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.
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
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…
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.
Dihybrid crosses cont’d
A person who is WwSs (widow’s peak
and short fingers) and a person who is
also WwSs have children:
1. Figure out the gametes for each
parent
• WS, Ws, wS, and ws for both
• Place these combinations on the edges of a
Punnett square
2. Figure out the genotypes of the
offspring (use a Punnett square)
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.
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
Dihybrid Crosses cont’d
Test crosses with two traits:
• Cross an individual with the
dominant phenotype for both traits
with 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
• If the individual
with dominant
traits is
heterozygous,
you’ll see a
1:1:1:1
phenotypic ratio
for the offspring.
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)?