Using Laws of Inheritance to Predict Genotypes and Phenotypes PDF

Unit 1: Laws of Inheritance Lesson 1.2 Using Laws of Inheritance to Predict Genotypes and Phenotypes Contents Introduction 1 Learning Objectives 2 Warm Up...

Unit 1: Laws of Inheritance Lesson 1.2 Using Laws of Inheritance to Predict Genotypes and Phenotypes Contents Introduction 1 Learning Objectives 2 Warm Up 2 Learn about It! 4 Using a Testcross to Determine Genotypes 4 Punnett Square 6 The Six Different Monohybrid Combinations 8 Forked-Line Method 10 Probability Method 12 Product Rule of Probability 13 Sum Rule of Probability 14 Key Points 25 Check Your Understanding 26 Challenge Yourself 27 Photo Credit 27 Bibliography 28 Key to Try It! 28 Unit 1: Laws of Inheritance Lesson 1.2 Using Laws of Inheritance to Predict Genotypes and Phenotypes Introduction Have you tried playing games where your probability of winning relies purely on chance? You may have played Snakes and Ladders with your siblings or friends, and your luck depends on two factors—the number of moves that will take you ahead and the random chance of encountering a snake or a ladder. Each of the dice that you throw in this game gives six possible outcomes. Also, you may have settled the priority of turns in games through a coin toss. This time, the outcome may either be heads or tails. 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 1 Unit 1: Laws of Inheritance Our inheritance also works similar to some chance games. Because of the law of segregation and the law of independent assortment, different combinations of traits occur at various probabilities in the offspring. In relation to these laws, we can liken the two sides of a coin to the two alleles that you may inherit from your mother. Understandably, if she is heterozygous for a trait, the probability of getting the dominant or the recessive allele is similar to a coin toss, landing with either heads or tails. However, if she happens to be homozygous, it will be like tossing a coin with both sides as heads. In this chapter, you will apply the laws of inheritance in predicting the possible outcomes of fertilization. Learning Objectives DepEd Competency In this lesson, you should be able to do the Predict genotypes and phenotypes of parents and offspring using the laws following: of inheritance Apply the laws of inheritance in (STEM_BIO11/12-IIIa-b-1). identifying the possible outcomes of genetic crosses. Compute for offspring probabilities in genetic problems. Warm Up Probing Probabilities: 15 minutes Rock vs. Paper vs. Scissors When do you usually play the Rock, Paper, Scissors game? Is it when you settle who will make the first move in a tournament? Is it to determine who will win a game when it ends in a draw? How lucky are you in winning this game? In this activity, you will determine your chances of winning the Rock, Paper, Scissors, and relate it to biological inheritance. Materials two dice (per pair) worksheet 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 2 Unit 1: Laws of Inheritance Procedure 1. Access the link below and print a copy of the worksheet. You may do this before the class. Alternatively, if a printer is not available, you may draw and complete the worksheet table in a blank sheet of paper. BIO2 0102 Warmup_Probing Probabilities Quipper Limited, “Probing Probabilities: Data Sheet,” https://drive.google.com/file/d/1Ebf8RJta1bLJGsTfk7SFxpa1it8 XXVHv/view?usp=sharing, last accessed on April 22, 2020. 2. Look for a partner with whom you will play Rock, Paper, Scissors. Designate a player one and a player two between you and your partner. 3. At the start of the activity, you will be asked by your teacher: Is there a way to win the Rock, Paper, Scissors game? Try to briefly establish an answer to this question with your chosen partner. 4. Play the game for 30 rounds with your partner. For every round, record the results by placing a checkmark on the winning option (rock, paper, or scissors) and the winning player (player one or player two) in the table found in the worksheet. 5. Some rounds will end in a tie or a draw. If this occurs, repeat that round until the tie is settled and someone wins. 6. After completing all rounds, count the number of checkmarks for each column. 7. On the next page of the worksheet, analyze the table on the possible outcomes of a Rock, Paper, Scissors game. Based on that table, compute for the possibilities of player 1 and player 2 to win by calculating for the percentages. ○ Take note that this table is independent of the previous table. This step requires you to calculate for the possible outcomes of a hypothetical Rock, Paper, Scissors game. 8. According to the results of your computation, try to discuss with your partner whether there is a strategy to win Rock, Paper, Scissors. 