AA. Brown hair BB. BB CC. Xx DD. Green eyes

 Genotype:  TWO Alleles  TWO Letters  3 “types”  AA, Aa, aa  Phenotype:  Trait you see  Determined by the genotype  Ex: Red socks, brown hair, tall, short, round eyes

Punnett Square: A tool used to predict the probability of genotypes and phenotypes of offspring Why would we need/want to know this info? Genetic Counselor = Analyzes parent’s DNA for possible traits/diseases Would you??

#6 How would a child have a trait NEITHER parent has? How would a child have a trait NEITHER parent has? (Hint: Think of a genotype) (Hint: Think of a genotype) Answer must include a minimum of: Answer must include a minimum of: 2 green words (in back) 2 green words (in back) 2 yellow words (in back) 2 yellow words (in back)

#6 How would a child have a trait NEITHER parent has? How would a child have a trait NEITHER parent has? (Hint: Think of a genotype) (Hint: Think of a genotype) Parents are HETEROZYGOUS Parents are HETEROZYGOUS One dominant & 1 recessive allele One dominant & 1 recessive allele Child inherits 2 recessive alleles Child inherits 2 recessive alleles

Punnett Square: A tool used to predict the probability of genotypes and phenotypes of offspring Why would we need/want to know this info? Genetic Counselor = Analyzes parent’s DNA for possible traits/diseases Would you??

 1. Dominant allele  2. Add a recessive allele  3. Is this a genotype or phenotype?  4. How would you abbreviate??

 1. Dominant allele  2. Add a dominant allele  3. Is this a genotype or phenotype?  4. How would you abbreviate??

 Genotype OR Phenotype?

 Mom has freckles (F) and Dad has no freckles (f). Each parent has a homozygous genotype.  Mom:  Dad:

 Step 1…..  Determine the genotypes of both parents.  Ex: Parent 1 is homozygous dominant for brown eyes (A) and Parent 2 is heterozygous  Homozygous = Same  Dominant = Capital  Parent 1 = AA  Heterozygous = Different  Parent 2 = Aa

 Step 2….  Make a Punnett Square – Leave room on top & bottom

 Step 3….  Write the first genotype (AA) next to box # 1 and box # 3 of the Punnett Square  IMPORTANT!!! Only one allele next to each box! 12 34

 Step 4….  Write the second genotype (Aa) on top of the box # 1 & # 2 of the Punnett Square 12 34

 Step 5….  “Multiply” the alleles together starting with box #1  Outside  In  Top  Bottom 12 34

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GENOTYPESPHENOTYPES  Box #1 = AA  Box #2 = Aa  Box #3 = AA  Box #4 = Aa  Brown eyes are dominant to blue eyes.  AA = Brown eyes  Aa = Brown eyes  AA = Brown eyes  Aa = Brown eyes  2 of 4: AA  50% Brown Eyes  2 of 4: Aa  50% Brown Eyes

 The 4 boxes = 100%  What % does box #1 represent?  25% or 1 of 4 probability of brown eyes  What % does box #2 represent?  25% or 1 of 4 probability of brown eyes  What % does box #1 & #2 represent?  50% or 2 of 4 probability of brown eyes

 Genotype  Phenotype  Ratio  AA  Brown eyes  50% OR 2:4  Aa  Brown eyes  50% or 2:4  Total = 100% BROWN eyes

 The dominant allele (T) is long whiskers & the recessive allele (t) is for short whiskers.  a) Both parents are heterozygous (hybrid)  Step 1: Tt and Tt

a) Both parents are heterozygous TT Tt tt T t T t TT: Long whiskers 25% OR 1:4 Tt: Long whiskers 50% OR 2:4 tt: Short whiskers 25% OR 1:4 Total = 75% long 25% short

 In purple people eaters, one-horn is dominant (H) and no horns is recessive (h).  Cross a heterozygous purple people eater with a purple people eater that has no horns.  Step 1: Tt and tt (no horns is recessive)

Cross a purple people eater that is heterozygous with a purple people eater that has no horns. Hh hh Hh hh Hh h h Hh: One horn 50% OR 2:4 hh: No horns 50% OR 2:4