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Western University
Biology 2581B
Jim Karagiannis

Chapter 2 1 Chapter 2 Mendel's Breakthrough: Patterns, Particles and Principles of Heredity Synopsis: Chapter 2 covers the basic principles of inheritance, first described by Mendel, that form the foundation of the Laws of Segregation and Independent Assortment. You will see in chapter 4 how these laws relate to chromosome segregation during meiosis. Chapter 2 contains most of the essential terminology used to describe inheritance. You should become very familiar with and fluent in the use of these terms because they will be used in increasingly sophisticated ways in subsequent chapters. A good way to assure that you have a solid grasp of the meanings of the new terms is to pretend you are describing each word or phenomenon to a friend or relative who is not a science major. Often giving an example of each term is useful. The first problem at the end of the chapter is also a useful gauge of how well you know these terms. A few of the terms defined in this chapter are very critical yet often misunderstood. Learn to be precise about the way in which you use these terms: genes and alleles of genes - a gene determines a trait; and there are different alleles or forms of a gene. The color gene in pea has two alleles: the yellow allele and the green allele; genotype and phenotype - genotype is the genetic makeup of an organism (written as alleles) and phenotype is what the organism looks like; homozygous and heterozygous - when both alleles of a gene are the same, we say the organism is homozygous for that gene; if the two alleles are different, the organism is heterozygous; dominant and recessive - the dominant allele is the one that controls the phenotype in the heterozygous genotype. Significant Elements: After reading the chapter and thinking about the concepts you should be able to:  Remember there are two alleles of each gene when describing genotypes of individuals. If you are describing gametes remember there is only one allele of each gene per gamete.  Recognize if a trait is dominant or recessive by considering the phenotype of the F generation. 1  Recognize ratios: (i) in the Study Guide monohybrid ratio refers to any ratio involving one gene, e.g. the phenotypic monohybrid ratio from a cross between two individuals heterozygous for one gene (3 dominant :1 recessive); 2 Chapter 2 (ii) in the Study Guide dihybrid ratio refers to any ratio involving two genes, for e.g. a phenotypic dihybrid ratio from a cross between two individuals heterozygous for two genes (9:3:3:1).  Recognize the need for and be able to set up a test cross.  Determine probabilities using the basic rules of probability: (i) Product rule: If two outcomes must occur together, the probability of one outcome AND the other occurring is the product of the two individual probabilities. (The final outcome is the result of two independent events.) So, the probability of getting a 4 on one die AND a 4 on the second die is the product of the two individual probabilities. (ii) Sum rule: If there is more than one way in which an outcome can be produced, the probability of either one OR the other occurring is the sum of the individual probabilities. In this case, the outcomes are mutually exclusive.  Draw and interpret pedigrees as in Figures 2.21and 2.22 using the helpful information and hints found in Figure 2.20 and Table 2.2.  Set up Punnett squares by determining the gametes produced by the parents and the probabilities of the potential offspring as in Figures 2.11 and 2.15.  Use and interpret branched line diagrams as in Figure 2.17 and Problem 2-24a. Problem Solving - How to Begin:  As you work on problems in the first few chapters, you will begin to recognize some similarities in the type of problem. The majority of problems in the first few chapters involve genetic crosses. It is best to BE CONSISTENT in your approach and formatting to solve such problems. To begin, rewrite the information given in a format that will be useful in solving the problem: in other words, learn to DIAGRAM THE CROSS. phenotype of one parent x phenotype of other parent → phenotype(s) of progeny The goal is to assign genotypes to the parents then use these predicted genotypes to generate genotypes, phenotypes and ratios of progeny. If the predicted progeny match the observed data you were given in the problem then the genetic explanation you have created is correct.  