Did you do your homework? Happy Monday Bio-Ninjas! Please READ ALL sections below! Journals: NO NEW NOTES TODAY!!! Table of Contents Right Page 45: Dihybrid Crosses Left Page 44 Dihybrid Practice EQ: How are genetic outcomes predicted and when two traits are involved? Due today: Alien Punnett Square Homework One Day Late: (70) Two days Late: (50) Three Days Late: (0) Did you do your homework? HAVE ALL HOMEWORK, LATE WORK/EXTRA CREDIT OUT READY TO BE CHECKED!

Homework Check…Aliens!

What are the trait(s)?

What are the traits?

What are the traits?

Dihybrid Cross: a cross that shows the possible offspring for two traits Fur Color: B: Black b: White Coat Texture: R: Rough r: Smooth In this example, we will cross two individuals that are both heterozygous for each trait. What will their genotypes be? BbRr x BbRr

BbRr heterozygous for both traits. Notice how the genotypes are written when working with dihybrid scenarios… BbRr heterozygous for both traits. FFTt Homozygous dominant for trait 1 heterozygous for trait 2 rrXx Homozygous recessive for trait 1 hhgg Homozygous recessive for both traits

How do you set up a dihybrid cross? Before setting up a dihybrid cross/Punnett square one must understand the following concept… The Principle of Independent Assortment: the separation and assortment of genes during meiosis is independently random and gives traits equal opportunity of appearing together. *Remember “shuffling the deck”

Dihybrid Crosses…the steps BbRr x BbRr Step 1: Find ALL possible gamete combinations that can be made from each parent’s genotype. Remember, each parent gives one allele for each trait so each gamete must have one “B” allele and one “R” allele. Parent 1 Parent 2 You will ONLY use the following steps if you are ever asked to “find the possible gametes” -more than likely on a handout -rarely on a test but you will still need to know just in case.

BbRr x BbRr F O I L BR Br bR br Possible combinations: Parent 2 Step 2: FOIL Method Combine the FIRST, OUTER, INNER, and LAST letters of the genotype.

Each allele from trait 1 must “dance” with each allele from trait 2

Remember…you will only use FOIL if you are asked to “find the possible gametes” Also, you will more than likely never have to fill out an entire dihybrid Punnett Square but you will definitely need to recognize and know how to read one.

Step three: Set up the Punnett square using the gametes (possible genotypes) from each parent. BR bR br Br BBRR BbRR BbRr BBRr BBrr Bbrr bbRR bbRr bbrr

Dihybrid Crosses: a cross that shows the possible offspring for two traits BBRR BbRR BbRr BBRr BBrr Bbrr bbRR bbRr bbrr BbRr x BbRr Fur Color: B: Black b: White Coat Texture: R: Rough r: Smooth Glue this into your journals!

Phenotypic Ratio= 9:3:3:1 BR bR br Br BBRR BbRR BbRr BBRr BBrr Bbrr bbRR bbRr bbrr How many of the offspring would have black, rough coat? How many of the offspring would have a black, smooth coat? How many of the offspring would have a white, rough coat? How many of the offspring would have a white, smooth coat? Phenotypic Ratio= 9:3:3:1 Fur Color: B: Black b: White Coat Texture: R: Rough r: Smooth Write down and answer these questions on the left side under what you glued in.

SAVE YOURSELF SOME TIME!! When crossing two individuals that are heterozygous for both traits on a dihybrid cross, the phenotypic ratio will ALWAYS be 9:3:3:1

Is the FOIL method the only way to solve dihybrid probabilities?

Fractions Method BbRr x BbRr Bb Bb Rr Rr To figure out probabilities of dihybrid crosses we will use the “fractions method” for multiple reasons: It’s faster You’re doing what you already know You’ll get the same answer! You don’t have to FOIL You don’t have to fill in a 16 box square Umm…it’s faster Step 1: Match up the alleles for both individuals Parent 1 Parent 2 BbRr x BbRr Bb Bb Rr Rr

Step 2: Create a Monohybrid Punnett square for each of the traits Parent 1 Parent 1 B b R r B BB Bb R RR Rr b Bb r Rr rr bb Parent 2 Parent 2 Step 3: Fill in the punnett squares for each trait

Step 4: Find the probability for each trait then multiply Parent 1 Parent 1 Fur Color: B: Black b: White B b R r B Coat Texture: R: Rough r: Smooth BB Bb R RR Rr b Bb Rr rr bb r Parent 2 Parent 2 How many of the offspring would have a black, rough coat? 3 3 9 X = 4 4 16

Step 4: Solve probabilities by multiplying fractions Parent 1 Parent 1 Fur Color: B: Black b: White B b R r B Coat Texture: R: Rough r: Smooth BB Bb R RR Rr b Bb Rr rr bb r Parent 2 Parent 2 How many of the offspring would have a black, smooth coat? 3 1 3 X = 4 4 16

Step 4: Solve problems by multiplying fractions Parent 1 Fur Color: B: Black b: White Parent 1 B b R r Coat Texture: R: Rough r: Smooth B BB Bb R RR Rr b Bb Rr rr bb r Parent 2 Parent 2 How many of the offspring would have a white, rough coat? 1 3 3 X = 4 4 16

