1. Saturday Study Session 2 Theme of the day: Information Transfer
Session 4 - Mendelian Genetics

2. Clue: Highest percentage of Roan = Rr
Question 1 Answer A
Clue: Highest percentage of Roan = Rr
Please help students see the combination of two opposing pure breeds yields the highest number of hybrids by working through the matings in Punnett square analysis. Also point out that answers C and D are essentially the same.

3. Clue: recessive = must be homozygous recessive
Question 2 Answer D
Clue: recessive = must be homozygous recessive
Please stress to students that probability has no memory. If each is a heterozygous carrier, there is a 75% chance of normal versus a 25% of suffering. It was maybe bad karma two have 2 for 2.

4. Clue: all traits are homozygous and opposites
Question 3 Answer B
Clue: all traits are homozygous and opposites
Please help students see that this situation will result in all heterozygous offspring for both traits; therefore, they will all be tall and pink flowered.

5. Clue: two different genes affect one trait
Question 4 Answer C
Clue: two different genes affect one trait
Students need to understand that epistasis is the result of gene interaction. One gene is influencing the phenotypic display of another gene.

6. Epistasis
BbCc
BbCc
Sperm 1 BC 1 bC 1 Bc 1 4 4 4 4 bc 1 BC 4
BBCC
BbCC
BBCc
BbCc 1 bC
BbCC 4
bbCC
BbCc
bbCc 1 Bc
BBCc
BbCc
BBcc
Bbcc 4
Please help students understand that epistasis also does not present in the normal Mendelian ratios for a dihybrid. Collegeboard loves to ask epistasis questions on the exam. In epistasis, the ratio is usually a 9:3:4 ratio instead of the Mendelian 9:3:3:1 (next slide) 1
bbcc bc
BbCc
bbCc
Bbcc 4 9 3 4 16 16 16

7. Dihybrid – Normal 9:3:3:1 ratio
Please point out the normal 9:3:3:1 ration that Mendel recognized. The difference was Mendel was see non-interacting gene traits; but epistasis shows gene interaction.

8. Clue: AB; but both parents are B
Question 5 Answer C
Clue: AB; but both parents are B

9. Codominance (AB) & Multiple Alleles
Please help students see the blood types as they relate to the genotype. Please stress to students how blood genes are written in column b.

10. Clue: one gene with several phenotypic effects
Question 6 Answer B
Clue: one gene with several phenotypic effects

11. Pleiotropy (Sickle Cell)
Please be sure to discuss the effects of possessing this allele, such as RBC shape, decreased ability to carry O2, resistance to Malaria. Cystic Fibrosis is another example of Pleiotropy.

12. Math Grid In Answer The correct answer is: 1/32
Father – A a B b C c D d
   
Mother -A a B B c c D d
Desired - A a B B C c d d
#arrows = 2/ / / ¼
Reduce ½ x ½ x ½ x ¼ = 1/32
Please help sttudents see each trait as an individual Punnett square with four possible squares. This way has proven to be the quickest and easiest in working these probability problems. Tell students to put the two parent genomes one under the other, as above. Then draw a line and put the desired underneath. Then draw arrows for the possible ways the desired could be achieved from the given genomes. Then count the arrows and show as a fraction. There should only be ¼, 2/4, or 1 in working these problems. Then just multiple the fractions in order as above. This can be done in well under a minute, if practiced.

13. Short Free Response 1 (4 points possible)
Possible points awarded for:
Discussion of this being an X-linked trait (1pt.)
Discussion of the trait “skipping a generation”
(1 pt.)
Discussion of only males being affected and females being shown as carriers. (1 pt.)
Discussion of a female inheriting two recessive alleles; one on each of her twoX chromosomes. (1pt.)

14. Pedigree
Please point out that the affected individuals are all males, but stress that females can also get this condition by inheriting to recessive alleles (one on each X chromosome). Also remind them that colorblindness can also be a X-linked recessive trait.

15. Short Free Response 2 (3 points possible)
Possible points awarded for: Stating the order of genes being: vg-cn-rb-b (1 pt.) Discussion of the highest frequency being farthest apart. (1 pt.) Discussion of lowest frequency being closest together. (1 pt.)
Please help students construct the chromosome by using the percentages (next slide)

16. Crossover frequencies