Download LECTURE 2: MENDEL`S LAWS and EXTENSIONS TO MENDEL

LECTURE 2: MENDEL’S LAWS and EXTENSIONS TO MENDEL
Reading: Ch. 2, p. 13-28; Ch. 3, p. 43-53
Supplemental Reading: Mendel’s 1865 paper (use the ESP link off the 140 website)
Problems: Ch. 2, solved problems I, II; 2-4, 2-5, 2-7, 2-8, 2-13, 2-14, 2-16 – 2-20; Ch. 3, 3-2 –
3-5, 3-7 – 3-9, 3-11, 3-13, 3-14
Website: http://mcb.berkeley.edu/courses/mcb140/
In the late 1850’s when Mendel was making his discoveries, no one had yet described
meiosis or genes or even chromosomes. Two theories of inheritance were prevalent at the time;
(1) blended inheritance, in which traits from the parents become mixed and forever changed and
(2) “uni-parental” inheritance, in which one parent contributes most to the inherited features of
progeny. Mendel performed a very careful, controlled, and quantitative analysis to examine the
rules of inheritance. In his own words, he sought “to deduce the law according to which [traits]
appear in successive generations.
Why did Mendel succeed?
- He used an experimentally tractable system, Pisum sativum (garden peas). Peas have short
generation time and can be self- or cross-fertilized.
- He studied easily scorable (unambiguous) traits (discrete vs continuous traits).
- He used pure-breeding lines.
- He carefully controlled his matings, so that he could be sure that the progeny observed were
really a result of the cross he performed. He realized that “accidental” pollinations would
completely confuse his analyses.
- Quantitative analysis! Mendel worked with large numbers of plants and used statistics to
analyze the data.
The monohybrid cross
A cross between plants that exhibit differences in one distinct trait (seed color, seed shape,
flower color, etc), i.e. plants that produce yellow peas x plants that produce green peas.
Generation
Parental (P)
First filial (F1)
Second filial (F2)
yellow x green
all yellow
3 yellow: 1 green
(cross-fertilize)
(self-fertilize)
What did Mendel deduce from these crosses?
- “Blending” did not occur. Green peas were recovered in the F2 generation.
- Yellow was dominant to green.
- There must be two types of yellow peas, those that breed true (like the yellow P generation
plants) and those that carry latent information for green peas (like the F1 plants). He
confirmed this by selfing some of the F2 plants that gave yellow peas: 1/3 of these breed true
(all yellow) and 2/3 gave a 3:1 ratio of yellow to green.
- To account for the 3:1 ratio, Mendel deduced that each plant carries two “differentiating
characters” (genes), one inherited from the maternal parent and one from the paternal parent.
Each gene can have alternative forms (alleles); if the two alleles are the same, the parental
type results and if the two alleles are different, a hybrid type results.
LAW OF SEGREGATION: “The two alleles for each trait segregate during gamete formation,
then unite at random, one from each parent, at fertilization”.
To represent the monohybrid cross a different way:
Genotype
Phenotype
P
YY x yy
yellow x green
F1
Yy
yellow
F2
1 YY: 2 Yy: 1 yy
3 yellow: 1 green
The Punnett Square allows us to visualize a cross by examining the possible combinations of
gametes from the parents. A cross between two Yy heterozygotes:
Y
y
Y
YY
(yellow)
yY
(yellow)
y
Yy
(yellow)
yy
(green)
The Law of the Product states “The probability of two or more independent events occurring
together is the product of the probabilities that each event will occur by itself.”
-- e.g, probability of yy progeny from a cross of two Yy heterozygotes = 1/2 x 1/2 = 1/4
The Law of the Sum states “The probability of either of two such mutually exclusive events
occurring is the sum of their individual probabilities”.
-- What’s the probability of yellow progeny from a cross of two Yy heterozygotes?
Probability of YY + Yy + yY = 1/4 + 1/4 + 1/4 = 3/4
THE LAW OF INDEPENDENT ASSORTMENT: “During gamete formation, different pairs
of alleles segregate independently of one another.”
The dihybrid cross:
Generation
P
F1
F2
Phenotype
Genotype
yellow; round x green; wrinkled
YYRR x yyrr
all yellow; round
YyRr
yellow; round (9 parental type)
Y-Ryellow; wrinkled (3 recombinant type)
Y-rr
green; round (3 recombinant type)
yyRgreen; wrinkled (1 parental type)
yyrr
The 9:3:3:1 ratio observed derives from two separate 3:1 phenotypic ratios: the ratio of yellow to
green is 12:4 (or 3:1) and of round to wrinkled is 12:4 (or 3:1).
How could you determine the genotype of the yellow; round plants? You could perform a Test
Cross by crossing an individual showing a dominant phenotype (Y-R-) to an individual with the
recessive phenotype (yyrr) to reveal the genotype behind the dominant phenotype. The test cross
is a very important tool for the geneticist; this is especially true when analyzing inheritance in
organisms that cannot self-fertilize!
