Punnett Squares Review

1. Punnett Squares Review By: ScarlettNeuberger, Barbara Zamilatskaya, Ishmath Cellion

2. Monohybrid Cross • A monohybrid cross is a genetic cross using a single trait with two alleles.

3. Do It Yourself !

4. Monohybrid (Sex Linked) PROBLEM:  Cross a female carrier for hemophilia with a male with hemophilia.  H=normal, and h=hemophilia STEP 1:  Determine what kind of problem you are trying to solve.   Does it involve simple dominant and recessive traits,  Incomplete dominance, or Co-dominance? Is it a monohybrid or dihyrid? In this case there is only one trait..........this is a monohybrid cross involving sex linked traits. STEP 2:  Determine letters you will use to specify traits.   In this case it is a sex-linked problem.  Remember that XX is female, and XY is male.  H=normal and h=hemophilia.  Normally you would not write the capital letters on the genotypes, only the small case (the recessive gene responsible for the disorder) STEP 3:  Determine parents genotypes.   In this case you were told the parents were: Female carrier = XhX, and a male with hemophilia = XhY.   The Cross is:  XhXxXhY STEP 4:  Make your punnet square and make gametes (these go on the top and side of your punnett square. XhXwould make a Xhand X   XhYwould make a Xhand Y   NOTE:  The females gametes always go on top of the punnett square and the males on the side.  STEP 5:  Complete cross and determine possible offspring. STEP 6:  Determine genotypic and phenotypic ratios. Genotypic ratio:  Make a list of all the different genotypes (the letter combinations) and determine how many of each you have. In your problem this would be:  XhX  = 1,  XhXh  = 1, and XY= 1, and XhY = 1 The genotypic ratio would therefore be 1 : 1 : 1 : 1  Phenotypic ratio:  Make a list of all the different phenotypes (physical characteristics. In you problem this would be:  Female Carrier = 1, Female w/ hemophilia = 1, Normal male = 1, and Male w/ hemophilia = 1.   The phenotypic ratio would be 1 : 1 : 1 : 1 

5. The BIG Punnett Square • Used when performing a dihybrid cross (the crossing of two traits) • Each letter of one trait is paired with a letter from the other • Example: AaBb x AaBa • AB • Ab • aB • ab

6. What it looks like…Cross: AaBb x AaBb AB AbaB ab AB Ab aB ab

7. You try it…Cross: FFgg x FfGG _____ ______ ______ _____ _____ _____ _____ _____

8. The easy way out… • To look for the probability of a specific genotype: • Separate the two traits: • AaBb x AaBbAa x Aa and Bb x Bb • Make two separate Punnett squares • Find the probability for each part of the genotype • Ex. ¼ chance of AA and ¼ chance of BB • Multiply the two probabilities • ¼ x ¼ = 1/16 A a B b B b A a

9. Example: Find the probability of genotype DdHH in the offspring of parents who have genotypes DDHH and ddHh • DDHH x ddHh DD x dd and HH x Hh • Probability of Dd: 4/4 • Probability of HH: 2/4 • Probability of DdHH: 4/4 x 2/4 = 8/16 = 1/2 d d H h D D H H

10. You try it… • Find the probability of the genotype QQEE in the offspring of parents who have genotypes QqEe and QQEE _____ _____ _____ _____ _____ _____ ____ ____

11. Blood Type Punnett Square • Used when trying to figure out the blood type of the offspring based on the blood type of the parents. • Solved like regular monohybrid cross Punnett square once the genotype for the blood is figured out Blood type: ABABO Genotype: AA,AOBB,BOABOO

12. The mother is type A homozygous, the father is type AB. What is the possible genotype for their children? A A • B Answer: A A A B • B

13. Now you try it…The mother it type O, the father is type A heterozygous. Find the possible genotype of the children. O Answer O A O