Download Basic Principles of Heredity I. Mendel`s impact

I. Mendel's impact
Chapter 3
A. Prior to Mendel, scientists were concerned
with continuously varying traits.
Mendel concentrated on discontinuous traits.
Basic Principles of Heredity
B. Mendel worked with peas because they had discrete
traits, a short generation, an ability to either self- or
cross-pollinate, and a large number of offspring.
C. The key to Mendel's crosses is that he started with
homogeneous, true-breeding plants.
What does that mean genetically?
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D. Reciprocal crosses showed that the results were
independent of parental sex.
Round ♀ x wrinkled ♂
shows the same results as
wrinkled ♀ x Round ♂
What does that mean genetically?
E. In crosses, one form of the trait disappeared in the
hybrid (recessive), while the other form remained
(dominant).
What does that mean genetically?
F. F2 results indicate that the recessive trait is not
lost, only masked.
What is the evidence for this?
II. Alleles segregate from each other to form
gametes (Rule of Segregation).
What cellular phenomenon does this reflect?
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A. In a cross of a dominant trait and a recessive
trait, the recessive trait disappears in the F1
generation but reappears in one quarter of the
F2 generation.
1.
2.
3.
Capital letters are used for dominant alleles, and
lowercase letters are used for recessive alleles.
A cross between a tall, individual and a short
individual (from truebreading lines) can be
indicated by TT x tt.
Since each gamete gets only one allele, the F1
hybrid is Tt. The F2 offspring are
¼ TT: ½ Tt: ¼ tt
NOTE: These alleles are usually designated D and d, since alleles are
most frequently named for the mutant form.
B. ➔ Mendelian Ratios
The dominant and recessive phenotypes are in a 3:1
ratio in the F2 generation.
3 tall : 1 dwarf
The genotypic ratio is 1:2:1
1 TT: 2 Tt: 1 tt
C. Mendel tested individuals from the F2 generation.
Of the dominant tall --•
•
one-third were found to be true-breeding
two-thirds were found to be heterozygous and like
the F1 generation
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D. A testcross crosses an individual showing a
dominant trait with a recessive homozygous
individual. If the dominant individual is
homozygous, all progeny show the dominant
trait. If it is heterozygous, ½ of the progeny
show the recessive trait and ½ show the
dominant trait.
• T– x tt
• All tall OR ½ tall ½dwarf
Testcross: What genotype is a tall plant (T–)?
TT x tt
All tall
Tt x tt
½ tall ½ dwarf
E. In partial dominance, the F1 offspring have a
phenotype different from either parent; this
phenotype is often intermediate between
those of the parents.
1. A cross of two heterozygotes produces ¼
offspring like one parent, ¼ like the other
parent, and ½ intermediate.
2. Dominance/recessiveness depends upon the
phenotypic level tested.
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A note: Wild-type is the phenotype of the
organism in nature; wild-type alleles are
indicated by “+.”
Dominant mutant alleles are represented by
capital letters; recessive mutant alleles are
represented by lowercase letters.
Many genes can affect the same phenotype
Genes are usually named after the mutant,
rather than wild-type phenotype.
WARNING: Your textbook does not always
follow this convention consistently.
II. Independent assortment of two or more
genes
A. The F2 offspring in a dihybrid cross show a
9:3:3:1 phenotypic ratio. 9/16 show both
dominant traits, and 1/16 show both recessive
traits.
F2 genotypic ratio: 1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1
B. Genotypes and phenotypes can be
determined by using a Punnett square.
Remember that each gamete must get one
copy of each gene.
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round yellow x wrinkled green
RRYY x
gametes:
F2 phenotypic ratio:
rryy
RY
ry
9
:3
:3
:1
F1: RrYy
RY
F2 :
Ry
rY
RY
ry
Ry
rY
ry
RY
RY
Ry
Ry
rY
rY
ry
ry
F2 genotypic ratio: 1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1
1 RRYY: 2 RRYy: 1 RRyy:
C. Rule of Independent Assortment: Each
gene behaves independently of other genes.
