Chapter 14 (Part 2) DIHYBRIDS AP Biology Ms. Day

DIHYBRID CROSSES: Assuming genes follow Mendelian Genetics (complete dominance)

Dihybrid Crosses  crosses involving crossing 2 DIFFERENT traits at one time Example: Mate 2 parents and look at the probability of seeing 2 traits, such as:  eye color AND hair color  freckles AND dimples

How do You Do Dihybrid Crosses?  2 separate monohybrid crosses (in your head)  1 square for EACH trait  use PROBABILITY RULES and MULTIPLY

The Multiplication and Addition Rules Applied to Genetic Crosses Probability Rule #1 The multiplication rule (“ and ”) product (x ing) States that the probability that 2 or more independent events will occur together is the product (x ing) of their individual probabilities

Probability Rule #2  The rule of addition (“or”) States that the probability that any 1 of 2 or more exclusive events will occur is calculated by adding (+ ing) together their individual probabilities

How do You Do Dihybrid Crosses? 2 ways 1. one BIG Punnett Square (16 boxes instead of 4) 2. Make 2 separate Punnett Squares  1 square for EACH trait  Then use PROBABILITY RULES and MULTIPLY

One BIG Punnett Square 1. Find the possible gametes using F.O.I.L method AaBb x AABbABAb aBAB abAb

Place the gametes in the Punnett Square & fill in boxes AB Ab aB ab AB Ab AB Ab AABBAABbAaBBAaBb AAbBAAbbAabBAabb AABBAABbAaBBAaBb AAbBAAbbAabBAabb

Genotypic Ratio AAbB = 8/16 or 1/2 AaBB = 2/16 or 1/8 Aabb = 2/16 or 1/8 AABB = 2/16 or 1/8 AAbb = 2/16 or 1/8

Phenotypic Ratio A= freckles, a= no freckles B = Dimples, b= no dimples AAbB = 8/16 (50%) freckles, dimples AaBB = 2/16 (12.5%) freckles, dimples Aabb = 2/16 (12.5%) freckles, no dimples AABB= 2/16 (12.5%) freckles, dimples AAbb = 2/16 (12.5%) freckles, no dimples So…12/16 (75%) freckles AND dimples 4/16 (25%) freckles AND no dimples

Solving a Dihybrid Cross  Write out a key for EACH trait AA, Aa= Freckles, aa= No freckles BB, Bb= Dimples, bb= No dimples  Figure out the genotypes of EACH parent and put them together  Look at the alleles for EACH gene SPERATELY & figure out the FRACTION ratio of possible genotype and phenotypes trait #1  genotype/phenotype possibilities trait #2  genotype/phenotype possibilities  PUTTING IT ALL TOGETHER…Figure out genotype/phenotype possibilities of offspring by multiplying fractions

Probability Rules and Dihybrid Crosses  A dihybrid or other multicharacter cross Is equivalent to two or more independent monohybrid crosses occurring simultaneously  In calculating the chances for various genotypes from such crosses Each character first is considered separately and then the individual probabilities are multiplied together

But…You STILL do NOT need to use a Punnett Square  Every organism ALWAYS receives 1 allele for EACH gene from MOM and DAD  For Example: H= tall & h= short, F=purple & f = white In a cross HHFf x hhff, what will is the phenotype possibilities?  4/ 4 Tall (HH) x ½ Purple (Ff)=  PR= 50% Tall & Purple In a cross HhFf x hhFf, what will is the possibility that the offspring will have the 1 st genotype?  HhFf? Hh x hh  what is the probability of getting Hh? ½½ Ff x Ff  what is the probability of getting Ff? ½½  So..the probability of getting HhFf is ½ x ½ = ¼

Now…Let’s do Multicharter Problems  What is the probability of producing an offspring with the phenotype AaBBCcDDeeFf in a cross between 2 parents with the following genotypes? AABbCcDDeeFf X AaBbCcDdeeFf  ½ x ¼ x ½ x ½ x 4/ 4 x ½  = 4/256 = 1/64 or 1.5% chance

How can you find possible gametes combinations? Find the possible gametes using F.O.I.L method AaBb (male) x AABb (female) AB AB Ab Ab aB AB repeat ab Ab repeat

Would you like to know a few SHORT CUTS?

Short Cuts for MONOHYBRID CROSSES  Every parent “donates” only 1 allele to each offspring Law of Segregation  When crossing 2 heterozygous individuals in complete dominance, you will ALWAYS get 1:2:1 GENOTYPE ratio  1 homozygous dominant: 2 heterozygous: 1 recessive 3:1 PHENOTYPE ratio  3 dominant phenotype: 1 recessive

Short Cuts for DIHYBRID CROSSES  When crossing 2 heterozygous individuals in complete dominance, you will ALWAYS get 9:3:3:1 PHENOTYPE ratio  9 dominant, dominant phenotype  3 dominant, recessive phenotype  3 recessive, dominant phenotype  1 recessive, recessive phenotype NOTE: The genotypes have to be ALL heterozygous Ex: HhFf x HhFf