Bari Kashif A08218440 Section B09 BICD 100 (Genetics) Homework 1
1. Yellow (Y) Round (R) Green (y) Wrinkled (r)
A: Y_R_ x yyrr 50% Yellow Round; 50% Green Round YyRR x yyrr 2 YyRr (Yellow Round): 2 yyRr (Green Round) Genotype of Pea A= YyRR B: Y_R_ x yyrr 100% Yellow Round YYRR x yyrr 4 YyRr (Yellow Round) Genotype of Pea B= YYRR C: Y_R_ x yyrr 25% Yellow Round; 25% Yellow Wrinkled; 25% Green Round; 25% Green Wrinkled YyRr x yyrr 1 YyRr (Yellow Round): 1 yyRr (Green Round): 1 Yyrr (Yellow Wrinkled): 1 yyrr (Green Wrinkled) Genotype of Pea C= YyRr 2. If the fruit fly was of phenotype `A', then that particular allele is dominant relative to another allele. In order to determine if the genotype was homozygous dominant or heterozygous, I would cross the fruit fly with another fruit fly of phenotype `a', which would be homozygous recessive. If the original fruit fly was homozygous dominant, then the phenotype of the offspring would all show that of allele `A'. If the original fruit fly was heterozygous, then half progeny would display phenotype `A' and the other half would display phenotype `a'.
3. Multiplication Rule: Probability of two independent events occurring Addition Rule: Probability of any two or more mutually exclusive events occurring together a) 1/6 x 1/6 x 1/6 = 1/216
Kashif Bari A08218440 Section B09 b) 1/6 x 1/6 x 1/6 = 1/216
c) 1/6 x 1/6 x 1/6 = 1/216
d) 5/6 x 5/6 x 5/6 = 125/216
e) (1/6 + 1/6) x 1/6 x 1/6 = 1/108
f) (1/6 + 1/6) x 1/6 x 1/6 = 1/108
g) (1/6 + 1/6 + 1/6 + 1/6 + 1/6 +1/6) x 1/6 x 1/6 = 1/36
h) (1/6 + 1/6 + 1/6 + 1/6 + 1/6 +1/6) x 5/6 x 4/6 =20/36 =5/9
4. Achondroplastic Dwarfism (A), Neurofibromatosis (N) Male: aaNn x Female: Aann 1 AaNn : 1 aaNn : 1 Aann : 1 aann in Phenotypically the progeny, there is a 50% chance of Achondroplastic dwarfism and a 50% chance of Neurofibromatosis. There is a 25% chance of the progeny having both Achondroplastic dwarfism and Neurofibromatosis. The phenotypic ratio can be presented as follows: 1 (Achondroplastic dwarfism and Neurofibromatosis) : 1 (Achondroplastic dwarfism) : 1 (Neurofibromatosis) : 1 (Neither)
5. Aa Bb Cc Dd Ee x aa Bb cc Dd ee 50% Aa 25% BB 50% Cc 25% DD 50% Ee 50% aa 50% Bb 50% cc 50% Dd 50% ee 25% bb 25% dd
Kashif Bari A08218440 Section B09 a) Phenotypical resemblance (Multiplying the probability for each specific genetic combination to occur) a. First Parent (Aa Bb Cc Dd Ee)= x x x x = 9/128 b. Second Parent (aa Bb cc Dd ee) = x x x x = 9/128 c. Either Parent = 1 x x 1 x x 1 = 9/16 d. Neither Parent = 0 x x 0 x x 0 = 0 (none) b) Genotypical resemblance(Multiplying the probability for each genetic combination to occur) a. First Parent = x x x x = 1/32 b. Second Parent = x x x x = 1/32 c. Either Parent =1 x x x x =1/16 d. Neither Parent= 0 x x 0 x x 0 = 0 (none)
6. If a cell undergoes a process of mitosis, then the genotype of the daughter cells are exactly the same. The daughter cells should have the genotype AaBbCc.
7. Must be Sex-linked XX (Female) XY (Male) System of Inheritance
XGXg x XgY 1 XGXg : 1 XGY : 1 XgXg : XgY XGXG x XgY 2 XGY (Graceful Male): 2 XGXg (Graceful Female) Results indicate that it cannot be XX-XY System of Inheritance ZZ (Male) ZW (Female) System of Inheritance ZgZg x ZGW 2 ZGZg (Graceful Male) : 2 ZgW (Gruesome Female) Results indicate that it is a ZZ-ZW System of Inheritance with the parents having the Genotypes represented in the cross above.