2 RrYY: 4 RrYy: 2 Rryy:
In dihybrid crosses, single genes will
segregate as always; for example,
AaBb X AaBb
yields 3/4 A-: 1/4 aa
and 3/4 B-: 1/4 bb.
1 RRyy: 2 rrYy: 1 rryy
RY
Ry
rY
ry
RY
RRYY
RRYy
RrYY
RrYy
Ry
RRYy
RRyy
RrYy
Rryy
rY
RrYY
RrYy
rrYY
rrYy
ry
RrYy
Rryy
rrYy
rryy
round yellow x wrinkled green
RrYy x rryy
Incomplete dominance in a dihybrid cross will yield
a ratio of 1:2:1:2:4:2:1:2:1.
1
:1
:1
RY
Test crosses of dihybrids yield a 1:1:1:1 ratio.
:1
Ry
rY
ry
ry
ry
ry
ry
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D. In a multihybrid crosses, with n dominant
genes, there will be:
In a multihybrid crosses with n dominant genes:
2n gametes
2n phenotypes
3n genotypes
(½n)2 frequency of homozygous recessives
Why is the number of genotypes less than the number of
cells in the Punnett square?
In multihybrid crosses, with n genes and no
dominance, we expect 3n phenotypes.
The number of phenotypes = the number of genotypes.
Homozygous recessives
N
1
2
3
4
Gametes
Genotypes
Phenotypes
2n
3n
2n
III. Problems solving for multihybrid crosses:
(½n)2
• Punnett square
•
How many cells for 2, 3, 4, 5 genes?
•
WAY TOO SLOW!
• Stick-fork diagrams
– Use when you need to calculate an entire
genotypic of phenotypic distribution
• Probability calculation
– Use when you need to calculate an one or a
few genotypes or phenotypes
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Laws of Probability
Probability is defined as
the likelihood of a particular event occurring.
Always between 0 and 1.
0 means the event will never happen
1 means the event will always happen
The probability that several mutually
independent events will happen (this
AND that) is the PRODUCT of their
individual probabilities.
For example, in a cross of RrCc x RrCc,
what is the probability of wrinkled and
white?
1/4 (rr--) x 1/4 (--cc) = 1/16.
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The probability of one OR another of
mutually exclusive events is the SUM of
their individual probabilities.
For example, in a cross of Ww x Ww,
what is the probability of round?
1/4 (WW) + 2/4 (Ww) = 3/4
The binomial expansion
• Used to find the probability of some
combination of events
(a + b)n
where a is the probability of one
outcome, b is the probability of the
alternate outcome, and n is the number
of events
Consider the recessive human trait PKU
(¾ + ¼)5
(¾)5 +5(¾)4(¼) +10(¾)3(¼ )2 +10(¾)2(¼)3 +5(¾)(¼) 4 +(¼)5
Consider the recessive human trait PKU
• In a cross between two heterozygotes, the
probability of a normal child is ¾ and the
probability of PKU is ¼. If a family has 5
children, what is the probability that all of
them will have PKU?
(a + b)n
a is the probability of normal (¾)
b is the probability of PKU (¼)
and n is the number of children (5)
How do you expand the binomial quickly?
Exponents for the a term start at n and go
down (5, 4, 3, 2, 1, 0)
Exponents for the b term start at 0 and go up
(0, 1, 2, 3, 4, 5)
Remember that you don’t need to show an exponent
of 1, and anything to the 0 power is 1.
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To find the coefficient of each term,
use Pascal’s triangle
http://mathforum.org/dr.math/faq/
Look under P for Pascal
In a cross between two heterozygotes, the probability of
a normal child is ¾ and the probability of PKU is ¼. If a
family has 5 children, what is the probability that all of
them will have PKU?
(a + b)n
(¾)5 +5(¾)4(¼) +10(¾)3(¼ )2 +10(¾)2(¼)3 +5(¾)(¼) 4 +(¼)5
1/1024
.000098
χ2 (chi square) test for goodness of fit
How well do the data fit the hypothesis?
What is the probability that differences
between the observed data and the expected
values are due to chance?
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The larger the χ2 value, the more the data
deviate from the expected value.
We predetermine a level of significance
(usually .05) at which we will reject the
hypothesis that any deviation can be
explained by chance.
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