BSC 2010L
Human Chromosomes

Humans have 46 chromosomes, 23
homologous pairs
 Remember that the # of chromosomes
differs for different species
 When looking at each pair of homologous
chromosomes:
○ 1 came from mom, one came from dad
○ Each contains the same genes, however
an individual can have 2 alternate forms of
that gene (alleles)
Alleles
o
Alleles can be dominant or recessive
o Let’s look at eye color
o Brown eyes are dominant – B
o Blue eyes are recessive – b
o
Homozygous – having the same alleles
o BB or bb
o
Heterozygous – having different alleles
o Bb
o
Genotype vs Phenotype
o Genotype – what alleles does that person have for that gene?
o BB or bb or Bb
o Phenotype – refers to an individual’s appearance, what is their eye
color?
o BB or Bb – Brown eyes
o bb – blue eyes
Law of Segregation

Each organism contains 2 alleles for
each trait, and the alleles segregate
during the formation of gametes. Each
gamete then contains only 1 allele for
each trait. When fertilization occurs, the
new organism has 2 alleles for each
trait, one from each parent
 This is why it is important that 1 of each
homologous pair of chromosomes ends up
in the gamete at the end of meiosis
Monohybrid cross

Eye color
 Bb
x
Bb
(what is the eye color of both these parents?)
 Gametes of both these parents:
 Looking at the Punnett Square below, these parents
have a 75% chance of having a brown eyed child
and 25% chance of having a blue eyed child
- Genotypic Ratio – 1: 2: 1 (BB:Bb:bb)
- Phenotypic Ratio – 3:1 (brown:blue)
Dihybrid Cross

Let’s look at corn
 P = purple kernel
 p= yellow kernel
 S = smooth
 s = wrinkled kernel

Think of each kernel as being an
individual offspring
Dihybrid Cross

PpSs
x
PpSs
 What are the gametes of these 2
individuals? Remember, one of each letter
MUST be in each gamete
○
 Now let’s do a Punnett Square
Dihybrid Cross
o
Resulting possible offspring:
o
o
o
o
o
o
9 – purple, smooth
3 – purple, wrinkled
3 – yellow, smooth
1 – yellow wrinkled
= 16 possibilities (count # of boxes in cross)
Dihybrid cross between 2 heterozygous
individuals results in a 9:3:3:1 phenotypic ratio
Chi-Square Analysis

When doing these Punnett Squares, will
data obtained from sample always follow
the calculated ratios?
 NO! These are predictions. However,
looking at a larger population, one can test
and see if deviations from the expected
values are just due to chance
○ We do that by using a statistical tool such as
the Chi-Square
Before we go on:

Chi-Square versus Punnett Square
 Don’t get these confused!!!!
○ Punnett squares are actual squares that we
use to help us match up chromosomes that
are in gametes
○ Chi-Square
 Statistical test – the number is referred to as “x2=“
Chi-Square Analysis

Let’s look at the dihybrid cross we did with the peas a couple of slides back
 We expect a 9:3:3:1 ratio
 We take an ear of corn and count:
- 201 purple, smooth
- 70 purple, wrinkled
- 89 yellow, smooth
- 22 yellow, wrinkled
Chi-Square Analysis
 X2 = 5.289
○ C-1=3 (# of phenotypes-1, 4-1=3)
○ Scan across row 3
 p value for x2 value of 5.289 falls between
0.20 and 0.10
○ Hypothesis is supported
 Our sample fit 9:3:3:1 ratio, differences are due to
just random chance
X-linked crosses

With the sex chromosomes, some
alleles only occur on the X chromosome
 Therefore, males with a recessive gene on
their X chromosome will express that trait
 Let’s take a look at hemophilia:
○ XhX
x
XY
X-linked Crosses

This couple has the following chances
with their offspring:
 50% chance girl, 50% boy
 75 % chance of having child with no
hemophilia
 25% chance of having boy with hemophilia
 25% chance of having girl who is carrier of
hemophilia
Today’s Lab
Continue looking at slides of mitosis and
meiosis under microscope from last week
 Continue looking at models of mitosis and
meiosis
 Practice Punnett squares
 Observe corn examples

 Monohybrid cross
 Dihybrid cross

Perform Chi-Square analysis using corn
examples