ALIEN GENETICS - Mater Lakes
ALIEN GENETICS (ANSWERS)
1) Answer: 0%
Trait: Body Color
Alleles: (Y) Yellow or (y) orange
Cross: Heterozygous male (yellow, Yy) vs. Pure Yellow Female (YY)
| |Y |Y | |
| | | |Orange = Recessive Phenotype = Recessive Genotype = yy |
| | | |Hybrid vs. pure (Test Cross) --> Same as parents!! |
| | | |Genotypes / Genotype Ratio = Same as parents, in this case: 50% Yy & 50% YY (1 : 1) |
| | | |Phenotypes / Phenotype Ratio = Same as parents, in this case: 100% Yellow! |
| | | |Therefore, chances of an orange baby = 0% |
| | | |(As per Punnett Square to the side) |
|Y |YY |YY | |
|y |Yy |Yy | |
| | | | |
2) Answer: 25%
Trait: Eye Color
Alleles: (P) Purple or (p) White
Cross: Purple-Eyed Hybrid (Pp) vs. Purple-Eyed Hybrid (Pp)
| |P |p | Homozygous recessive = pp |
| | | |Hybrid vs. Hybrid (F1 Cross) --> |
| | | |Genotype: 25% Pure Dominant; 50% Hybrids; 25% Pure Recessive (1 : 2 : 1) |
| | | |Phenotype: 75% Dominant; 25% Recessive (3 : 1) |
| | | |Therefore, chance of homozygous recessive = 1/4 or 25% |
| | | |(As per Punnett Square to the side) |
|P |PP |Pp | |
|p |Pp |pp | |
| | | | |
3) Answer: 50%
Trait: # of Body Rings
Alleles: (R) 3 or (r) 5
Cross: Heterozygous female (3 rings, Rr) vs. 5-ringed male (recessive phenotype = rr)
| |R |r | Mother look or phenotype = Dominant, or 3 rings (RR or Rr genotype will be expressed as 3 rings) |
| | | |Hybrid vs. pure (Test Cross) --> Same as parents!! |
| | | |Genotypes / Genotype Ratio = Same as parents, in this case: 50% Rr & 50% rr (1 : 1) |
| | | |Phenotypes / Phenotype Ratio = Same as parents, in this case: 50% 3 ring; 50% 5-ring |
| | | |Therefore, chance of looking like mother (Rr or RR) is 2/4 = 1/2 = 50% |
| | | |(As per Punnett Square to the side)) |
|r |Rr |rr | |
|r |Rr |rr | |
| | | | |
4) Answer: 100%
Trait: Body Color
Alleles: (Y) Yellow or (y) orange
Cross: Pure dominant male (yellow, YY) vs. Homozygous recessive female (green, yy)
| |Y |Y | Hybrid gene = Yy |
| | | |Remember Pure vs. Pure (P Cross) --> 100% Hybrids |
| | | |Therefore, 100% change of hybrid |
| | | |(As per Punnet Square to the side) |
|y |Yy |Yy | |
|y |Yy |Yy | |
| | | | |
5) Answer: 50%
Trait: # of Antennae
Alleles: (A) 2 or (a) 1
Cross: 1-antennae male (recessive trait, aa) vs. Heterozygous 2-antennae female (Aa)
| |A |a | Recessive phenotype = Recessive Genotype = aa |
| | | |Hybrid vs. pure (Test Cross) --> Same as parents!! |
| | | |Genotypes / Genotype Ratio = Same as parents, in this case: 50% Aa & 50% aa (1 : 1) |
| | | |Phenotypes / Phenotype Ratio = Same as parents, in this case: 50% 1-antennae; 50% 2 |
| | | |Therefore, chance of a recessive phenotype (aa) = 2/4 = 1/2 = 50% |
| | | |(As per Punnet Square to the side) |
|a |Aa |Aa | |
|a |aa |aa | |
| | | | |
6) Answer: 75%
Trait: Eyesight
Alleles: (E) Glasses needed or (e) Glasses not worn
Cross: Heterozygous male (Ee) vs. Phenotypically dominante female carrying recessive allele (same as a hybrid, Ee)
| |E |e | Wear glasses = Dominant Phenotype = EE or Ee |
| | | |Hybrid vs. Hybrid (F1 Cross) --> |
| | | |Genotype: 25% Pure Dominant; 50% Hybrids; 25% Pure Recessive (1 : 2 : 1) |
| | | |Phenotype: 75% Dominant; 25% Recessive (3 : 1) |
| | | |Therefore, chance of a Dominante Phenotype (EE or Ee) = 3 to 1 or 75% |
| | | |(As per Punnet Square to the side) |
|E |EE |Ee | |
|e |Ee |ee | |
| | | | |
7) Answer: 50%
Trait: # of Body Rings
Alleles: (R) 3 or (r) 5
Cross: 3-ringed female carrying both dominant and recessive (hybrid, Rr) vs. Recessive phenotype male (rr)
| |R |r | Genetically pure baby = Homozygous = RR or rr |
| | | |Hybrid vs. pure (Test Cross) --> Same as parents!! |
| | | |Genotypes / Genotype Ratio = Same as parents, in this case: 50% Rr & 50% rr (1 : 1) |
| | | |Phenotypes / Phenotype Ratio = Same as parents, in this case: 50% 3-rings; 50% 5-rings |
| | | |Therefore, chance of a pure baby = 2/4 = 1/2 = 50% |
| | | |(As per Punnet Square to the side) |
|r |Rr |rr | |
|r |Rr |rr | |
| | | | |
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