11-3 Notes: Chromosomal Basis of Inheritance
The Chromosome Theory of Inheritance

Mendelian factors or genes are located on chromosomes

Chromosomes separate and independently assort

Genes = sections of a chromosome that code for a particular trait
Thomas Hunt Morgan (1909) and the fruit fly, Drosophila melanogaster

These are useful because they:
1. Are easily cultured in the lab
2. Are prolific breeders
3. Have a short generation time
4. Have only 4 pairs of chromosomes which are easily seen with a microscope
-

3 pairs of autosomes and one pair of sex chromosomes
For fruit flies, instead of using A or a to designate alleles, we use the following:
Sex Chromosomes = chromosomes that determine the gender of the individual

May carry other traits as well

In most species, the females’ are identical (XX) = “homogametic” and the males’ differ (XY) =
“heterogametic”
o
Some organisms, like grasshoppers have homogametic (XX) females, but males that are
hemigametic = XO (Only have one sex chromosome)

In humans, after meiosis, each haploid gamete has 22 autosomes and 1 sex chromosome
o
Females can only give an X
o
Males can give either an X or a Y

50% chance of either

determines the gender of the child
Sex Linkage: Morgan’s Discovery
Initial Cross:
P
Red-eyed female
x
White-eyed male

F1
all red-eyed offspring

F2
mostly red-eyed : some white-eyed
EXPECTED 3:1 ratio, but GOT closer to 4:1!!!!!
What happened?
Breakdown of results:


Red Females: 2459
White Females: 0 (huh?!?!)
Red Males: 1011
White Males: 782
Is it possible to get a white-eyed female, or is it lethal to them?
Crossed white-eyed male with F1 female and got an F1 generation with:
Red Females:
129
White Females:
88
Expected a 1:1:1:1 ratio and was
Red Males:
132
pretty close to these results
White Males:
106
Morgan’s Hypothesis: Gene for eye color is carried only on the X chromosome and red is dominant
over white  That is, they are X-LINKED = carried on the X chromosome
P Generation
F1 Generation
F2 Generation
(do in class)

When solving X-linked crosses, you must include the sex chromosomes and designate the
genotype as a gene carried on each
o
Heterogametic sex (usually males) are said to be hemizygous for sex-linked traits
(because they only have 1 allele instead of 2 for a that trait)

Humans X-linked traits include: Red-Green Color-blindness, Menkes Syndrome, Duchenne
Muscular Dystrophy, Hemophilia, ALD (Lorenzo’s Oil), and Male Pattern Baldness
LINKED GENES: Another problem with Mendel’s Law of Independent Assortment…
Ex.
P
BOTH DOMINANT
BOTH RECESSIVE
(purebreeding) tan & long x (purebreeding) black & short
F1
All tan & long
↓ bred together
F2
Two Possibilities: If unlinked: will get a 9:3:3:1 ratio
If linked: will get a 3:1 ratio with ¾ (75%) tan & long
and ¼ (25%) black & short
Results: 95 tan & long (Like Parent #1)
31 black & short (Like Parent #2)
4 tan & short
6 black & long
Close to the expectation for linked, but a few exceptions. WHY?
Answer: CROSSING OVER
-
when crossing over occurs between positions of two different genes on the same
homologous chromosomes, the genes may become “unlinked”
Parental Type = Progeny that have the same phenotype as one or the other of the parents
Recombinants = Progeny whose phenotypes differ from either parent
-
Arise from Genetic Recombination = the production of offspring with new
combinations of traits different from those combinations found in the parents; results
from the events of meiosis and random fertilization
-
Can be used to determine how far apart two genes are on a chromosome
MAPPING CHROMOSOMES
Loci (singular: Locus) = the particular position on a chromosome where a gene is located
-
same for all homologous chromosomes
Sturtevandt (1913) – determined that the percentage of recombinants was probably related to the physical
distance between two gene loci
-
Postulated:
1) genes are arranged in linear order on chromosomes
2) genes closer together separated by crossing over less frequently than those farther
apart
3) should be possible to plot the sequence and relative distance of gene loci by
determining the frequencies of recombination
-
defined 1 map unit = distance that would give one recombinant organism per 100
fertilized eggs
-
That is, 1 map unit = 1% recombination between genes
EXAMPLE:
Loci
Recombination Frequencies
Approximate Map Units
b vg
17%
17
cn b
7%
7
cn vg
10%
10
So, if mapping on a chromosome:

7

10

│______________│_____________________│
b

cn
17
vg