9. Answer the guide questions below to deepen your understanding afterward. 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 3 Unit 1: Laws of Inheritance Guide Questions 1. What is the theoretical probability of player 1 to win? How about player 2? 2. Given your actual and theoretical probabilities, is there a definite strategy to win the game? 3. How do you think this game is similar to gene transmission from a heterozygous parent to its offspring? Learn about It! Transmission genetics, as discussed in the previous chapter, focuses on the patterns of inheritance of biological characteristics. Both the laws of segregation and independent assortment are essential in predicting the genotypes and phenotypes of parents and offspring in genetic crosses and human reproduction. If you are given a plant with a dominant trait, how would you know if it is homozygous or heterozygous? Using a Testcross to Determine Genotypes For some genes, one allele completely masks the expression of the other allele. This mode of inheritance is called complete dominance. Thus, if the genotype of an individual is heterozygous, the dominant allele will completely mask the expression of the recessive allele. In peas, for example, the gene for seed shape has two alleles. Smooth or round pea is completely dominant over wrinkled pea. In Fig. 1.2.1, if the individual has the recessive phenotype, its genotype will always be homozygous or true-breeding. By contrast, an individual with the dominant trait may either be homozygous or heterozygous. If you are given a mature plant with round seeds, would you be able to determine whether the plant is homozygous or heterozygous for seed shape? One can identify the genotype of an individual with the dominant trait through a testcross. A testcross is a simple technique wherein the individual with the dominant phenotype is crossed with or mated with a recessive individual. The results of the cross will help you determine parental genotype, particularly the one with the dominant trait. 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 4 Unit 1: Laws of Inheritance Fig. 1.2.1. The inheritance of seed shape in Pisum sativum follows complete dominance. Given the example above, the pea plant with smooth seeds must be mated with a plant that has wrinkled seeds. One of the two possible outcomes of mating can be generated, as shown in Table 1.2.1. In both crosses, the second allele of the dominant individual is left blank. Regardless, the phenotype is still dominant because only one allele is needed to express round seeds. After mating with the recessive (i.e., wrinkled-seeded peas) individual, there are two possible results. a. Case 1: If only one phenotype appears in F1, which is the dominant trait, the genotype must be homozygous. This result is due to the fact that the first parent will only contribute the dominant allele to all offspring. b. Case 2: Two phenotypes appear in the offspring—both dominant and recessive individuals are present. The presence of a recessive offspring in the progeny is an indication that both parents must have contributed recessive alleles. Thus, the first parent must be heterozygous in this scenario, as shown in the table below. 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 5 Unit 1: Laws of Inheritance Table 1.2.1. The two possibilities for the testcross of a round-seeded pea Cases Crosses Results in F1 Conclusion The round-seeded individual round × wrinkled: is homozygous. 1 all with smooth seeds A__ × aa Therefore, the correct cross is: AA × aa The round-seeded individual round × wrinkled: is heterozygous. some with round seeds, 2 some with wrinkled seeds A__ × aa Therefore, the correct cross is: Aa × aa Is the cross between a tall pea and a dwarf pea an example of a testcross? Why? Both the laws of segregation and independent assortment are fundamental to the analysis of matings in both plants and animals. These laws are used to determine the genotypes and phenotypes of both parents and offspring in crosses by using the Punnett square, forked-line method, and probability method. Punnett Square The previous chapter already demonstrated the laws of inheritance by using the Punnett square, which was named after the British geneticist Reginald C. Punnett. In this method, the alleles of both parents must be first determined. Thereafter, these alleles are fused to determine the possible genotypes in the offspring. Then, the principle of dominance is applied to determine the corresponding phenotypes of each genotype. This straightforward method is applicable when one or two genes are involved in the cross. Fig. 1.2.2 below applies the Punnett Square to the two possible cases involving the testcross of a violet-flowered pea. 