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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? 1 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? 2 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 3 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. 4 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. 5 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 6 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 7 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. 8 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. 9 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? 10 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. 11