Use the THREE ESSENTIAL QUESTIONS to determine basic information about the genotypes and/or phenotypes of the parents and/or offspring. These questions are distilled from many years of helping students figure out how to solve problems. They are designed to force you to focus on the underlying genetic basis of the information in a problem. Each of the questions has an identifying characteristic that helps you answer the question - see the Hints for an idea of Chapter 2 3 what sort of information allows you to formulate specific answers. As you go through further chapters the Hints will be further refined. THREE ESSENTIAL QUESTIONS (3EQ): 1. How many genes are involved in the cross? 2. For each gene involved in the cross: what are the phenotypes associated with the gene? Which phenotype is the dominant one and why? Which phenotype is the recessive one and why? [3. For each gene involved in the cross: is it X-linked or autosomal?] At this point, only questions 3EQ #1 and 3EQ #2 may be applied. The material that is the basis of question 3EQ #3 will be covered in Chapter 4. Hints: For 3EQ #1 look for the number of different phenotypes or phenotypic classes in the progeny. In Chapter 2 each gene has only 2 phenotypes. For 3EQ #2 if the parents of a cross are true-breeding, look at the phenotype of th1 F individuals. Their genotype must be heterozygous, and their phenotype is thus controlled by the dominant allele of the gene. Also, look at the F progeny – the 3/4 portion of the 3:1 phenotypic 2 monohybrid ratio is the dominant one. Solutions to Problems: Vocabulary 2-1. a. 4; b. 3; c. 6; d. 7; e. 11; f. 13; g. 10; h. 2; i. 14; j. 9; k. 12; l. 8; m. 5; n. 1. Section 2.1 - Background 2-2. People held two basic misconceptions about inheritance. Firstly, it was believed that one parent contributes the most to an offspring's inherited features. Second was the idea of blended inheritance - the parental traits become mixed and forever changed in the offspring. 2-3. There are several advantages to using peas for the study of inheritance. (1) Peas have a fairly rapid generation time (at least two generations per year if grown in the field, three or four generations per year if grown in greenhouses). (2) Peas can either self-fertilize or be artificially crossed by an experimenter. (3) Peas produce large numbers of offspring (hundreds per parent). (4) Peas can be maintained as pure-breeding lines, simplifying the ability to perform subsequent 4 Chapter 2 crosses. (5) Because peas have been maintained as inbred stocks, two easily distinguished, discrete forms of many phenotypic traits are known. (6) Peas are easy and inexpensive to grow. In contrast, studying genetics in humans has several disadvantages. (1) The generation time of humans is long (roughly 20 years). (2) There is no self-fertilization in humans, and it is not ethical to manipulate crosses. (3) Humans produce only a small number of offspring per mating (usually one) or per individual (almost always less than 20). (4) Although people that are homozygous for a trait (analogous to pure-breeding) exist, homozygosity cannot be maintained because mating with another individual is needed to produce the next generation. (5) Because human populations are not inbred, most human traits show a continuum of phenotypes; only a few traits have two very distinct forms. (6) People require a lot of expensive care to "grow". There is nonetheless one advantage to the study of genetics in humans. Because many inherited traits result in disease syndromes, and because the world's population now exceeds 6 billion, a very large number of individuals with variant phenotypes can be recognized. Thus, the number of genes identified in this way is rapidly increasing. Section 2.2 – Genetic Analysis According to Mendel 2-4. a. Two phenotypes are seen in the second generation of this cross, normal and albino. Thus there is one gene controlling the phenotypes in this cross (3EQ#1). To answer 3EQ#2 note that the phenotype of the first generation progeny is normal color and that in the second generation there is a ratio of 3 normal : 1 albino. Both of these prove that allele controlling the normal phenotype is dominant to the allele controlling the albino phenotype. b. Test crosses are crosses between an organism in which there is interest and an organism that is homozygous recessive for all the genes of interest. The male parent is albino and must have the aa genotype. The normally colored offspring must receive an A allele from the mother, so their genotype is Aa; the albino offspring must receive an a allele from the mother, so their genotype is aa. Therefore the female parent is heterozygous Aa. 2-5. Because two phenotypes result from the mating of two cats of the same phenotype, the short- haired parent cats must have been heterozygous. The phenotype expressed in the heterozygotes (the parent cats) is the