Step 4: Solve problems by multiplying fractions Parent 1 Parent 1 Fur Color: B: Black b: White B b R r B Coat Texture: R: Rough r: Smooth BB Bb R RR Rr b Bb Rr rr bb r Parent 2 Parent 2 How many of the offspring would have a white, smooth coat? 1 1 1 X = 4 4 16

Practice in your notes… In pea plants, yellow seeds (Y) are dominant over green seeds (y), and rounded peas (R) are dominant over wrinkled peas (r). Cross a plant that is heterozygous for both traits with a plant that is homozygous recessive for both traits. Genotypes: YyRr and yyrr What is the probability that the plant will be yellow and wrinkled?

FOIL (dance) the parents… YyRr and yyrr

Isolate the genotypes for trait 1 and trait 2 YyRr and yyrr Parent 1: Yy Rr Parent 2: yy rr Yy x yy Rr x rr

Y y R r y Yy yy r Rr rr y Yy Rr rr yy r 2 2 4 X = 4 4 16 How many of the offspring would have a yellow, wrinkled seed? 2 2 4 X = 4 4 16

YyRr X yyrr YR yR yr Yr YyRr yyRr yyrr Yyrr 4 1 or 16 4

1. *Hybrid* 2. *Purebred* What is the probability that the baby Reebop will have 2 antennae and 3 body segments?

A a B b A AA Aa b Bb bb AA Aa A b Bb bb 4 2 8 = X 4 4 16 What is the probability that the baby Reebop will have 2 antennae and 3 body segments? 1. *Hybrid* 2.*Purebred* A a B b A AA Aa b Bb bb AA Aa A b Bb bb 4 2 8 = X 4 4 16

Genotype / phenotype code 1. *Hybrid* 2. Hybrid eyes Characteristic Genotype / phenotype code eyes EE = 2 eyes Ee = 2 eyes ee = one eye humps HH = 1 hump Hh = 1 hump hh = 3 humps What is the probability that the baby Reebop will have 2 eyes and 3 humps?

Genotype / phenotype code Characteristic Genotype / phenotype code eyes EE = 2 eyes Ee = 2 eyes ee = one eye humps HH = 1 hump Hh = 1 hump hh = 3 humps What is the probability that the baby Reebop will have 2 eyes and 3 humps? 1. *Hybrid* 2.hybrid eyes E e H h E EE Ee h Hh hh Ee ee e h Hh hh 3 2 6 = X 4 4 16

Genotype / phenotype code Characteristic Genotype / phenotype code eyes EE = 2 eyes Ee = 2 eyes ee = one eye humps HH = 1 hump Hh = 1 hump hh = 3 humps What would the phenotypic ratio be? E e H h 1. *Hybrid* 2.hybrid eyes EE E Ee h Hh hh 2 eyes, 1 hump: 2 eyes, 3 humps: 1 eye, 1 hump: 1 eye, 3 humps: ¾ x ¾ = 9/16 Ee ee e h Hh hh ¾ x 2/4 = 6/16 ¼ x 2/4 = 2/16 9:6:2:2 ¼ x 2/4 = 2/16

Work on your Dihybrid practice for the rest of class Work on your Dihybrid practice for the rest of class. We’ll finish tomorrow 

 HAPPY TUESDAY  On the left side of your pedigree notes: Bellwork: On the left side of your pedigree notes: Sketch the pedigree chart Determine the genotypes for numbers 1-5 using the information given.  HAPPY TUESDAY  5 Tt tt tt Tt Shaded = Tongue Rolling Not Shaded = Non Tongue Rolling Tongue Rolling (T) is Dominant to Non Tongue Rolling (t) Bonus: Is there anything particularly interesting about this family? Have your Pedigree Homework and any Late Work out on your desk! tt

Punnett Square Homework (FC) Due This Week: Tuesday: Pedigree HW (FC) Family Tree (FC) Punnett Square HW (70) Wednesday: Pedigree HW (70) Family Tree (70) Punnett Square HW (50) Thursday: Pedigree HW (50) Family Tree (50) Friday: Notecards (picture w/color) Have out: Pedigree Homework Due today: Turn in or show me Graded Quiz Family Tree Part 1 Notecards (50) Punnett Square Homework (FC) I have PLENTY of extra Punnett Square EXTRA CREDIT if you need or want it.  Extra Credit can be turned in all the way up until the day of the test.

Pedigree HW check

Case Study #1 -2 points each shape A = Normal -2 points each genotype a = albinism Aa Aa aa Aa A- A- Aa Aa Aa aa

H = Huntington's Disease -2 points each shape -2 points each genotype Case Study #2 H = Huntington's Disease h = Normal hh Hh hh hh hh hh hh H- Hh hh hh hh hh hh Hh

Heterozygous Case Study #3 Nn No -2 points each genotype N = Neurofibromatosis n = Normal -2 points each Nn nn -2 points each genotype nn nn Nn nn Nn Nn Nn nn nn nn nn N- N- nn