Download Lecture 11 Beyond Mendel

CHAPTER 13
Beyond Mendel
continued
Modifications of Dominance
Relationships
• Incomplete dominance
• Codominance
Gene Interactions and Modified
Mendelian Ratios
• Phenotypes result from complex interactions of
molecules under genetic control. Using genetic analysis
one can often detect the patterns of these interactions.
For example:
• a. In the dihybrid cross AaBb´ x AaBb, nine
genotypes will result. If each allelic pair controls a
distinct trait and exhibits complete dominance, a 9;3;3;1
phenotypic ratio results.
• b. Deviation from this ratio indicates that interaction
of two or more genes is involved in producing the
phenotype.
• Two types of interactions occur:
• a. Different genes control the same general
trait, collectively producing a new phenotype.
• b. One gene masks the expression of others
(epistasis) and alters the phenotype.
• The examples that we will study here are
dihybrid, but in the “real world” larger numbers
of genes are often involved in forming traits.
Gene Interactions That Produce
New Phenotypes
• Nonallelic genes that affect the same
characteristic may interact to give novel
phenotypes, and often modified phenotypic
ratios.
• For example: Comb shape in chickens,
influenced by two gene loci to produce four
different comb types. Each will breed true if
parents are homozygous (Figure 13.8).
RRpp shows Rose comb.
RRPP has walnut comb.
rrPP has pea comb
rrpp has single comb
• In a cross between a homozygous rose-combed
RRpp bird and a single-combed rrpp bird:
• The F1 Rrpp will all have rose combs.
• The F2 will be 3 rose R-pp;1 single rrpp
• Crossing true-breeding rose RRpp and pea rrPP
results in (Figure 13.9):
• An F1 with all walnut combs RrPp.
• An F2 showing a ratio of 9 walnut R-P-;3 rose
R-pp;3 pea rrP-;1 single rrpp.
• These interactions fit the expected ratios for a
Mendelian dihybrid cross. The molecular basis
for each phenotype is unknown, but it appears
that the dominant alleles R and P each produce a
factor that modifies comb shape from single to a
more complex form.
• The molecular basis for each phenotype is
unknown, but it appears that the dominant
alleles R and P each produce a factor that
modifies comb shape from single to a more
complex form.
• Fruit shape in summer squash shows a 9:6:1 ratio. In some
varieties of summer squash, two genes are involved in the
expression of the shape. Each gene is completely dominant.
Interaction between the two loci produces a new phenotype
(Figure 13.10).
• Long fruit aabb are always true-breeding.
Sphere-shaped fruit A-bb or aaB- are not always truebreeding, and sometimes produce long aabb or disk-shaped
A-B- fruit.
• A cross between true-breeding spherical strains AAbb x aaBB
produces a disk-shaped F1. The F2 will be 9⁄16 disk-shaped
A-B-, 6⁄16 spherical A- bb or aaB-, and 1⁄16 long aabb. This
modification of the Mendelian ratio indicates that two loci are
involved.
• The precise molecular basis of these phenotypes is unknown.
Epistasis
•In epistasis, one gene masks the expression of another, but no
new phenotype is produced.
A gene that masks another is epistatic.
A gene that gets masked is hypostatic.
Several possibilities for interaction exist and all of them will
modify the 9;3;3;1 dihybrid ratio:
Epistasis may be caused by recessive alleles, so that aa masks
the effect of B (recessive epistasis).
Epistasis may be caused by a dominant allele, so that A masks
the effect of B.
Epistasis may occur in both directions between genes,
requiring both A and B to produce a particular phenotype
(duplicate recessive epistasis).
Recessive epistasis.
• Coat color determination in labrador retriever dogs
(Figure 13.13)
• Gene B- makes black pigment, while bb makes brown.
• Another gene, E- allows expression of the B gene,
while ee does not.
• Genotypes and their corresponding phenotypes:
• E-B- is black
• E-bb is brown (chocolate)
• ee-- produces yellow with nose and lips either dark eeBor pale eebb
What would be the F2
result of a Cross
between a black dog
and a golden dog ?
Recessive Epistasis
E Allele
B_ allele
Black
Pigment
Pigment Precursor
No Pigment
bb allele
ee alleles
9/16 C_B_ = black
3/16 C_bb = brown
3/16 ccB_ = white
1/16 ccbb = white
Brown
Pigment
9 black: 3 brown: 4 white
• Summary: In recessive epistasis, eeBand eebb have the same phenotype,
producing an F2 ratio of 9;3;4.
• Does this biochemical explanation ring any
bells? What other gene did you study
today that shows similar behavior?
Dominant epistasis
• In dominant epistasis A-B- and A-bb have the same
phenotype, producing an F2 ratio of 12:3:1.
• For example, in summer squash fruit have 3 common
colors, white, yellow, and green (Figure 13.14).
• Yellow is recessive to white but dominant to green.
• Gene pairs are W/w and Y/y.
• W- are white no matter the genotype of the other locus.
• ww are yellow in Y- and green in yy.
• What are all the genotypes and phenotypes?
Lets suppose a cross between two dihybrids. What
is their color? Solve the cross. And determine
the proportions for dominant epistasis
W_Y_= 9/16
W_yy = 3/16
12/16 white phenotype
ww
wwY_ = 3/16 yellow phenotype
Dominant Epistasis
wwyy = 1/16 green phenotype
12 white: 3 yellow: 1 green