dominant phenotype. Therefore, short-hair is dominant to long-hair. Chapter 2 5 2-6. a. 3EQ #1 is answered in the problem (1 gene); 3EQ #2 is the question! Two affected individuals have an affected child and a normal child. This is not possible if the affected individuals were homozygous for a recessive allele conferring piebald spotting. Therefore, the piebald trait must be the dominant phenotype. b. If the trait is dominant, the piebald parents could be either homozygous (PP) or heterozygous (Pp). However, because the two affected individuals have an unaffected child (pp), they both must be heterozygous (Pp). Diagram the cross: Spotted x spotted → 1 spotted : 1 normal Pp x Pp→ 1 Pp : 1 pp 2-7. Do a test cross between your normal winged fly (W-) and a short winged fly that must be homozygous recessive (ww). The possible results are diagrammed here; the first genotype in each cross is that of the normal winged fly whose genotype was originally unknown: WW x ww → all W- (normal wings) or Ww x ww → 1/2 W- (normal wings) : 1/2 ww (short wings). 2-8. Diagram the crosses: closed x open → F op1n → F 145 o2en : 59 closed F1open x closed → 81 open : 77 closed The results of the crosses all fit the pattern of inheritance of a single gene, with the closed trait being recessive. The first cross is a cross like Mendel did with his pure-breeding plants, although we don't know from the information provided if the starting plants were pure-breeding or not. The F 1 result of this cross shows that open is dominant. The closed parent must be homozygous for the recessive allele. Because only one phenotype is seen the 1 the open parent must be the homozygous dominant genotype. Thus the parental cucumber plants were indeed true-breeding homozygotes. The self-fertilization of th1 F resulting in a ratio of 3 open : 1 closed ratio shows that these phenotypes are controlled by one gene and that the F1plants are all heterozygous. The second cross is a test cross. It confirms that1the F plants from the first cross are heterozygous hybrids, because there is a 1:1 ratio of open and closed progeny. Thus, all the data is consistent with one gene with two alleles and open being the dominant trait. 6 Chapter 2 2-9. The dominant trait (short tail) is easier to eliminate from the population by selective breeding. You can recognize every animal that has inherited the short tail allele, because only one such dominant allele is needed to see the phenotype. Short-tailed mice can be prevented from mating. The recessive dilute coat color allele, on the other hand, can be passed unrecognized from generation to generation in heterozygous mice (carriers). The heterozygous mice do not express the phenotype, so they cannot be distinguished from homozygous dominant mice with normal coat color. Only the homozygous recessive mice express the dilute phenotype and could be prevented from mating; the heterozygotes could not. 2-10. a. In this problem, the first of the 3 Essential Questions (3EQ #1) is answered for you (there is only 1 gene), as is 3EQ # 2 (the dominant allele is dimple D and the recessive allele is nondimple d; this dominance relationship will be symbolized in the form dimple D > nondimple d from now on.) Next, diagram the cross. In all cases, the male parent is written first for consistency. nondimple x dimpled → proportion F with dimple? 1 Therefore: dd x Dd* → 1/2 dimple : 1/2 nondimple (*Note that the dimpled woman in this cross had a dd [nondimple] mother, so the woman's genotype MUST be heterozygous.) b. dimpled x nondimple → nondimple F 1 D? x dd → dd Because they have a nondimple child (dd), the husband must also have a d allele to contribute to the offspring. The husband has genotype Dd. c. dimpled x nondimple → eight F ,1all dimpled D? x dd → 8 D- The D allele in the children must come from their father. The father could be either DD or Dd but it is more probable that his genotype is DD. We cannot rule out the heterozygous genotype (Dd). However, the probability that all 8 children would inherit the D allele from a Dd parent is 8 only (1/2) or 1/256. 2-11. a. The only unambiguous cross is: homozygous recessive x homozygous recessive → all homozygous recessive The only cross that fits this criteria is: dry x dry → all dry. Therefore, dry is the recessive phenotype (ss) and sticky is the dominant phenotype (S-). Chapter 2 7 b. A 1:1 ratio comes from a testcross of heterozygous sticky (Ss) x dry (ss). However, the sticky x dry matings here include both the Ss x ss AND the homozygous sticky (SS) x dry (ss). A 3:1 ratio comes from crosses between two heterozygotes, Ss x Ss. However, the SS individuals are also sticky. Thus the sticky x sticky matings in this human population are a mix of matings between two heterozygotes (Ss x Ss), between two homozygotes (SS x SS) and between a homozygote and heterozygote (SS x Ss). The 3:1 ratio of the heterozygote cross is therefore obscured by being combined with results of the two other crosses. 