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 6 Unit 1: Laws of Inheritance Fig. 1.2.2. Punnett square is a simple technique that is used to determine all possible offspring of a cross. A male and a female parent must be designated in this method. The testcross of a violet-flowered pea gives 100% violet-flowered offspring if the dominant parent is homozygous (left). By contrast, F1 consists of 50% violet- and 50% white-flowered peas if the violet-flowered parent is heterozygous (right). Some Punnett squares do not necessarily have to assume the shape of a square. There are cases when the parents of a cross will not produce the same number of allelic combinations. For example, your objective is to subject a doubly heterozygous pea plant (AaBb) with round and yellow seeds to a testcross (as shown in Fig. 1.2.3). Previously, it was discussed that the round seed is dominant over the wrinkled seed, while the yellow seed is dominant over the green seed. 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 7 Unit 1: Laws of Inheritance If a testcross is to be done, this means that the other parent must be recessive for both traits, i.e., it should have wrinkled and green seeds. If we assign gene A for seed shape (then, A for round and a for wrinkled seeds) and gene B for seed color (then, B for yellow and b for green seeds), we can generate the cross AaBb × aabb. ○ The first parent can produce gametes AB, Ab, aB, and ab. ○ The second parent only produces the gamete ab. AaBb parent AB Ab aB ab AaBb Aabb aaBb aabb aabb ab round, round, wrinkled, wrinkled, parent yellow green yellow green Fig. 1.2.3. The Punnett square of the cross AaBb × aabb will give the same genotypic and phenotypic ratios, which is 1:1:1:1 (same as 1/4:1/4:1/4:1/4). The Six Different Monohybrid Combinations The next two techniques (forked-line method and probability method), where the laws of inheritance are applied, will require the use of Punnett squares for every gene pair of a cross. Solving genetic crosses requires mastery of the six possible parental combinations in a monohybrid cross, as shown in Fig. 1.2.4 and Table 1.2.2. below. Case 1: AA × AA Case 2: AA × Aa Case 3: Aa × Aa A A A A A a A AA AA A AA AA A AA Aa A AA AA a Aa Aa a Aa aa GR: 100% or all AA GR: 1/2 AA: 1/2 Aa GR: 1/4 AA: 2/4 Aa: 1/4 aa PR: 100% or all dominant PR: 100% or all dominant PR: 3/4 dominant: 1/4 recessive 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 8 Unit 1: Laws of Inheritance Case 4: AA × aa Case 5: Aa × aa Case 6: aa × aa A A A a a a a Aa Aa a Aa aa a aa aa a Aa Aa a Aa aa a aa aa GR: 100% or all Aa GR: 1/2 AA: 1/2 aa GR: 100% or all aa PR: 100% or all dominant PR: 1/2 dominant: 1/2 recessive PR: 100% or all recessive Fig. 1.2.4. The basic monohybrid cross can come in six different combinations by using a hypothetical gene A. Each has a characteristic GR (genotypic ratio) and PR (phenotypic ratio). These combinations are further explained in Table 1.2.2. Table 1.2.2. The six possible parental combinations in a monohybrid cross is described below. In each cross, since no specific trait is assigned, the dominant phenotype can be represented with A__, while the recessive phenotype with aa. However, note that for other problems, there is a need to specify the traits or phenotypes of the offspring. Case Crosses Result The dominant phenotype with a homozygous 1 Two homozygous dominant genotype is expected in the offspring. Two genotypes (homozygous and heterozygous), and only the dominant Homozygous dominant and phenotype in the offspring. 2 heterozygous This is because the first parent is not a bearer of the recessive allele. Three genotypes and both dominant and 3 Two hybrids recessive phenotypes Two true-breeding individuals Produces hybrids in the F1 generation 4 with different traits Heterozygous individual and Yields two genotypes and two phenotypes in its 5 an individual with the offspring recessive phenotype Two homozygous recessive The recessive phenotype with a homozygous 6 individuals genotype is expected in the offspring. 