n 1. To familiarize yourself with Mendel\'s protocol, study at Fig. 2.15 (Title:
ID: 144186 • Letter: N
Question
n 1. To familiarize yourself with Mendel's protocol, study at Fig. 2.15 (Title: A dih types and recombinant types) carefully (we use this protocol repeatedly in Chapters 2-5). Answer the following questions: (a) How many genes are being discussed in this cross? (b) How many alleles are being discussed and what are they? (c) Are they on the same chromosome or different chromosomes-how do you know? (d) What are the parental genotypes? (e) What are the genotypes of the Fl dihybrids? What kind of a cross is the F1 cross? (1) What are the genotypes of the gametes? (g) How many independent F2 genotypes are possible? (h) How many unique genotypes are possible? (i) How many phenotypic classes are there, what are their proportions, and what is their corresponding genotypic category? G) Show how the product rule of probability can be used to understand the mating of dihybrids? Question 2. A loss-of-function recessive mutation in gene brown (bw) makes the normal red eye color brown. This gene resides on the second chromosome. A loss-of-function recessive mutation in the gene ebony (e) makes body color of the fly darker. It resides on the third chromosome of Drosophila. I give you flies with wild type eye color and normal gray body color. Set up a cross to test if the flies I gave you are (1) true-breeding wild type for both genes, (2) heterozygous just for the bw locus, (3) heterozygous just for the e locus, or (4) double heterozygous. Clearly write out your crosses with genotypic and phenotypic consequences. Question 3. Both Manx anury (taillessness, designated M) and polydactylia (extra digits, designated Pd) in cats are cause by dominant mutations. The M and Pd genes assort independently. A heterozygous Manx female (genotype Mi+) is crossed with a heterozygous polydactyl male (genotype Pdl+). a) What proportion of the offspring are expected to be tailless and have normal digits? b) What proportion of the offispring are expected to have extra digits and normal tails? c) What proportion of the offspring are expected to be tailless and have extra digits? Question 4. Consider a trihybrid self cross Aa Bb Cc X Aa Bb Cc. How many total and different genotypes are expected? How many different phenotypic classes are expected? Explain your answers clearly Question 5, If you roll a die (singular of dice), what is the probability you will roll: (a) a 6? (b) an even number? (c) a number divisible by 3? (d) If you roll a pair of dice, what is the probability that you will roll two 6s? (e) an even number on one and an odd number on the other? Question 6. In a standard deck of playing cards, there are four suits (red suits hearts and diamonds, black suits spades and clubs). Each suit has thirteen cards: Ace (A), 2,3, 4, 5,6,7, 8, 9, 10, and the face cards Jack (J), Queen (Q), and King (K). In a single draw, what is the probability that you will draw a face card? A red card? PS 3 Page1Explanation / Answer
Question 1.
(a) Two genes: (i) pea color gene, and (ii) pea shape gene.
(b) Total four alleles (two for each gene): (i) yellow pea color (Y), (ii) green pea color (y), (iii) round pea shape (R), and (iv) wrinkled pea shape (r). Yellow and round are dominant alleles while green and wrinkled are recessive alleles.
(c) They are on different chromosomes. It is obvious because they show independent assortment and are not linked. It means they are not located on the same chromosome.
(d) Parental genotypes: YYRR (all dominant) and yyrr (all recessive).
(e) Genotypes of F1 dihybrid: YyRr, both the genes are heterozygous. This type of cross is called dihybrid cross or fertilization.
(f) Genotypes of the gametes: YR, yR, Yr, and yr.
(g) A total of 16 genotypes are possible.
(h) A total of 9 unique genotypes are possible: YYRR, YyRR, yyRR, YYRr, YyRr, yyRr, YYrr, Yyrr, yyrr.
(i) There are 4 phenotypic classes. Yellow round : yellow wrinkled : green round : green wrinkled :: 9:3:3:1.
Yellow round => dominant homozygous.
Yellow wrinkled => heterozygous.
Green round => heterozygous.
Green wrinkled => recessive homozygous.
(j) The Product Rule is applied to calculate the outcome with two independent events. In dihydrid cross, law of segregation (for alleles) and law of independent assortment (for non-linked genes) both are applied. Thus, product rule of probability can be applied to determine the outcome of the cross.
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