Mendel’s Laws cont’d

1. Mendel’s Laws cont’d 2. The Law of Independent Assortment: • So far, we’ve talked about single traits being inherited. • Real crosses involving many genes being transferred to offspring at once. • Every pair of alleles separates independently of every other pair when meiosis is occurring.

2. Mendel’s Laws cont’d Law of Independent Assortment cont’d: • Eg: If one pair of homologous chromosomes has a ‘W’ on one chromosome and ‘w’ on the other, and a second pair has an ‘R’ on one chromosome and an ‘r’ on the other, when gametes form all combinations of these letters are possible: WR, Wr, wR, and wr

4. Mendel’s Laws cont’d • The law of independent assortment states: • Each pair of alleles separates independently (without regard to how the other pairs separate) • All possible combinations of alleles can occur in the gametes • This law allows us to carry out crosses with more than one trait…

5. Dihybrid Crosses • A dihybrid cross, like monohybrid crosses, involves two parents, but tracks the inheritance of two pairs of alleles at the same time. • Eg: Consider two traits: hairline (straight VS widow’s peak) and finger length (long VS short). • We’ll use W and w for hairline as before; S = short fingers, s = long fingers.

6. Dihybrid crosses cont’d A person who is WwSs(widow’s peak and short fingers) and a person who is also WwSs have children: • Figure out the gametes for each parent • WS, Ws, wS, and ws for both • Place these combinations on the edges of a Punnett square • Figure out the genotypes of the offspring (use a Punnett square)

8. Dihybrid crosses cont’d • From the Punnett square: • Phenotypic ratio- 9 widow’s peak, short fingers; • 3 widow’s peak, long fingers; • 3 straight hairline, short fingers; • 1 straight hairline, long fingers (9:3:3:1) • Also note: ¾ have a widow’s peak, ¼ have a straight hairline; ¾ have short fingers, ¼ have long fingers.

9. Dihybrid crosses cont’d Product Rule of Probability: • To calculate the chances of having two traits, you must multiply the probabilities of having each individual trait and report the result. • Eg: From the previous problem, what are the chances of having a child with short fingers and a widow’s peak? • Chance of short fingers = ¾ • Chance of widow’s peak = ¾ • Chance of having both = ¾ x ¾ = 9/16

10. Dihybrid Crosses cont’d Test crosses with two traits: • Cross an individual with the dominant phenotype for both traitswith an individual with the recessive phenotype for both traits • If the individual with dominant traits is homozygous for both traits, all offspring will show the dominant phenotypes

11. If the individual with dominant traits is heterozygous, you’ll see a 1:1:1:1 phenotypic ratio for the offspring.

12. Dihybrid Cross Practice Problems Practice problems • Attached earlobes are recessive. What genotype do children have if one parent is homozygous recessive for earlobes and homozygous dominant for hairline, and the other is homozygous dominant for unattached earlobes and homozygous recessive for hairline? • If an individual from this cross reproduces with another of the same genotype, what are the chances that they will have a child with a straight hairline and attached earlobes? • A child who does not have dimples or freckles is born to a man who has dimples and freckles (both dominant) and a woman who does not. What are the genotypes of all the individuals in the problem (father, mother, and child)?