If Y-R- is YYRR, then all test cross progeny will be yellow; round.
If Y-R- is YYRr, then half will be yellow; round and half will be yellow; wrinkled.
If Y-R- is YyRR, then half will be yellow; round and half will be green; round.
If Y-R- is YyRr, then all four combinations occur with equal probability (1:1:1:1).
Bottom line of Independent Assortment:
During Meiosis I, different alleles of two genes on different chromosomes will move to opposite
poles independently of one another.
- Only true for genes that lie on different chromosomes or for genes that lie very far apart on
the same chromosome.
- Genes on the same chromosome exhibit “linkage”, that is they tend to assort together. We’ll
discuss linkage in more detail next week.
EXTENSIONS TO MENDEL
In determining the laws of inheritance, Mendel used a set of guidelines: (1) one of the two alleles
of a given gene showed complete dominance over the other, (2) there are only two alleles of any
given gene, (3) genes determine one specific trait, and (4) all genotypes are equally viable. Let’s
look at some exceptions to these general “rules”.
What if the alleles show incomplete dominance? If this is the case, a novel phenotype, unlike
that of either parent shows up in the F1. Color in snapdragons is an excellent example.
P:
pure-breeding red flowers x pure-breeding white flowers
F1:
all pink -- NEW PHENOTYPE (NOT LIKE EITHER PARENT!)
F2:
red:pink:white (1:2:1). This ratio is a good indication of the lack of complete dominance.
The phenotypic ratios are an exact reflection of the genotypic ratios. Molecularly, one can think
of the combinations as follows: CR CR gives 2 doses of gene expression and CW CW gives no
dose. CR CW has one dose of gene expression and this intermediate dose gives a novel phenotype.
CR
CW
CR
CR CR
(red)
CW CR
(pink)
CW
CR CW
(pink)
CW CW
(white)
What if the alleles show co-dominance? If this is the case, then the F1 progeny display both
parental traits. The IA and IB blood types in humans are a good example. These two alternative
alleles produce a slightly different form of an enzyme that controls the presence of a sugar
polymer that sits on the red blood cell membrane. The complex sugar differs depending upon
whether the IA or IB form of the enzyme is present.
P:
IA IA
x
IB IB
A
B
F1:
all I I (and express both kinds of sugars on the RBC membrane)
F2:
IA IA : IA IB : IB IB (1:2:1). Again, this ratio is a good indication of the lack of complete
dominance. The phenotypic ratios are an exact reflection of the genotypic ratios.
IA
IB
IA
IA IA
(“A”-type sugar)
IB IA
(“A” and “B” sugars)
IB
IA IB
(“A” and “B” sugars)
IB IB
(“B”-type sugar)
What if there are multiple alleles (more than 2) for a given gene? Mutations are the source of
new alleles. For human blood type, there is an allele of the I gene (i) that produces a nonfunctional enzyme. IA and IB alleles are co-dominant, but the i allele is recessive to both IA and
IB alleles. Because each person can only carry two alleles, there are six possible genotypes and 4
possible blood types.
Genotype
Type of sugar
(Blood type)
Antibodies in serum
IA IA
IA i
A
against B
IB IB
IB i
B
against A
IA IB
AB
none
ii
O
against A and B
Now, we’ll consider a situation where a gene contributes to multiple traits (pleiotropy),
specifically in the case where a particular allele causes recessive lethality.
In mice, coat color is determined by the agouti gene. Unlike the traits that Mendel
described in peas, the agouti gene has multiple alleles. Multiple alleles arise by mutation (a
change in genetic sequence). Mice homozygous for the wild-type agouti allele (AA) have a dark
grey coat color. One of the mutant alleles of agouti gives rise to mice that have a yellow coat.
When pure-breeding wild-type agouti mice (AA) are mated with yellow mice, one observes dark
grey and yellow progeny in equal numbers (1:1). This suggests that (1) a single gene with 2
alleles determines the two phenotypes, (2) yellow mice must carry the agouti gene (heterozygous
AyA), and (3) yellow must be dominant to agouti. The Punnett Square looks like this:
A
Ay
A
AyA
(yellow)
AA
(dark grey)
When two yellow mice are bred together one observes yellow and grey progeny in a ratio
of 2:1. Yellow mice must indeed then be heterozygotes (AyA). But why the 2:1 ratio, not 3:1, as
we’ve come to expect from cross between heterozygotes? The Punnett Square looks like this:
Ay
A
Ay
A
Ay Ay
(lethal)
AyA
(yellow)
AyA
(yellow)
AA
(dark grey)
The AyAy homozygous combination is lethal and AyAy mice die in utero. Thus even
though the Ay allele is dominant to the wild-type allele for coat color, the same Ay allele is
recessive to the wild-type allele for survival. This explains why pure-breeding yellow mice can
never be obtained.