYY |
r rYy |
| ry | RrYy |
Rryy |
r rYy |
r ryy |
| Genotypes | Genotypic Ratios | Phenotypes | Phenotypic Ratios |
| RRYY | 1 | Round yellow seed |
9 |
| RRYy | 2 | ||
| RrYY | 2 | ||
| RrYy | 4 | ||
| RRyy | 1 | Round green seed |
3 |
| Rryy | 2 | ||
| r rYY | 1 | Wrinkled yellow seed |
3 |
| r rYy | 2 | ||
| r ryy | 1 | Wrinkled green seed |
1 |
Problems: Choose two other pea plant traits and work the P1 and F1 dihybrid crosses. Be sure to show the trait, alleles, genotypes, phenotypes, and all ratios.
Results of Mendel’s Experiments:
| Trait: Pod Color | |
| Genotypes: | Phenotype: |
| GG | Green Pod |
| Gg | Green Pod |
| gg | Yellow Pod |
| Rr | |
| R | r |
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RrYy |
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| RY | Ry | rY | ry |
Other Patterns of Inheritance:

| Genotype | Phenotype |
| IOIO | Type O |
| IAIO | Type A |
| IAIA | Type A |
| IBIO | Type B |
| IBIB | Type B |
| IAIB | Type AB |
| Genetics of Drosophila melanogaster | ![]() |
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Introduction:
Gregor Mendel revolutionized the study of genetics. By studying genetic inheritance in pea plants, Gregor Mendel established two basic laws of that serve as the cornerstones of modern genetics: Mendel’s Law of Segregation and Law of Independent Assortment. Mendel’s Law of Segregation says that each trait has two alleles, and that each gamete contains one and only one of these alleles. These alleles are a source of genetic variability among offspring. Mendel’s Law of Independent Assortment says that the alleles for one trait separate independently of the alleles for another trait. This also helps ensure genetic variability among offspring.
Mendel’s laws have their limitations. For example, if two genes are on the same chromosome, the assortment of their alleles will not be independent. Also, for genes found on the X chromosome, expression of the trait can be linked to the sex of the offspring. Our knowledge of genetics and the tools we use in its study have advanced a great deal since Mendel’s time, but his basic concepts still stand true.
Drosophila melanogaster, the common fruit fly, has been used for genetic experiments since T.H. Morgan started his experiments in1907. Drosophila make good genetic specimens because they are small, produce many offspring, have easily discernable mutations, have only four pairs of chromosomes, and complete their entire life cycle in about 12 days. They also have very simple food requirements. Chromosomes 1 (the X chromosome), 2, and 3 are very large, and the Y chromosome – number 4 – is extremely small. These four chromosomes have thousands of genes, many of which can be found in most eukaryotes, including humans.
Drosophila embryos develop in the egg membrane. The egg hatches and produces a larva that feeds by burrowing through the medium. The larval period consists of three stages, or instars, the end of each stage marked by a molt. Near the end of the larval period, the third instar will crawl up the side of the vial, attach themselves to a dry surface, and form a pupae. After a while the adults emerge.
Differences in body features help distinguish between male and female flies. Females are slightly larger and have a light-colored, pointed abdomen. The abdomen of males will be dark and blunt. The male flies also have dark bristles, sex combs, on the upper portion of the forelegs.
Hypothesis:
After performing a dihybrid cross between males with normal wings and sepia eyes and females with vestigial wings and red eyes, we expect to see only hybrids with normal wings and red eyes in the first filial generation. Then we expect to observe a 9:3:3:1 ratio of phenotypes in the second filial generation.
Materials and Methods:
The materials used for this lab were: culture vial of dihybrid cross, isopropyl alcohol 10%, camel’s hair brush, thermo-anesthetizer, petri dish, 2 Drosophila vials and labels, Drosophila medium, fly morgue.
A vial of wild-type Drosophila was thermally immobilized and the flies were placed in a petri dish. Traits were observed. A vial of prepared Drosophila was immobilized and then observed under a dissecting microscope. Males and females were separated and mutations were observed and recorded. The parental generation was placed in the morgue. The vial was placed in an incubator to allow the F1 generation to mature.
The F1 generation was immobilized and examined under a dissecting microscope. The sex and mutations of each fly were recorded. Five mating pairs of the F1 generation were placed into a fresh culture vial, and the vial was placed in an incubator. The remaining F1 flies were placed in the morgue. The F1 flies were left in the vial for about a week to mate and lay eggs. Then the adults were removed and placed in the morgue. The vial was placed back in the incubator to allow the F2 generation to mature. The F2 generation was immobilized and examined under a dissecting microscope. The sex and mutations of each fly were recorded.
Results:
Table 1 Phenotypes of the Parental Generation
| Phenotypes | Number of Males | Number of Females |
| Normal wings/red eyes | 0 | 0 |
| Normal wings/sepia eyes | 3 | 0 |
| vestigial wings/red eyes | 0 | 4 |
| vestigial wings/sepia eyes | 0 | 0 |
Table 2 Phenotypes of the F1 Generation
| Phenotype | Number of Males | Number of Females |
| Normal wings/red eyes | 78 | 95 |
| Normal wings/sepia eyes | 0 | 0 |
| vestigial wings/red eyes | 0 | 0 |
| vestigial wings/sepia eyes | 0 | 0 |
Table 3 Phenotypes of the F2 Generation
| Phenotypes | Number of Males | Number of Females |
| Normal wings/red eyes | 4 | 7 |
| Normal wings/sepia eyes | 4 | 5 |
| vestigial wings/red eyes | 0 | 1 |
| vestigial wings/sepia eyes | 0 | 0 |
| normal red/mutated body shape | 2 | 0 |
| normal sepia/mutated body shape | 1 | 0 |
Questions
Discussion and Conclusion:
The results of our parental cross turned out just as expected, but our F2 generation was not normal. Some sort of mutation must have occurred that caused the strange body shape seen in several individuals of our F2 generation.
