Lab Report Questions - Color Blindness, Pedigree Analysis, Blood Typing, and Bacterial Transformation

Lab Report 12 - Experiment 1: Color Blindness

Color Blindness Phenotypes and Genotypes

• Phenotype: Colorblind or Not Colorblind
• Genotype:
  • Colorblind Male: Xc Y
  • Carrier Female: Xc X

Question 1: Genotype of a Colorblind Male

• The genotype of a colorblind male is Xc Y, where Xc represents the X chromosome carrying the colorblindness allele.

Question 2: Genotype of a Carrier Female

• The genotype of a carrier female is Xc X, where one X chromosome carries the colorblindness allele (Xc) and the other X chromosome is normal (X).

Question 3: Punnett Square: Colorblind Man and Non-Colorblind Woman

• Cross: Colorblind Man (Xc Y) x Non-Colorblind Woman (XX)

• Punnett Square:

  | | X | X |
  | :---- | :- | :- |
  | Xc | | |
  | Y | | |

• Genotypes:

  • Females: X Xc
  • Males: X Y

• Phenotypes:

  • Females: Carriers (Non-Colorblind)
  • Males: Non-Colorblind

• Probabilities:

  • Females (Carriers): 100%
  • Males (Non-Colorblind): 100%

Question 4: Punnett Square: Colorblind Woman and Non-Colorblind Man

• Cross: Colorblind Woman (Xc Xc) x Non-Colorblind Man (XY)
  *Punnett Square:

• Genotypes:
  • Females: X Xc
  • Males: Xc Y
• Phenotypes:
  • Females: Carriers (Non-Colorblind)
  • Males: Colorblind
• Probabilities:
  • Females (Carriers): 100
  • Males (Colorblind): 100

Experiment 2: Pedigree Analysis

Determining Mode of Inheritance

• Analyze pedigrees to determine the mode of inheritance.
• Modes of Inheritance:
  • Autosomal Dominant
  • Autosomal Recessive
  • X-Linked Recessive

Genotype Analysis

• Determine genotypes for individuals A, B, and C in each pedigree based on the identified mode of inheritance.

Pedigree Examples:

• Pedigree #1:
  • Mode of Inheritance: X-Linked Recessive
  • Genotype for A: Xc Y
  • Genotype for B: X X
  • Genotype for C: X Xc
• Pedigree #2:
  • Mode of Inheritance: Autosomal Recessive
  • Genotype for A: Aa
  • Genotype for B: Aa
  • Genotype for C: aa
• Pedigree #3:
  • Mode of Inheritance: Autosomal Dominant
  • Genotype for A: AA
  • Genotype for B: Aa
  • Genotype for C: aa
• Pedigree #4:
  • Mode of Inheritance: X-Linked Recessive
  • Genotype for A: XY
  • Genotype for B: X X
  • Genotype for C: Xc Xc
• Pedigree #5:
  • Mode of Inheritance: Autosomal Recessive
  • Genotype for A: Aa
  • Genotype for B: Aa
  • Genotype for C: aa

Experiment 3: Blood Typing

Blood Type and Genotype

• Blood Types: A, B, AB, O
• Rh Factor: Positive (+), Negative (-)
• Possible Genotypes:
  • A: I^A I^A or I^A i
  • B: I^B I^B or I^B i
  • AB: I^A I^B
  • O: ii
  • Rh Positive: Rh^+ Rh^+ or Rh^+ Rh^-
  • Rh Negative: Rh^- Rh^-

Sample Analysis

• Andy Smith:
  • Blood Type: A+
  • Possible Genotypes: I^A I^A Rh^+ Rh^+ , I^A I^A Rh^+ Rh^-, I^A i Rh^+ Rh^+, I^A i Rh^+ Rh^-
• Mrs. Jane Green:
  • Blood Type: O+
  • Possible Genotypes: ii Rh^+ Rh^+, ii Rh^+ Rh^-
• Mr. John Green:
  • Blood Type: B+
  • Possible Genotypes: I^B I^B Rh^+ Rh^+, I^B I^B Rh^+ Rh^-, I^B i Rh^+ Rh^+, I^B i Rh^+ Rh^-
• Jimmy Green:
  • Blood Type: A-
  • Possible Genotypes: I^A I^A Rh^- Rh^-, I^A i Rh^- Rh^-

Question 1: Paternity Analysis

• Scenario: Jane Green is Jimmy Green's mother. John Green doubts he is Jimmy's father and suspects Andy Smith.

Part a) Could John Green be Jimmy's father?

• John Green (B+) vs. Jimmy Green (A-).
• For Jimmy to be A-, he must have the genotype I^A I^A Rh^- Rh^- or I^A i Rh^- Rh^-.
• John Green can contribute either I^B or i, and Rh^+ or Rh^-.
• If John is I^B I^B, he cannot contribute the I^A allele to Jimmy. So, he could not be the father.

Part b) Could Andy Smith be Jimmy's father?

• Andy Smith (A+) vs. Jimmy Green (A-).
• For Jimmy to be A-, he must have the genotype I^A I^A Rh^- Rh^- or I^A i Rh^- Rh^- .
• Andy Smith can contribute either I^A or i, and Rh^+ or Rh^-.
• If Andy were I^A I^A Rh^- Rh^- or I^A i Rh^- Rh^- it would be possible for him to be the father.

Question 2: Blood Type Probabilities

• Scenario: Jane Green is heterozygous for both traits (A+), and Andy Smith is homozygous (A+).

• Jane Green: I^A i Rh^+ Rh^-

• Andy Smith: I^A I^A Rh^+ Rh^+

• Punnett Square:

  	I^A Rh^+	I^A,Rh^+	I^A,Rh^+	I^A,Rh^+
  I^A,Rh^+
  I^A,Rh^-
  i,Rh^+
  i,Rh^-

• Possible Phenotypes: A+

Question 3: Offspring Blood Types with John Green

• Scenario: Jane Green is heterozygous for both traits. What blood types could her offspring with John Green have?

• Jane Green: I^A i Rh^+ Rh^-

• John Green: I^B i Rh^+ Rh^-

• Gametes:

  • Jane Green: I^A Rh^+, I^A Rh^-, i Rh^+, i Rh^-
  • John Green:I^B Rh^+, I^B Rh^-, i Rh^+, i Rh^-

• Punnett Square:

  	I^B Rh^+	I^B Rh^-	i Rh^+	i Rh^-
  I^A Rh^+
  I^A Rh^-
  i Rh^+
  i Rh^-

• Possible Blood Types for offspring: A+, A-, B+, B-, AB+, AB-, O+, O-

Experiment 4: Bacterial Transformation

Question 1: Definition of a Plasmid

• A plasmid is a small, circular DNA molecule that is physically separated from chromosomal DNA and capable of replicating independently. This is an extra piece of DNA that can hold different genes.

Question 2: Charge of E.coli Membrane

• The membrane of E.coli has a negative charge due to the lipopolysaccharides and phospholipids present in the cell membrane.

Question 3: Charge of a Plasmid

• A plasmid has a negative charge due to the phosphate groups in the DNA backbone.

Question 4: Role of Calcium Chloride

• Calcium chloride (CaCl2) is used to make the cell membrane more permeable. The positively charged calcium ions neutralize the negative charges on the cell membrane and the plasmid DNA, facilitating the entry of the plasmid into the cell.

Question 5: Diagram of Plates After Incubation

• P- LB/AMP Plate:

  • LB/AMP Plate: No colonies

• P+ LB/AMP Plate:

  • LB/AMP Plate: White colonies

• P+ LB/AMP/ARA Plate:

  • LB/AMP/ARA Plate: Red colonies