Transcript Chapter 8 Lesson 3

Chapter 8
Lesson 3
“Chromosome Theory of Heredity”
I. Other Notables…
A. Sutton
1. Looked at meiosis in grasshoppers
-Genes are carried on chromosomes
-hundreds of genes per chromosome
B. Morgan
1. Sex Determination
-male f.f. had different characteristics than
female f.f. (fig. 8-18)
*Males have the “Y” chromosome*
Possible
Zygotes
X
Y
X
XX
XY
X
XX
XY
50% chance of a male / 50% chance of a female
2. Sex Linked
Inheritance
-a trait that is linked
to the sex of a organism
-X chromosome
carries most genetic info
-Y chromosome
carries little/no genetic
info
-anything “Y” is a
male
-R = red eyes / r =
white eyes
Male
Possible
Zygotes
(F1)
XR
Xr
Y
XRXr
XRY
XR
XRXr
XRY
Possible Zygotes
(F2)
Xr
Y
XR
XRXr
XRY
Xr
XrXr
XrY
3. Example
The gene for a black coat which is “B” is
dominant over an orange coat “b” and the
gene is sex linked. Determine the expected
genotypic and phenotypic ratios expected
when a heterozygous female mates with a
black male
Possible Zygotes
XB
Y
XB
XBXB
X BY
Xb
XBXb
X bY
Genotype…1 : XBXB, 1 : XBXb, 1 : XBY, 1 : XbY
Phenotype…Female: 2 black coated / Male: 1 black coated and 1 orange coated
In humans, the genes for colorblindness and
hemophilia are both located on the X chromosome
with no corresponding gene on the Y. These are both
recessive alleles. If a man and a woman, both with
normal vision, marry and have a colorblind son, draw
the Punnett square that illustrates this. If the man dies
and the woman remarries to a colorblind man, draw a
Punnett square showing the type(s) of children could
be expected from her second marriage. How
many/what percentage of each could be expected?
1. A man with normal vision is XY. What kind(s) of
gametes (sperm) can he produce?
2. Any woman with normal vision could be XX or XX'.
Since this woman has a colorblind son (genotype
X'Y), she has to be XX' (a carrier).
What kind(s) of gametes (eggs) can she produce?
3. What would the children of the first marriage look
like?
4. In order to be colorblind, her second husband must be
X'Y (like her son). What kind(s) of gametes
(sperm) can he produce?
5. What would the children of the second marriage look
like?
6. If the woman has eight children (4/4) what would
their phenotypes be?
Answers to colorblindness example
1.
2.
3.
4.
5.
6.
X and Y
X and X’
XX / XY / XX’ / X’Y
X’ and Y
XX’ / XY / X’X’ / X’Y
4 males: 2 colorblind and 2 non-colorblind
4 females: 1 colorblind and 3 non-colorblind
II. Many Genes – One Effect
A. Human Traits
1. Height
a. Tall or short (complete dominance)
2. Controlled by more than one pair of alleles
a. Tall, medium, short (incomplete
dominance)
3. Continuous Variation – The presence of
many different phenotypes. (5’5”- 6’9”…and
everything in between)
Example
Cross AaBb with AaBb…Capital letters tall /
lower case represents short)
AB
Ab
aB
ab
AB
AABB
AABb
AaBB
AaBb
Ab
AABb
AAbb
AaBb
Aabb
aB
AaBB
AaBb
aaBB
aaBb
ab
AaBb
Aabb
aaBb
aabb
1 tallest / 4 tall / 6 normal / 4 short / 1 shortest
# of People (16 Total)
Height of All People
8
6
4
2
0
Tallest
Tall
Normal
Heights
Short
Shortest
4. Other Factors
a. Modifier Genes
-influence a trait one way or another
-green…hazel…blue…gray
b. Temperature (fruit flies pg. 218)
c. Our Environment…soil conditions,
altitude, weather…