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Exam II Lectures and Text Pages
• I. Cell Cycles
–
Mitosis (218 – 228)
–
Meiosis (238 – 249)
• II. Mendelian Genetics (251 – 270)
• III. Chromosomal Genetics
• IV. Molecular Genetics
–
Replication
–
Transcription and Translation
• V. Microbial Models
• VI. DNA Technology
Monohybrid Crosses
• When Mendel crossed contrasting, truebreeding white-flowered and purple-flowered
pea plants
– All of the F1 offspring were purple-flowered
• When Mendel crossed the F1 plants
– Many of the F2 plants had purple flowers, but
some had white flowers
– The traits did NOT blend
Large Samples and Accurate Quantitative Records
• Mendel hypothesized that if the inherited factor for white
flowers had been lost, then a cross between F1 plants
should produce only purple-flowered plants in the F2.
EXPERIMENT True-breeding purple-flowered pea plants and
white-flowered pea plants were crossed. The resulting F1 hybrids
were all purple-flowered. They were allowed to self-pollinate or
were cross-pollinated with other F1 hybrids. Flower color was then
observed in the F2 generation.
P Generation
(true-breeding
parents)

Purple
flowers
White
flowers
F1 Generation
(monohybrids)
All plants had
purple flowers
RESULTS Both purple-flowered plants and whiteflowered plants appeared in the F2 generation. In Mendel’s
experiment, 705 plants had purple flowers, and 224 had white
flowers, a ratio of about 3 purple : 1 white.
Figure 14.3
F2 Generation
Genes
• Mendel reasoned that since the inheritable factor
for white flowers was not lost in the F1 generation,
it must be masked by the presence of the purpleflower factor.
• Mendel's factors are now called genes; and in
Mendel's terms, purple flowers is the dominant
trait and white flowers is the recessive trait.
Repeated Experiments
• Mendel observed the
same pattern in
many other pea plant
characters
Table 14.1
Mendel’s Model
• Mendel developed a hypothesis
– To explain the 3:1 inheritance pattern that he
observed among the F2 offspring
• Four related concepts make up this model
Alleles
• First, alternative versions of genes
– Account for variations in inherited characters,
which are now called alleles
Allele for purple flowers
Locus for flower-color gene
Figure 14.4
Allele for white flowers
Homologous
pair of
chromosomes
Alleles Occur in Pairs in Diploid Organisms
• Second, for each character
– An organism inherits two alleles, one from
each parent
– A genetic locus is actually represented twice
– Homologous loci may have identical alleles
as in Mendel's true-breeding organisms, or the
two alleles may differ, as in F1 hybrids.
Dominance vs. Recessiveness
• Third, if the two alleles at a locus differ
– Then one, the dominant allele, is
completely expressed (designated by a
capital letter)
– The other allele, the recessive allele, is
completely masked (designated by a
lowercase letter)
Law of Segregation
• Fourth, the law of segregation
– The two alleles for a heritable character separate
(segregate) during gamete formation and end up in
different gametes
• Without any knowledge of meiosis, Mendel deduced that a
gamete carries only one allele for each inherited
characteristic, because the alleles of a pair separate
(segregate) from each other during gamete production.
• Gametes of true-breeding plants will all carry the same
allele.
• If different alleles are present in the parent, there is a 50%
chance that a gamete will receive the dominant allele, and
a 50% chance that it will receive the recessive allele.
Law of Segregation, Probability and the Punnett Square
• Does Mendel’s segregation model account for the 3:1 ratio
he observed in the F2 generation of his numerous crosses?
Each true-breeding plant of the
parental generation has identical
alleles, PP or pp.
Gametes (circles) each contain only
one allele for the flower-color gene.
In this case, every gamete produced
by one parent has the same allele.
P Generation
Appearance:
Purple flowers White flowers
Genetic makeup:
PP
pp
Gametes:
p
P
Union of the parental gametes
produces F1 hybrids having a Pp
combination. Because the purpleflower allele is dominant, all
these hybrids have purple flowers.
F1 Generation
When the hybrid plants produce
gametes, the two alleles segregate,
half the gametes receiving the P
allele and the other half the p allele.
Gametes:
This box, a Punnett square, shows
all possible combinations of alleles
in offspring that result from an
F1  F1 (Pp  Pp) cross. Each square
represents an equally probable product
of fertilization. For example, the bottom
left box shows the genetic combination
resulting from a p egg fertilized by
a P sperm.

