Transcript Problem PowerPoint Concepts in Genetics End of Chapter 4 Honors Genetics - Exam.

Problem PowerPoint
Concepts in Genetics
End of Chapter 4
Honors Genetics - Exam
Type of Problem

Epistasis
Concepts in Genetics
16
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Pigment in mouse fur is only
produced when the C allele is
present. Individuals of the cc
genotype are white. If c olor
is present it may be
determined by the A,a alleles
AA, results in agouti color,
while aa results n black fur.
16(a)
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What F1 and F2 genotypic and
phenotypic ratios are obtained
from a cross between AACC
and aacc mice
F1 = AaCc = agouti
F2 = 9 agouti,4 white, 3 black
16(b)
1. How do you get 8 Agouti and 8
white
AACc x AAcc
 How do you get 9 agouti, 10 black
aaCC x AaCC
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How do you get 4 agouti, 5 black,
and 10 white
AaCc x aacc
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Problem 23.

Five human matings( 1-5)
including both maternal and
paternal phenotypes for ABO
and MN blood group anitgen
status, are shown in the
following table:
Parental Phenotypes
( 1) A, M x A, B
(2) B,M x B, M
(3) O,N x B, N
(4) AB,M x O,N
(5) AB, MN x AB,MN
Offspring – Matches?
a.
b.
c.
d.
e.
A, N
O,N
O, MN
B, M
B, MN
Matches
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( 1) = c
(2) = d
( 3) = b
( 4) = e
(5) = a
Color Vision- Problem 24
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RG – normal vision
Rg- color blind
A husband and wife have normal
vision , although both of their
fathers are red-green color
blind, an inherited X linked
recessive condition. What is
the probability that their
first child will be
Results
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Mother’s father: Xrg/Y
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Father’s Father: Xrg/Y
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Mother: XRGXrg
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Father: XRG/Y
Explanation
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Notice that the mother must
be heterozygous for the rg
allele( being normal and having
inherited an Xrg from her
father and the father because
he has normal vision, must be
XRG
Problems - 25
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In humans, the ABO blood
type is under the control of
autosomal alleles. Color
blindness is a recessive Xlinked trait. It two parents
who are both type S and have
normal vision produce a son
who is color blind and is type
O, what is the probability that
their next child will be female
with normal vision?
Based on the son who is
colorblind and the blood
type O
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XRGXrg, IAIO ,mother
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Xrg/Y , father
Solutions - 25
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The probability of having a
female child is ½
The probabilty that she has
normal vision is 1 ( because the
father’s X is normal)
Probability of type blood is
1/4
Problem 10
Fur color
C ( full color)
cch ( chinchilla)
ch ( himalayan )
ca ( albino )
Problem 10
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Multiple alleles and rabbit fur
Himalayan x Himalayan
Albino
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chca x chca
Problem 10
Full color x albino
chinchilla
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Ccch
x caca
cchca
Problem 1 – Incomplete
dominance
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In shorthorn cattle, coat color
may be red, white, or roan.
Roan is an intermediate
phenothype expressed as a
mixture of re and white hairs
Red x white
RR x WW
roan
RW
Mixed Problem –
Incomplete dominance and
dominant and recessive
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In radishes, flower color may be
red, pruple or white. The edible
protion of the raidsh may be long
or oval. When only flower color is
studied not dominance is evident.
Red x White yields all Purple. If
these F1 purples are interbred, the
F2 generation consists of ¼
red:1/2 purple:1/3 white.
Regarding radish shape , long is
dominant to oval in a Mendelian
fashion
Genotypes
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RR = red
RW = purple
WW = white
O_ = long
Oo = oval
Problem based on radish
facts
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If a red X unknown produced
204 red and 198 purple what is
the genotype of the unknown
plant?
What would the cross of a
RWOo plant x RW oo plant
produce( just phenotypes)
Problem 17
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In some plants, a red pigment
cyanidin is synthesized by a
colorless precursor. The
addition of a hydroxyl group (
OH) to the molecule causes it
to become purple
Epistasis
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Colorless is aa
Purple is A_ B_
Red is A_bb
In a cross between two purple
plants the following results
were obtained
Results
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94 purple
31 red
43 white
This is approximately what
epistatic ratio?
What does this tell you about
the genotype of the original
purple plants?
Answer
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The plants were both
A_B_ x A_B_
Problem 26
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The X linked recessive
mutation, scalloped causes
irregular wing margins in
Drosophila melanogaster, the
fruit fly.
What occurs if a scalloped
female is crossed with a
normal male in the F1.
F1
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Symbol for singed in Fruit
flies is sd
XsdXsd x X+Y – Parental
F1=1/2 XsdX+ This is a female
with normal long wings
1/2 XsdY – This is a male
with scalloped wings that have
irregular edges
F2 
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XsdX+ x
XsdY
Results in
¼ X+X+ ( normal female)
¼ XsdXsd ( scalloped wing female)
¼ X+Y (normal male)
¼ XsdY( male scalloped wings)