ࡱ> xzw'` 4bjbj 4l4 v-v-v-v-4-e<2.....///;;;;;;;$=h?b <2//22 <..<6662..;62;666.. p㇭v-V5675<0e<6a@5a@6a@6T/06?1t1/// < <6///e<2222v-v-$  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) YY 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)YYYYYyYyYy 2) Answer: 25% Trait: Eye Color Alleles: (P) Purple or (p) White Cross: Purple-Eyed Hybrid (Pp) vs. Purple-Eyed Hybrid (Pp) Pp 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) PPPPppPppp 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) Rr 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))rRrrrrRrrr 4) Answer: 100% Trait: Body Color Alleles: (Y) Yellow or (y) orange Cross: Pure dominant male (yellow, YY) vs. Homozygous recessive female (green, yy) YY Hybrid gene = Yy Remember Pure vs. Pure (P Cross) --> 100% Hybrids Therefore, 100% change of hybrid (As per Punnet Square to the side)yYyYyyYyYy5) Answer: 50% Trait: # of Antennae Alleles: (A) 2 or (a) 1 Cross: 1-antennae male (recessive trait, aa) vs. Heterozygous 2-antennae female (Aa) Aa 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)aAaAaaaaaa6) 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) Ee 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)EEEEeeEeee7) 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) Rr 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)rRrrrrRrrr &)-}      & ) 5 6 7 : A L X Y [ n w ì՝vddddd#h B*CJOJPJQJaJphh hdjCJOJQJaJhdjCJOJQJaJh`CCJOJQJaJh h`CCJOJQJaJ,h`Ch`C5B*CJOJPJQJaJph#h`CB*CJOJPJQJaJph)h h B*CJOJPJQJaJph)h h.%B*CJOJPJQJaJph"),?b n $Ifgd`Cd$Ifgd2 g$d$Ifa$gd $d$Ifa$gdL`$d$Ifa$gd dgd $da$gd 4  I8$d$Ifa$gd`Ckd$$IfTl4J\S*c`6# t0*644 laytL`T $Ifgd2 g      4kd$$IfTl4-\S*c6# t0*644 laytL`T$d$Ifa$gd $d$Ifa$gd    ! 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