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 9 Unit 1: Laws of Inheritance Forked-Line Method The forked-line method is also a technique that can be used to determine the genotypic and phenotypic ratios of a genetic cross involving two or more genes. This method does not require the identification and enumeration of the alleles of each parent and combining them in square-like units. Rather, each of the monohybrid crosses in the problem is analyzed for their outcomes. This method can also separately identify the genotypic ratio from the phenotypic ratio. From the previous lesson, the cross between two dihybrids, AaBb × AaBb, 16 square units as needed. The process also requires a close assessment of which of the genotypes are identical. The forked-line method, by contrast, can directly give you offspring ratios without counting them individually. For example, a cross between doubly heterozygous tall pea with violet flowers is made. Fig. 1.2.5. The forked-line method is an ideal method to enumerate all possible genotypes of a cross and to directly obtain their ratios without counting genotypes. 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 10 Unit 1: Laws of Inheritance 1. Assigning the respective alleles, M–tall, m–dwarf, N–violet, and n–white, our cross will be MmNn × MmNn. 2. Fig. 1.2.5 shows the straightforward determination of the genotypic ratio through the forked-line method. To begin this technique, you should first determine the result of each of the monohybrid crosses. 3. Two monohybrid crosses are involved: Mm × Mm and Nn × Nn. Both of these crosses are similar to case 3 in Fig. 1.2.4 earlier, where three genotypes are produced in a ratio of 1:2:1. 4. The first column of Fig. 1.2.5. represents the outcomes of Mm × Mm cross. List the three resulting genotypes first in a vertical fashion. 5. Then, work on Nn × Nn cross in the second column, where three genotypes are also possible. 6. Branch each of the three outcomes of gene N to that of gene M in the first column. 7. Lastly, combine the genotypes from both the first and second columns and multiply their probabilities. Without having to count the genotypes as in the Punnett square, we can directly generate the genotypic ratio of the dihybrid cross. Fig. 1.2.6. The forked-line method can also directly determine the phenotypic ratio of the offspring of a genetic cross without identifying the genotypic ratio. 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 11 Unit 1: Laws of Inheritance In Punnett squares, one should first identify all genotypes to determine the phenotypes of a cross. By contrast, the fork-line method can also be used to determine the phenotypic ratio of a cross without identifying its genotypic ratio (as shown in Fig. 1.2.6). 1. The same principle applies. One should have a mastery of the six possible combinations in a monohybrid cross to conveniently perform this technique. 2. In the first column, the cross of Mm × Mm will give 3/4 tall (M_) and 1/4 dwarf (mm). 3. The same ratio applies in the second column where the cross of Nn × Nn gives 3/4 violet (N_) and 1/4 white (nn). 4. Again, branch the results of cross Nn × Nn to each of the results of cross Mm × Mm. How do the sum and product rules of probability apply to the laws of inheritance? Probability Method Another method to solve genetic problems is the probability method, which is considered easier and more convenient than both the Punnett square and the fork-line method. There will be cases, for example, when you are provided with a cross, and you will be asked to just provide the probability of obtaining a type of offspring instead of the entire genotypic and phenotypic ratios. This is when the probability method comes in handy. Imagine if you are given the cross AaBbCcDdEe × AaBbCcDdEe, and you are just asked to determine the probability of getting AABbccDdEe genotype in the progeny. Punnett square and forked-line methods are considered unideal for cases like this. Probability refers to the mathematical measurement of likelihood or chance. Probabilities are represented by quantitative values, which show how likely events will occur. The highest probability value is 1 (or 100%), which means that an event is guaranteed to occur. If an event has a 0 probability, it is guaranteed to not happen. As earlier mentioned, Mendelian segregation is similar to a coin toss. If an individual is heterozygous for a gene, e.g., Aa, the probability of a gamete to obtain a dominant allele is 1/2. The same is true for the recessive allele, as shown in Table 1.2.3. In a coin toss, likewise, the probability of getting heads or tails is 1/2. 