Appearance:
Genetic makeup:
Purple flowers
Pp
1/
1/
2 P
F1 sperm
P
p
PP
Pp
F2 Generation
P
F1 eggs
p
pp
Pp
Figure 14.5
Random combination of the gametes
results in the 3:1 ratio that Mendel
observed in the F2 generation.
2 p
3
:1
Genetic Vocabulary
• An organism that is homozygous for a gene
– Has a pair of identical alleles (PP or pp)
– All gametes carry that one type of allele
– Exhibits true-breeding
• An organism that is heterozygous for a gene
– Has a pair of alleles that are different (Pp)
– Half the gametes carry one allele and half
carry the other
– Is not true-breeding
Phenotype versus genotype
The phenotype is
expressed traits
- In the flower color
experiment, the F2
generation had a 3:1
phenotypic ratio of
purple-flowered to
white-flowered plants.
Phenotype
Purple
3
Purple
Genotype
PP
(homozygous)
1
Pp
(heterozygous)
2
The genotype is
genetic makeup
Pp
(heterozygous)
Purple
- The genotypic ratio
of the F2 generation
was 1:2:1
Figure 14.6
1
White
pp
(homozygous)
Ratio 3:1
Ratio 1:2:1
1
The Testcross
• In pea plants with purple flowers
– The genotype is not immediately obvious
– It may be homozygous dominant (PP) or
heterozygous (Pp).
• To determine whether such an organism is
homozygous dominant or heterozygous, we
use a testcross.
The Testcross
• Crossing an individual of unknown genotype with a
homozygous recessive
• Example: If a cross between a
purple-flowered plant of unknown
genotype (P_) produced only purpleflowered plants, the parent was
probably homozygous dominant since
a PP x pp cross produces all purpleflowered progeny that are
heterozygous (Pp).
If the progeny of the testcross
contains both purple and white
phenotypes, then the purple-flowered
parent was heterozygous since a Pp X
pp cross produces Pp and pp progeny
in a 1:1 ratio.

Dominant phenotype,
unknown genotype:
PP or Pp?
Recessive phenotype,
known genotype:
pp
If PP,
then all offspring
purple:
If Pp,
then 2 offspring purple
and 1⁄2 offspring white:
p
p
P
p
p
Pp
Pp
pp
pp
P
Pp
Pp
P
p
Pp
Figure 14.7
1⁄
Pp
The Law of Independent Assortment
• The law of segregation was derived
– From monohybrid crosses using F1
monohybrids heterozygous for one character
• The Law of Independent Assortment
requires
– Using dihybrid crosses between F1 dihybrids
• Crossing two, true-breeding parents differing in
two characters
– Produces F1 dihybrids, heterozygous for both
characters
The Dihybrid Cross
•
–
Illustrates the inheritance of two characters
–
Produces four phenotypes in the F2 generation
When the F1 dihybrid progeny self-pollinate.
–
If the two characters segregate together, the F1 hybrids can only produce the same two
classes of gametes (RY and ry) that they received from the parents, and the F2 progeny will
show a 3:1 phenotypic ratio.
–
If the two characters segregate independently, the F1 hybrids will produce four classes of
gametes (RY, Ry, rY, ry), and the F2 progeny will show a 9:3:3:1 phenotypic ratio.
EXPERIMENT Two true-breeding pea plants—
one with yellow-round seeds and the other with
green-wrinkled seeds—were crossed, producing
dihybrid F1 plants. Self-pollination of the F1 dihybrids,
which are heterozygous for both characters,
produced the F2 generation. The two hypotheses
predict different phenotypic ratios. Note that yellow
color (Y) and round shape (R) are dominant.
P Generation
YYRR
yyrr
Gametes
F1 Generation
YR

Hypothesis of
dependent
assortment
yr
YyRr
Hypothesis of
independent
assortment
Sperm
1⁄ YR
2
RESULTS
CONCLUSION The results support the hypothesis of
independent assortment. The alleles for seed color and seed
shape sort into gametes independently of each other. Note the
ratios are 3:1 for each monohybrid cross
Sperm
yr
1⁄
2
Eggs
1
F2 Generation ⁄2 YR YYRR YyRr
(predicted
offspring)
1 ⁄ yr
2
YyRr yyrr
3⁄
4
1⁄
4
1⁄
4
Yr
1⁄
4
yR
1⁄
4
yr
Eggs
1 ⁄ YR
4
1⁄
4
Yr
1⁄
4
yR
1⁄
4
yr
1⁄
4
Phenotypic ratio 3:1
YR
9⁄
16
YYRR YYRr YyRR YyRr
YYrr
YYrr YyRr
Yyrr
YyRR YyRr yyRR yyRr
YyRr
3⁄
16
Yyrr
yyRr
3⁄
16
yyrr
1⁄
16
Phenotypic ratio 9:3:3:1
Figure 14.8
315
108
101
32
Phenotypic ratio approximately 9:3:3:1
The Law of Independent Assortment
• Using the information from a dihybrid cross,
Mendel developed the law of independent
assortment
– Each pair of alleles segregates independently
from every other pair during gamete formation
Probability
• Segregation, independent assortment and
fertilization are random events and
– Reflect the rules of probability
• From the genotypes of parents, we can predict the
most likely genotypes of their offspring using
simple laws of probability.
Probability Scale
• The probability scale: ranges from 0 to 1; an event that is
certain to occur has a probability of 1, and an event that is
certain not to occur has a probability of 0.
–
The probabilities of all possible outcomes for an event must
add up to 1.
• Random events are independent of one another.
–
The outcome of a random event is unaffected by the
outcome of previous such events.
–
Example: it is possible that five successive tosses of a
normal coin will produce five heads; however, the
probability of heads on the sixth toss is still 1/2.
Two basic rules of probability
1. Rule of multiplication states that the probability that independent
events will occur simultaneously is the product of their individual
probabilities.
•
•
Question: In a monohybrid
cross between pea plants
(Rr), what is the probability
that the offspring will be
homozygous recessive?
Rr
Segregation of
alleles into eggs
Segregation of
alleles into sperm
Answer:
–
–
–
Sperm
Probability that an egg
from the F1 (Rr) will
receive an r allele = 1/2.
Probability that a sperm
from the F1 will receive an
r allele = 1/2.
Rr