2-12. Diagram the cross: black x red → 1 black : 1 red No, you cannot tell how coat color is inherited from the results of this one mating. In effect, this was a test cross – a cross between animals of different phenotypes resulting in offspring of two phenotypes. This does not indicate whether red or black is the dominant phenotype. To determine which phenotype is dominant, remember that an animal with a recessive phenotype must be homozygous. Thus, if you mate several red horses to each other and also mate several black horses to each other, the crosses that always yield only offspring with the parental phenotype must have been between homozygous recessives. For example, if all the black x black matings result in only black offspring, black is recessive. Some of the red x red crosses (that is, crosses between heterozygotes) would then result in both red and black offspring in a ratio of 3:1. To establish this point, you might have to do several red x red crosses, because some of these crosses could be between red horses homozygous for the dominant allele. You could of course ensure that you were sampling heterozygotes by using the progeny of black x red crosses (such as that described in the problem) for subsequent black x black or red x red crosses. 2-13. a. 1/6 because a die has 6 different sides. b. There are three possible even numbers (2, 4, and 6). The probability of obtaining any one of these is 1/6. Because the 3 events are mutually exclusive, use the sum rule: 1/6 + 1/6 + 1/6 = 3/6 = 1/2. c. You must roll either a 3 or a 6, so 1/6 + 1/6 = 2/6 = 1/3. d. Each die is independent of the other, thus the product rule is used: 1/6 x 1/6 = 1/36. e. The probability of getting an even number on one die is 3/6 = 1/2 (see part b). This is also the probability of getting an odd number on the second die. This result could happen either of 2 ways 8 Chapter 2 – you could get the odd number first and the even number second, or vice-versa. Thus the probability of both occurring is 1/2 x 1/2 x 2 = 1/2. f. The probability of any specific number on a die = 1/6. The probability of the same number on the other die =1/6. The probability of both occurring at same time is 1/6 x 1/6 = 1/36. The same probability is true for the other 5 possible numbers on the dice. Thus the probability of any of these mutually exclusive situations occurring is 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 6/36 = 1/6. g. The probability of getting two numbers both over four is the probability of getting a 5 or 6 on one die (1/6 + 1/6 = 1/3) and 5 or 6 on the other die (1/3). The results for the two dice are independent events, so 1/3 x 1/3 = 1/9. 2-14. The probability of drawing a face card = 12 face cards / 52 cards = 0.231. The probability of drawing a red card = 26 / 52 = 0.5. The probability of drawing a red face card = probability of a red card x probability of a face card = 0.231 x 0.5 = 0.116. 2-15. a. The Aa bb CC DD woman can produce 2 genetically different eggs that vary in their allele of the first gene (A or a). She is homozygous for the other 3 genes and can only make eggs with the b C D alleles for these genes. Thus, using the product rule (because the inheritance of each gene is independent), she can make 2 x 1 x 1 x 1 = 2 different types of gametes: (A b C D and a b C D). b. Using the same logic, an AA Bb Cc dd woman can produce 1 x 2 x 2 x 1 = 4 different types of gametes: A (B or b) (C or c) d. c. A woman of genotype Aa Bb cc Dd can make 2 x 2 x 1 x 2 = 8 different types of gametes: (A or a) (B or b) c (D or d). d. A woman who is a quadruple heterozygote can make 2 x 2 x 2 x 2 = 16 different types of gametes: (A or a) (B or b) (C or c) (D or d). 2-16. a. The probability of any phenotype in this cross depends only on the gamete from the heterozygous parent alone. The probability that a child will resemble the quadruply heterozygous parent is thus 1/2A x 1/2B x 1/2C x 1/2D = 1/16. The probability that a child will resemble the quadruply homozygous recessive parent is 1/2a x 1/2b x 1/2c x 1/2d = 1/16. The probability that a child will resemble either parent is then 1/16 + 1/16 = 1/8. This cross will produce 2 different phenotypes for each gene or 2 x 2 x 2 x 2 = 16 potential phenotypes. Chapter 2 9 b. The probability of a child resembling the recessive parent is 0; the probability of a child resembling the dominant parent is 1 x 1 x 1 x 1 = 1. The