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 12 Unit 1: Laws of Inheritance Table 1.2.3. Probability of obtaining dominant or recessive alleles by the gametes Genotype Probability of obtaining A Probability of obtaining a AA 1 or 100% 0 Aa 1/2 or 50% 1/2 or 50% aa 0 1 or 100% Product Rule of Probability In statistics, we can compute for the likelihood of two or more events to occur simultaneously. According to the product rule of probability, the chance of two or more independent events to occur together is equal to the product of their individual probabilities. For example, you are given a turn to roll two dice simultaneously, and you want to determine the likelihood of getting two and four. To get the probability, you must multiply their probabilities, i.e., (probability of obtaining 2) × (probability of obtaining four) = (1/4) × (1/4) = 1/16. In genetics, the application of the product rule is evident in the determination of probabilities of obtaining genotypes in a monohybrid cross. ♂ A a (1/2 chance of getting (1/2 chance of getting from the father) from the father) A AA Aa (1/2 chance of getting (1/2) × (1/2) = 1/4 (1/2) × (1/2) = 1/4 from the mother) ♀ a Aa aa (1/2 chance of getting (1/2) × (1/2) = 1/4 (1/2) × (1/2) = 1/4 from the mother) Fig. 1.2.7. Product rule can help compute for genotype probabilities. 1. For example, you want to determine the chance of getting an AA genotype from the 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 13 Unit 1: Laws of Inheritance cross Aa × Aa. 2. By looking at Fig. 1.2.7 below, each gamete has a 50% probability of obtaining the dominant or recessive allele. 3. Because the segregation of alleles between male and female gametogenesis are independent events (i.e., they do not influence each other), we should multiply probabilities. 4. Thus, AA probability is equal to (probability of an egg with A) × (probability of sperm with A) = (1/2) × (1/2) = 1/4 or 25%. Aside from simple monohybrid crosses, the product rule also significantly helps in the determination of offspring probability in crosses where the genes are independently assorting. For example, you are given the cross AaBbCc × aaBBCc, where genes A, B, and C are independently assorting. What will be the probability of the following genotypes below? Again, since the genes are independently inherited, we can multiply the probabilities. You can peek in Fig. 1.2.4 to recall the individual probabilities. 1. AaBbCc = (chance of Aa) × (chance of Bb) × (chance of Cc) = (1/2) × (1/2) × (1/2) = 1/8 2. aaBbcc = (chance of aa) × (chance of bb) × (chance of cc) = (1/2) × (1/2) × (1/4) = 1/16 3. aaBBCC = (chance of aa) × (chance of BB) × (chance of CC) = (1/2) × (1/2) × (1/4) = 1/16 Sum Rule of Probability Some genetic problems involve cases wherein two or more events do not occur simultaneously, but we have to determine the likelihood of either of them occurring. In cases like this, the sum rule of probability must be applied. According to this rule, the probability of either of two mutually exclusive events occurring is equal to the sum of their individual probabilities. For example, you are given one turn or attempt to roll a die, and you want to determine the chance of landing with two or four. The probability would be equal to the sum of their individual probabilities, i.e., (chance of getting a two) + (chance of getting a four) = 1/6 + 1/6 = 2/6 or 1/3 or 33.33%. These two events are mutually exclusive because either of them can happen, but not at the same time. The sum rule also applies in genetics, as shown in Fig. 1.2.8. 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 14 Unit 1: Laws of Inheritance Fig. 1.2.8. The probability of a dominant phenotype in the cross Rr × Rr is equal to the sum of probabilities of getting a dominant allele. 1. Let us use the probability of obtaining the dominant phenotype from the cross Rr × Rr to discuss this. 2. Given the cross, both maternal and paternal parents (or mother and father) will contribute dominant and recessive alleles. 3. During fertilization, three events may lead to the dominant phenotype in the offspring. a. First, both parents may contribute their dominant R allele. b. Second, the father contributes the dominant R, while the mother contributes the recessive r. c. Third, the mother contributes the dominant R, while the father contributes 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 15 Unit 1: Laws of Inheritance the recessive r. 4. As shown in the figure below, each of the three events has a probability of 1/4. 