Genetics Problems
ppt Questions
Independent Assortment
1. How many different kinds of gametes could the following individuals produce? Remember the formula 2n where n equals the number of heterozygotes.
a. aaBb
b. CCDdee
c. AABbCcDD
d. MmNnOoPpQq
e. UUVVWWXXYYZz
P1, F1, and F2 Monohybrid Crosses
2. In dogs, wire-haired is due to a dominant gene (W), smooth-haired is due to its recessive allele (w). Show the results of crossing a homozygous wire-haired dog with a smooth-haired dog.
3. What kind of cross is this?
4. What was the genotype of all of the puppies? the phenotype?
5. The puppies belong to the _________ generation.
6. How would you write the F1 cross for this trait?
7. Show the results of working the F1 cross for this trait.
6. What phenotypic ratio did you get from this F1 cross?
7. What genotypic ratio did you get from this F1 cross?
8. Two wire-haired dogs are mated. Among the offspring of their first litter is a smooth-haired pup. If these two dogs mate again, what are the chances of them having another smooth-haired pup?
9. What are the chances that the pup will be wire-haired?
10. A Wire-haired male is mated with a smooth-haired female. The mother of the wire-haired male was smooth-haired. What are the phenotypes and genotypes of the pups they could produce? Show how you got your results.
Incomplete Dominance
11. In snapdragons, red flower color (R) is incompletely dominant over white flower color (r). The hybrids or heterozygous plants (Rr) are pink in color. Show the genotype for a white flower and for a red flower.
12. If a red-flowered plant is crossed with a white-flowered plant, what are the genotypes and phenotypes of the F1 generation plants? Show your work.
13. What is the phenotype of the flowers? what is their genotype?
14. What genotypes and phenotypes will be produced in the F2 generation? Show your work.
15. How did the genotypic and phenotypic ratio compare to each other in this incomplete dominance cross?
16. What would the phenotypic ratio have been if this had been complete dominance?
17. What kind of offspring can be produced if a red-flowered plant is crossed with a pink-flowered plant? Show your work.
18. What kind of offspring is/are produced if a pink-flowered plant is crossed with a white-flowered plant? Show your work.
Sex-linked Traits
19. What is the genotype for female? for male?
20. In humans, colorblindness (Xc) is a recessive sex-linked trait. Two people with normal color vision (XC) have a colorblind son. What are the genotypes of the parents?
21. What are the genotypes and phenotypes possible among their other children? Show your work.
22. A couple has a colorblind daughter. What are the possible genotypes and phenotypes of the parents and the daughter?
Dihybrid Crosses
23. In humans, the presence of freckles is due to a dominant gene (F) and the non-freckled condition is due to its recessive allele (f). Dimpled cheeks (D) are dominant to non-dimpled cheeks (d). Two persons with freckles and dimpled cheeks have two children. One child has freckles but no dimples. The other child has dimples but no freckles. What is the genotypes of the parents? the children?
24. What are the possible phenotypes and genotypes of the children that they could produce? Show all your work.
25. What phenotypic ratio did you get?
26. What genotypic ratio did you get?
27. What are the chances that they would have a child whom lacks both freckles and dimples? What would be the child’s genotype?
28. A person with freckles and dimples whose mother lacked both freckles and dimples marries a person with freckles but no dimples whose father did not have freckles or dimples. What are the chances that they would have a child whom lacks both freckles and dimples? Show the genotypes of the parents and all the offspring.
29. In dogs, the inheritance of hair color involves a gene (B) for black hair and a gene (b) for brown hair. A dominant (C) is also involved. It must be present for the color to be synthesized (made). If this gene is NOT present, a blond condition results. Complete the following table:
| Genotype | Phenotype | Color Deposition gene |
| BB or Bb | CC or Cc | |
| bb | CC or Cc | |
| BB or Bb | cc | |
| bb | cc |
30. A brown-haired male, whose father was a blond, is mated with a black-haired female ,whose mother was brown-haired and her father was blond. What is the genotype of the man and woman? Show the genotypes and phenotypes of all of their offspring.
Population Genetics or Hardy-Weinberg Law
Sixteen percent (16%) of the human population is known to be able to wiggle their ears. This trait is determined to be a recessive gene. Use the following equations to answer this population genetics problem:
1 = p2 + 2pq + q2 then use 1 = p + q
p2 – frequency of homozygous dominants
2pq – frequency of heterozygotes
q2 – frequency of homozygous recessives
p – frequency of dominant allele
q – frequency of recessive allele
31. What percent of the population is homozygous dominant for this trait? Show your work.
32. What percent of the population is heterozygous for this trait? Show your work.
Multiple Alleles – ABO Blood Type
33. Henry Anonymous, a film star, was involved in a paternity case. The woman bringing the suit had two children. One child had blood type A and the other child had blood type B. Her blood type was O, the same as Henry’s. The judge in the case awarded damages to the woman, saying that Henry had to be the father of at least one of her children. was the judge correct in his decision? Show how you got your answer.