Download Station Activity: Cell Transport

Name:____________________________________
Date:______________
Period:_____
Sex-Linked Station Activity: Shrek Genetics Answer Key
Pre-Lab: (To be completed before station activity)
Trait
Baldness
Hemophilia
Color blindness
Cat Fur
Dominant Gene
Normal hairline (B)
NO disease (H)
Normal Vision (C)
Solid fur color (F)
Recessive Gene
bald (b)
Having hemophilia (h)
Color blindness (c)
Calico fur (black & orange) (f)
1. Write the phenotype (physical appearance) for each item. Remember to indicate if the person is
male or female in your answer:
(a) XBXB –female w/ normal hairline (b) XbY – male w/ baldness (c) XHY – male w/out disease
(d) XCXc – female w/ normal vision (e) XhXh - female w/ hemophilia (f) XFY - male w/ solid fur
2. Determine the genotypes (combination of alleles) for each trait:
(a) female carrier for hemophilia- XHXh (b) Male with color blindness – XcY
(c) Male with normal hairline- XBY
(d) Female homozygous for Solid Fur – XFXF
Station #1 – Shrek’s Bald Head
XB
1.
Xb
Y
Xb
XBXb
XbXb
XBY
XbY
2. Chance of baby that is bald – 50%
3. Chance of baby with normal hairline – 50%
Station #2 – Baby Baldness
1.
Xb
XB
Y
Xb
XBXb
XBXb
XbY
XbY
2. Possible Male Genotype – XBY
3. Carriers
4. Yes it is possible to have sons who are bald
Assignments
Name:____________________________________
Date:______________
Period:_____
Station #3 – Hemophilia
1.
XH
XH
Xh
XHXh
XHXh
Y
XHY
XHY
2. Explanation: No. There is a 0% chance of having offspring with the disease.
3. % Chance of carriers – 50%
Station #4 – Calico Cats
Xf
1.
XF
Y
Xf
XF
Xf
XFXf
XFXf
XF
XFXF
XFXf
XfY
XfY
Y
XFY
XfY
2. Possible Female Genotypes – XfXf and
XFXf
3. Correct Genotype of mother - XFXf
Station #5 – Color blindness
1. Explanation: It is impossible for the gingerbread man to pass this allele onto his son. This is
because sons inherit their X chromosome from mom only and the Y chromosome from dad.
2. Explanation: It is possible for the gingerbread man to pass this allele onto his daughters. This
is because females inherit one X chromosome from their mother and one from their father.
3.
XC
XC
Xc
XCXc
XCXc
Y
XCY
XCY
Possible Female Genotype - XCXC (homozygous normal vision)
Assignments