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, true- breeding white-flowered and purple-flowered pea plants – All of the F 1 offspring were purple-flowered When Mendel crossed the F 1 plants – Many of the F 2 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 F 1 plants should produce only purple-flowered plants in the F 2. Figure 14.3 P Generation (true-breeding parents) Purple flowers White flowers F 1 Generation (monohybrids) All plants had purple flowers F 2 Generation EXPERIMENT True-breeding purple-flowered pea plants and white-flowered pea plants were crossed. The resulting F 1 hybrids were all purple-flowered. They were allowed to self-pollinate or were cross-pollinated with other F 1 hybrids. Flower color was then observed in the F 2 generation. RESULTS Both purple-flowered plants and white- flowered plants appeared in the F 2 generation. In Mendel’s experiment, 705 plants had purple flowers, and 224 had white flowers, a ratio of about 3 purple : 1 white.
Genes Mendel reasoned that since the inheritable factor for white flowers was not lost in the F 1 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.
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 F 2 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 Figure 14.4 Allele for purple flowers Locus for flower-color gene Homologous pair of chromosomes Allele for white flowers
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 F 1 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 F 2 generation of his numerous crosses? Figure 14.5 P Generation F 1 Generation F 2 Generation P p P p P p P p PpPp PP pp Pp Appearance: Genetic makeup: Purple flowers PP White flowers pp Purple flowers Pp Appearance: Genetic makeup: Gametes: F 1 sperm F 1 eggs 1/21/2 1/21/2 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. Union of the parental gametes produces F 1 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. 3 : 1 Random combination of the gametes results in the 3:1 ratio that Mendel observed in the F 2 generation. This box, a Punnett square, shows all possible combinations of alleles in offspring that result from an F 1 F 1 (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.
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 Figure Phenotype Purple White Genotype PP (homozygous) Pp (heterozygous) Pp (heterozygous) pp (homozygous) Ratio 3:1 Ratio 1:2:1 The phenotype is expressed traits - In the flower color experiment, the F 2 generation had a 3:1 phenotypic ratio of purple-flowered to white-flowered plants. The genotype is genetic makeup - The genotypic ratio of the F 2 generation was 1:2:1
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 Figure 14.7 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 Pp pp Pp P p pp 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.
The Law of Independent Assortment The law of segregation was derived – From monohybrid crosses using F 1 monohybrids heterozygous for one character The Law of Independent Assortment requires – Using dihybrid crosses between F 1 dihybrids Crossing two, true-breeding parents differing in two characters – Produces F 1 dihybrids, heterozygous for both characters
YYRR P Generation GametesYRyr yyrr YyRr Hypothesis of dependent assortment Hypothesis of independent assortment F 2 Generation (predicted offspring) 1⁄21⁄2 YR yr 1 ⁄ 2 1⁄21⁄2 yr YYRRYyRr yyrr YyRr 3 ⁄ 4 1 ⁄ 4 Sperm Eggs Phenotypic ratio 3:1 YR 1 ⁄ 4 Yr 1 ⁄ 4 yR 1 ⁄ 4 yr 1 ⁄ 4 9 ⁄ 16 3 ⁄ 16 1 ⁄ 16 YYRR YYRr YyRR YyRr YyrrYyRr YYrr YyRR YyRr yyRRyyRr yyrr yyRr Yyrr YyRr Phenotypic ratio 9:3:3: Phenotypic ratio approximately 9:3:3:1 F 1 Generation Eggs YR Yr yRyr 1 ⁄ 4 Sperm 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 EXPERIMENT Two true-breeding pea plants— one with yellow-round seeds and the other with green-wrinkled seeds—were crossed, producing dihybrid F 1 plants. Self-pollination of the F 1 dihybrids, which are heterozygous for both characters, produced the F 2 generation. The two hypotheses predict different phenotypic ratios. Note that yellow color (Y) and round shape (R) are dominant. The Dihybrid Cross – Illustrates the inheritance of two characters – Produces four phenotypes in the F 2 generation When the F1 dihybrid progeny self-pollinate. – If the two characters segregate together, the F 1 hybrids can only produce the same two classes of gametes (RY and ry) that they received from the parents, and the F 2 progeny will show a 3:1 phenotypic ratio. – If the two characters segregate independently, the F 1 hybrids will produce four classes of gametes (RY, Ry, rY, ry), and the F 2 progeny will show a 9:3:3:1 phenotypic ratio. Figure 14.8
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 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 F 1 (Rr) will receive an r allele = 1/2. – Probability that a sperm from the F 1 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. Rr Segregation of alleles into eggs Rr Segregation of alleles into sperm R r r R R R R 1⁄21⁄2 1⁄21⁄2 1⁄21⁄2 1⁄41⁄4 1⁄41⁄4 1⁄41⁄4 1⁄41⁄4 1⁄21⁄2 r r R r r Sperm Eggs Figure Rule of multiplication states that the probability that independent events will occur simultaneously is the product of their individual probabilities.
Multiplication also applies to dihybrid crosses Question: For a dihybrid cross, YyRr x YyRr, what is the probability of an F 2 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 F 2 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 F 2 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.