5. To determine the chance of obtaining the dominant phenotype, we can just add their individual probabilities, i.e., (probability of obtaining R from both parents) + (probability of obtaining dominant maternal allele) + (probability of obtaining the dominant paternal allele) = (1/4) + (1/4) + (1/4) = 3/4 or 75%. If you want to determine the probability of getting a girl and a boy in consecutive pregnancies, which rule of probability is more applicable, the sum rule or the product rule? Why? Let’s Practice! Example 1 In humans, the deposition of melanin in the skin, eyes, and hair is under the control of a gene that is inherited through complete dominance. The recessive mutant allele is characterized by the impaired pigmentation, which results in the condition called albinism. If a normally pigmented couple, each of whom has an albino parent, had children, what is the expected genotypic and phenotypic ratios of their children with respect to the trait? Use a Punnett square to justify your answer. Solution Step 1: You are asked to provide the genotypic and phenotypic ratios of the children of a couple. Step 2: Normal skin pigmentation is dominant over the recessive albino condition. Therefore, we can assign the following alleles. A : normal pigmentation a : albino 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 16 Unit 1: Laws of Inheritance The genotypes with respect to the trait are as follows. AA : normal pigmentation Aa : normal pigmentation aa : albino Both members of the couple have normal skin pigmentation, but each has an albino parent. Step 3: Determine the genotypes of the couple. Both of them are normally pigmented. Thus we can initially assign the following. Male : A__ Female : A__ Since both of them have a parent with aa genotype, they shall automatically inherit a recessive allele. Male : Aa Female : Aa Step 4: Draw a Punnett square and assign the gametes of each parent. ♂ A a A ♀ a Step 5: Combine the alleles to determine offspring genotypes and phenotypes. 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 17 Unit 1: Laws of Inheritance ♂ A a AA Aa A pigmented pigmented ♀ Aa aa a pigmented albino The genotypic ratio of the cross is 1/4 AA: 2/4 Aa: 1/4 aa. The phenotypic ratio is 3/4 pigmented and 1/4 albino. 1 Try It! Given the cross AaBb × aaBb, where A–round seed, a–wrinkled seed, B–yellow seeds, and b–green seeds, what are the genotypic and phenotypic ratios of the offspring? Use Punnett square to solve for the answers. Example 2 In humans, the presence of dimples are controlled by dominant alleles while the presence of a hitchhiker’s thumb is recessive. The absence of dimples and the presence of a hitchhiker’s thumb in an individual requires two copies of the recessive alleles. By using the forked-line method, determine the genotypic and phenotypic ratios of the children of a couple, wherein the male is heterozygous for both traits, while the female is heterozygous for hitchhiker’s thumb, but has no dimples. Solution Step 1: You are asked to provide the genotypic and phenotypic ratios of the children of the given couple. Step 2: The presence of both the non-hitchhiker’s thumb (straight thumb) and dimples is dominant over their absence. Therefore, we can assign the following alleles. 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 18 Unit 1: Laws of Inheritance G : non-hitchhiker’s thumb (straight thumb) g : hitchhiker’s thumb H : with dimples h : without dimples Also, the following parental phenotypes are given. Male parent : non-hitchhiker’s thumb and with dimples, heterozygous for both Female parent : non-hitchhiker’s thumb (heterozygous), without dimples Step 3: Assign the genotypes of both parents given the alleles. Male parent : GgHh Female parent : Gghh Step 4: Draw the corresponding fork-line for genotypes. This step is what will determine phenotype ratios later on. 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 19 Unit 1: Laws of Inheritance Step 5: Draw the corresponding forked-line for phenotypes. The genotypic ratio is 1/8 GGHh: 1/8 GGhh: 2/8 (or 1/4) GgHh: 2/8 (or 1/4) Gghh: 1/8 ggHh: 1/8 gghh. The phenotypic ratio is 3/8 non-hitchhiker’s thumb, dimpled: 3/8 non-hitchhiker’s thumb, without dimples: 1/8 hitchhiker’s thumb, dimpled: 1/8 hitchhiker’s thumb, without dimples. 2 Try It! By using the fork-line method, determine the genotypic ratio of the offspring of the testcross of MmNnOo. Example 3 Two separate crosses were performed in peas. The first cross involved seed shape and seed color. The second cross involves height, inflorescence, and flower colors. The phenotypes of the parents and offspring are given below. Give the complete genotypes of both parents and offspring for both crosses. Use A for seed shape, B for seed color, C for height, D for inflorescence, and E for flower color. 