probability that a child will resemble one of the two parents is 0+1=1. Only 1 phenotype is possible in the progeny (dominant for all 4 genes), as (1) = 1. c. The probability that a child would show the dominant phenotype for any one gene is 3/4 in this sort of cross (remember the 3/4 : 1/4 monohybrid ratio of phenotypes), so the probability of resembling the parent for all four genes is (3/4) = 81/256. There are 2 phenotypes possible for each gene, so (2) = 16 different kinds of progeny. d. All progeny will resemble their parents because all of the alleles from both parents are identical, so the probability = 1. There is only 1 phenotype possible for each gene in this cross; because 4 (1) =1, the child can have only one possible phenotype when considering all four genes. 2-17. a. The combination of alleles in the egg and sperm allows only one genotype: aa Bb Cc DD Ee. b. Because the inheritance of each gene is independent, you can use the product rule to determine the number of different types of gametes that are possible: 1 x 2 x 2 x 1 x 2 = 8 (as in problem 2- 15). To figure out the types of gametes, consider the possibilities for each gene separately and then the possible combinations of genes in a consistent order. For each gene the possibilities are: a, (B : b), (C : c), D, and (E : e). The possibilities can be determined using the product rule. Thus for the first 2 genes [a] x [B : b]gives [a B : a b] x [C : c] gives [a B C : a B c : a b C : a b c] x [D] gives [a B C D : a B c D : a b C D : a b c D] x [E : e] gives [a B C D E : a B C D e : a B c D E : a B c D e : a b C D E : a b C D e : a b c D E : a b c D e]. This problem can also be visualized with a branch diagram: E a B C D E C D e a B C D e B D E a B c D E c D a D e a B c D e E a b C D E C D b D e a b C D e E a b c D E c D e a b c D e D 2-18. The first two parts of this problem involve the probability of occurrence of two independent traits: the sex of a child and galactosemia. The parents are heterozygous for galactosemia, so there is 10 Chapter 2 a 1/4 chance that a child will be affected (that is, homozygous recessive). The probability that a child is a girl is 1/2. The probability of an affected girl is therefore 1/2 x 1/4 = 1/8. a. Fraternal (non-identical) twins result from two independent fertilization events and therefore the probability that both will be girls with galactosemia is the product of their individual probabilities (see above); 1/8 x 1/8 = 1/64. b. For identical twins, one fertilization event gave rise to two individuals. The probability that both are girls with galactosemia is 1/8. For parts c-g, remember that each child is an independent genetic event. The sex of the children is not at issue in these parts of the problem. c. Both parents are carriers (heterozygous), so the probability of having an unaffected child is 3/4. The probability of 4 unaffected children is 3/4 x 3/4 x 3/4 x 3/4 = 81/256. d. The probability that at least one child is affected is all outcomes except the one mentioned in part c. Thus, the probability is 1 - 81/256 = 175/256. e. The probability of an affected child is 1/4 while the probability of an unaffected child is 3/4. Therefore 1/4 x 1/4 x 3/4 x 3/4 = 9/256. f. The probability of 2 affected and 1 unaffected in any one particular birth order is 1/4 x 1/4 x 3/4 = 3/64. There are 3 mutually exclusive birth orders that could produce 2 affecteds and 1 unaffected – unaffected child first born, unaffected child second born, and unaffected child third born. Thus, there is a 3/64 + 3/64 + 3/64 = 9/64 chance that 2 out of 3 children will be affected. g. The phenotype of the last child is independent of all others, so the probability of an affected child is 1/4. 2-19. Diagram the cross, where P is the normal pigmentation allele and p is the albino allele: normal x normal → albino P? x P? → pp An albino must be homozygous recessive pp. The parents are normal in pigmentation and therefore could be PP or Pp. Because they have an albino child, they must both be carriers (Pp). The probability that their next child will have the pp genotype is 1/4. 2-20. Diagram the cross: yellow round x yellow round → 156 yellow round : 54 yellow wrinkled The monohybrid ratio for seed shape is 156 round : 54 wrinkled = 3 round : 1 wrinkled. The parents must therefore have been heterozygous (Rr) for the pea shape gene. All the offspring are yellow and Chapter 2 11 therefore have the Yy or YY genotype. The parent plants were Y- Rr x YY Rr (that is, you know at least one of the parents must have been YY). 2-21. Diagram the cross: smooth black ♂ x rough white ♀ → F rough black 1 → F 8 smooth
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