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Exam II Lectures and Text Pages

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Exam II Lectures and Text Pages

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- 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

- When Mendel crossed contrasting, true-breeding 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

P Generation

(true-breeding

parents)

Purple

flowers

White

flowers

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.

F1 Generation

(monohybrids)

All plants had

purple flowers

F2 Generation

RESULTS Both purple-flowered plants and white-

flowered 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.

- 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.

Figure 14.3

- 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 purple-flower 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.

Table 14.1

- Mendel observed the same pattern in many other pea plant characters

- 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

Allele for purple flowers

Homologous

pair of

chromosomes

Locus for flower-color gene

Allele for white flowers

Figure 14.4

- First, alternative versions of genes
- Account for variations in inherited characters, which are now called alleles

- 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.

- 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)

- 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.

P Generation

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.

Appearance:Genetic makeup:

Purple flowersPP

White flowerspp

Gametes:

p

P

F1 Generation

Union of the parental gametes

produces F1 hybrids having a Pp

combination. Because the purple-

flower allele is dominant, all

these hybrids have purple flowers.

When the hybrid plants produce

gametes, the two alleles segregate,

half the gametes receiving the P

allele and the other half the p allele.

Appearance:Genetic makeup:

Purple flowersPp

1/2

Gametes:

1/2

p

P

F1 sperm

This box, a Punnett square, shows

all possible combinations of alleles

in offspring that result from an

F1 F1 (PpPp) 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.

F2 Generation

p

P

P

Pp

PP

F1 eggs

p

pp

Pp

Random combination of the gametes

results in the 3:1 ratio that Mendel

observed in the F2 generation.

3

: 1

Figure 14.5

- Does Mendel’s segregation model account for the 3:1 ratio he observed in the F2 generation of his numerous crosses?

- 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

Genotype

Purple

PP

(homozygous)

1

Pp

(heterozygous)

3

Purple

2

Pp

(heterozygous)

Purple

pp

(homozygous)

White

1

1

Ratio 3:1

Ratio 1:2:1

Figure 14.6

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.

The genotype is genetic makeup

- The genotypic ratio of the F2 generation was 1:2:1

- 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.

- 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 purple-flowered plants, the parent was probably homozygous dominant since a PP x pp cross produces all purple-flowered 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 1⁄2 offspring purple

and 1⁄2 offspring white:

p

p

p

p

P

P

Pp

Pp

Pp

Pp

P

p

Pp

pp

Pp

pp

Figure 14.7

- 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 F1dihybrids,heterozygous for both characters

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

YR

yr

F1 Generation

YyRr

Hypothesis of

independent

assortment

Hypothesis of

dependent

assortment

Sperm

Yr

1 ⁄4

1 ⁄4

YR

1 ⁄4

yR

yr

1 ⁄4

Sperm

Eggs

1⁄2

yr

1⁄2

YR

RESULTS

1 ⁄4

YR

Eggs

YyRr

YYRR

YYRr

YyRR

1 ⁄2

YR

F2 Generation

(predicted

offspring)

YYRR

YyRr

1 ⁄4

Yr

YYrr

YyRr

Yyrr

YYrr

yr

1 ⁄2

yyrr

YyRr

1 ⁄4

yR

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

YyRR

YyRr

yyRR

yyRr

3 ⁄4

1 ⁄4

yr

1 ⁄4

Phenotypic ratio 3:1

Yyrr

YyRr

yyRr

yyrr

1 ⁄16

3 ⁄16

3 ⁄16

9 ⁄16

Phenotypic ratio 9:3:3:1

315

108

101

Phenotypic ratio approximately 9:3:3:1

32

- Illustrates the inheritance of two characters
- Produces four phenotypes in the F2 generation

- 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.

Figure 14.8

- 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

- 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.

- 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.

Rr

Segregation of

alleles into eggs

Rr

Segregation of

alleles into sperm

Sperm

r

R

1⁄2

1⁄2

R

R

r

R

R

1⁄2

1⁄4

1⁄4

Eggs

r

r

R

r

r

1⁄2

1⁄4

1⁄4

Figure 14.9

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?
- Answer:
- 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.
- The overall probability that two recessive alleles will unite at fertilization: 1/2 x 1/2 = 1/4.

- 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.

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.

- So, the probability that the offspring will be heterozygous is the sum of the probabilities of those two possible ways:

- 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

- 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

- 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

- 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.