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 20 Unit 1: Laws of Inheritance Cross 1 Cross 2 round, yellow × wrinkled, green tall, terminal, white × dwarf, axial, violet 1/4 round, yellow 1/4 tall, axial, violet 1/4 round, green 1/4 dwarf, axial, violet 1/4 wrinkled, yellow 1/4 tall, terminal, violet 1/4 wrinkled, green 1/4 dwarf, terminal, violet Solution Step 1: You are asked to provide the genotypes of all individuals in the crosses. Step 2: The phenotypes of both parents and offspring are given. The allelic assignment for each characteristic is also given. Therefore, A : round C : tall a : wrinkled c : dwarf B : yellow D : axial b : green d : terminal E : violet e : white Step 3: Make an initial assignment for the alleles in both crosses. Cross 1: round, yellow × wrinkled, green A__B__ × aabb F1: 1/4 round, yellow A__B__ 1/4 round, green A__bb 1/4 wrinkled, yellow aaB__ 1/4 wrinkled, green aabb Cross 2: tall, terminal, white × dwarf, axial, violet C__ddee × ccD__E__ F1: 1/4 tall, axial, violet C__D__E__ 1/4 dwarf, axial, violet ccD__E__ 1/4 tall, terminal, violet C__ddE__ 1/4 dwarf, terminal, violet ccddE__ 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 21 Unit 1: Laws of Inheritance Step 4: Complete the missing alleles, if possible. Cross 1: round, yellow × wrinkled, green AaBb × aabb F1: 1/4 round, yellow AaBb 1/4 round, green Aabb 1/4 wrinkled, yellow aaBb 1/4 wrinkled, green aabb Cross 2: tall, terminal, white × dwarf, axial, violet Ccddee × ccDdEE F1: 1/4 tall, axial, violet CcDdEE 1/4 dwarf, axial, violet ccDdEE 1/4 tall, terminal, violet CcddEE 1/4 dwarf, terminal, violet ccddEE We can fill in the second allele of the individuals with a recessive one if a recessive phenotype appears in the offspring. By contrast, if no recessive phenotype appears in the offspring, such as in gene E, the dominant parent must be homozygous for the trait. 3 Try It! A cross between two peas was made. The first parent has smooth pods, white flowers, and yellow seeds. The second parent has smooth pods, violet flowers, and yellow seeds. Four offspring appeared in the offspring as follows: (1) smooth, violet, yellow; (2) wrinkled, violet, green; (3) smooth, violet, green; and (4) wrinkled, violet, yellow. Provide the genotypes of both parents and all of their offspring. Example 4 Given the multi-hybrid cross AaBbccDdEe × AabbCcDdEe, determine the probability of obtaining the following genotypes and phenotypes in the offspring. The allele assignments are given below. Assume the genes to be independently assorting. 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 22 Unit 1: Laws of Inheritance A – red B – thorny C – smooth leaf D – long pollen E – green fruit a – white b – thornless c – hairy leaf d – round pollen e – purple fruit 1. aaBbCcDdEE 2. AabbccDDee 3. AAbbCCDdEe 4. red, thornless, smooth, long, green 5. white, thorny, hairy, round, purple Step 1: You are asked to provide the probabilities of the provided offspring. Step 2: The allelic designation, dominance, and offspring genotypes and phenotypes are given. Step 3: Substitute probabilities for the genotypes in the problems. Use the patterns in the six monohybrid crosses given earlier in the discussion (Fig. 1.2.4). a. aaBbCcDdEE = (1/4) · (1/2) · (1/2) · (1/2) · (1/4) = 1/128 b. AabbccDDee = (1/2) · (1/2) · (1/2) · (1/4) · (1/4) = 1/128 c. AAbbCCDdEe = (1/4) · (1/2) · (0) =0 Step 4: Substitute probabilities for the phenotypes in the problems. Use the patterns in the six monohybrid crosses given earlier in the discussion (Fig. 1.2.4). a. red, thornless, smooth, long, green = A__bbC__D__E__ = (3/4) · (1/2) · (1/2) · (3/4) · (3/4) = 27/256 b. white, thorny, hairy, round, purple = aaB__ccddee = (1/4) · (1/2) · (1/2) · (1/4) · (1/4) = 1/256 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 23 Unit 1: Laws of Inheritance 4 Try It! Given the cross MmNnooPp × mmNnOoPp, what is the probability of obtaining the following in the F1 generation? (a) mmNnOoPp, (b) MmnnooPP, (c) M__N__O__P__, (d) mmnnoopp. Did You Know? Polydactyly, or the presence of extra fingers or toes, is a genetic condition in humans. Particularly, it is inherited as a dominant condition. Having at least one parent with polydactylism increases the chances of the trait to manifest among the children. This condition exhibits different degrees of expression. In some people, it may just be a flap of tissue, while in others, it may have bones but without joints. Also, more polydactylous individuals have their extra digit on the side of the little finger than on the side of the thumb. On very rare occasions, it is positioned in between fingers. Polydactylous individuals have at least one of their parents with the genetic condition. 