1⁄
R
2
The overall probability that
two recessive alleles will
unite at fertilization: 1/2 x
1/2 = 1/4.
1⁄
Eggs
r
1⁄
2
r
1⁄
4
R
1⁄
Figure 14.9
r
R
R
2
r
2
R
R
1⁄
1⁄
4
r
4
r
1⁄
4
Multiplication also applies to dihybrid crosses
• Question: For a dihybrid cross, YyRr x YyRr, what
is the probability of an F2 plant having the
genotype YYRR?
• Answer:
– Probability that an egg from a YyRr parent will
receive the Y and R alleles = 1/2 x 1/2 = 1/4.
– Probability that a sperm from a YyRr parent will
receive the Y and R alleles = 1/2 x 1/2 = 1/4.
– The overall probability of an F2 plant with the
genotype YYRR: 1/4 x 1/4 = 1/16.
Two Rules of Probability
2. Rule of addition states that the probability of an event that can
occur in two or more independent ways = sum of the separate
probabilities of the different ways.
• Question: In this cross between pea plants, Pp x Pp, what is the
probability of the offspring being heterozygous?
• Answer: There are two ways a heterozygote may be produced: the
dominant allele (P) may be in the egg and the recessive allele (p) in
the sperm, or vice versa.
–
–
So, the probability that the offspring will be heterozygous is the sum of
the probabilities of those two possible ways:
•
Probability that the dominant allele will be in the egg with the
recessive in the sperm is 1/2 x 1/2 = 1/4.
•
Probability that the dominant allele will be in the sperm and the
recessive in the egg is 1/2 x 1/2 = 1/4.
So, the probability that a heterozygous offspring will be produced is 1/4
+ 1/4 = 1/2.
Complex Genetics Problems
• A dihybrid or other multicharacter cross
– Is equivalent to two or more independent
monohybrid crosses occurring simultaneously
• In calculating the chances for various
genotypes from such crosses
– Each character first is considered separately
and then the individual probabilities are
multiplied together
Multiple Locus Problem
• Question: What is the probability that a trihybrid cross
between organisms with genotypes AaBbCc and AaBbCc will
produce an offspring with genotype aabbcc?
• Answer: Segregation of each allele pair is an independent
event, we can treat this as three separate monohybrid crosses:
Aa x Aa: probability for aa offspring = 1/4
Bb x Bb: probability for bb offspring = 1/4
Cc x Cc: probability for cc offspring = 1/4
•
The probability that these independent events will occur simultaneously is
the product of their independent probabilities (rule of multiplication).
•
The probability that the offspring will be aabbcc is: 1/4 aa x 1/4 bb x 1/4 cc =
1/64
Problem 2
•
Question: Using garden peas, where and assuming the cross is
PpYyRr x Ppyyrr: what is the probability of obtaining offspring with
homozygous recessive genotypes for at least two of the three traits?
•
Answer: Write the genotypes that are homozygous recessive for at
least two characters, (note that this includes the homozygous
recessive for all three). Use the rule of multiplication to calculate the
probability that offspring would be one of these genotypes. Then use
the rule of addition to calculate the probability of offspring in which at
least two of the three traits would be homozygous recessive.
•
Genotypes with at least two homozygous recessives
–
ppyyRr - 1/4 x 1/2 x 1/2 = 1/16
–
ppYyrr - 1/4 x 1/2 x 1/2 = 1/16
–
Ppyyrr - 1/2 x 1/2 x 1/2 = 2/16
–
PPyyrr - 1/4 x 1/2 x 1/2 = 1/16
–
ppyyrr - 1/4 x 1/2 x 1/2 = 1/16
= 6/16 or 3/8 chance of two recessive traits
Particulate Behavior of Genes
• Reviewing Mendel’s discoveries
• If a seed is planted from the F2 generation of a
monohybrid cross, we cannot predict with
absolute certainty that the plant will grow to
produce white flowers (pp). We can say that
there is a 1/4 chance that the plant will have
white flowers. Alternatively, we can say that if
there are several offspring, it is likely that 1/4
of them will have white flowers.
• Alleles are discrete units that segregate into
separate gametes at meiosis. Each gene pair
separates independently of all the other pairs.