Download Syllabus Notes 2-3-09

Syllabus Notes 2-3-09
Pedigree notes:
FYI: Females are circles, males are squares.
1. Find the 2 parents ‘the same’ that had a child
(one or more) that is ‘different’.
2. Label that child homozygous recessive -aa.
3. Label everyone like that child homozygous
recessive, give everyone aa.
4. Label the Parents heterozygous –Aa.
5. Label the rest (everyone else will have one
Dominant ‘A’ and we have to look at the parents
or children to see the rest)…
6. Note that the trait is SHADED in….
Trait is Dominant
Aa
aa
A__
Aa
A__
aa
aa
aa
aa
Well?
Epistasis
• One allele controls the expression of one or
more other alleles.
_ _cc (anything without color)
B_C_ (black with color)
Black is dominant to
Brown (B or b)
C codes for color,
c does not
bbC_ (brown with color)
Here is a dog that has a big stripe in the middle…
mitosis didn’t work very well, and only one part of
the dog got genes for ‘color’.
#23 A dominant allele B produces bristles in fruit flies when it is present
in the heterozygote. When it is homozygous (BB) it is lethal. Homozygous
recessive (bb) flies are nonbristled. Another gene S acts to suppress the
action of B, but it is also lethal when homozygous (SS). The ss genotype
has no effect on bristles. Two nonbristled flies in which the B allele is
being suppressed are crossed. What is the phenotypic ration of the F1? If
the bristled flies from the F1 are backcrossed to parental flies, what
phenotypic ratio would be predicted in the offspring?
BbSs X BbSs:
Dead: BB=dead=1/4,
SS=dead=1/4, BUT! BBSS happens, and you can’t count
them twice… so ¼ + ¼ (-probability of BBSS)1/16 = 7/16.
Living: bbss (1/16), bbSs (2/16), Bbss (2/16), BbSs (4/16)…
Bristled X Parent from first: Bbss X BbSs
Dead: BB = ¼, (no SS possible)
Living: bbss (1/4*1/2) , Bbss, bbSs, BbSs
#22 In guinea pigs, the gene for production of melanin is epistatic to the gene for the
deposition of melanin. The dominant allele M causes melanin to be produced. mm
individuals cannot product the pigment. The dominant allele B causes the deposition of
a lot of pigment and produces a black guinea pig, whereas only a small amount of
pigment is laid down in bb animals, producing a light-brown guinea pig. Without an M
allele, no pigment is produced so the allele B has no effect, and the guinea pig is white.
A homozygous black guinea pig is crossed with a homozygous recessive white: MMBB
X mmbb. Give the phenotypes of the F1 and F2 generations:
• F1 = MmBb =all black
• F2 = MmBb X MmBb
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Phenotype possibilities =
M_B_ =black
M_bb = brown
mm?? = white