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 24 Unit 1: Laws of Inheritance Key Points __________________________________________________________________________________________ Given the probability that an individual with the dominant phenotype may be homozygous or heterozygous, a testcross is used. When performing a testcross, the individual with the dominant trait is crossed to an individual with the recessive trait. Punnett square is the most basic technique in combining the gametes of parents to determine the possible genotypes and phenotypes of the offspring. The forked-line method is a more straightforward technique that also allows the determination of the genotypes and phenotypes of the progeny, as well as their corresponding ratios. The probability method is useful when enumerating all genotypes and phenotypes of the offspring is not needed. It is more applicable when directly solving for the probabilities of particular genotypes and phenotypes in the progeny. ○ The product rule of probability should be used when solving the likelihood of two independent events occurring simultaneously. It is applicable to problems involving independently assorting genes. ○ The sum rule of probability should be used when solving for the probability of either of two or more events occurring. __________________________________________________________________________________________ 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 25 Unit 1: Laws of Inheritance Check Your Understanding A. Determine the accuracy of each of the following statements. Write true if the statement is correct and false if otherwise. 1. A cross between two violet-flowered plants is an example of a testcross. 2. In a Punnett square, the gametes of the parents are combined. 3. In the cross Aa × aa, two possible phenotypes may appear in the offspring. 4. In the cross Aa × Aa, the phenotypic ratio in the offspring is 1:2:1. 5. In the cross aa × AA, all offspring will be hybrids. 6. When solving for the probability of obtaining the genotype AABbCc, the individual probabilities of obtaining AA, Bb, and Cc must be added. 7. In a Punnett square involving the cross Aa × Aa, the 1/4 chance to obtain AA is solved by multiplying the chances of getting an A allele from each parent. 8. A cross between a pea with round, yellow seeds, and another pea with wrinkled, green seeds is an example of a testcross. 9. When performing the forked-line method, the alleles must be first identified. 10. Two normally pigmented parents with an albino son imply that both parents are heterozygous for the trait. 11. In a testcross involving one characteristic, if two phenotypes are present in the offspring, the individual with the dominant phenotype must be heterozygous. 12. In the F1 intercross of Mendel, where AaBb × AaBb, six unique phenotypes are present in the offspring. 13. An individual with a recessive trait has ½ chance of transmitting his or her recessive allele to gametes. 14. The Punnett square was devised by Reginald Punnett. 15. Albinism is a recessive trait. B. Describe the application of item (a) to the provided item (b). Provide brief answers only. 1. (a) law of segregation in a (b) Punnett square 2. (a) law of independent assortment in (b) probability method 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 26 Unit 1: Laws of Inheritance 3. (a) testcross in (b) analyzing the genotype of a round-seeded pea 4. (a) sum rule in (b) identifying the probability of Bb from the cross Bb × Bb 5. (a) the forked-line method in (b) identification of genotypes of a cross Challenge Yourself Answer the following questions. 1. How is the genotype of an individual related to the number of gametes it can produce? 2. Determine how many types of allele combinations will each of the following individuals produce given their genotypes: (a) AABbCc, (b) aaBbCc, (c) AabbCc, (d) AaBbCc. 3. A woman possesses a rare eyelid abnormality called ptosis. This condition prevents her from completely opening her eyes, and it has been found to be controlled by a dominant allele T. This woman’s father has the condition, but her mother is normal. Also, the mother of her father had the condition. (a) Given the case, what are the possible genotypes of the woman, her father, and her mother? (b) If she marries a man with normal eyelids, what is the probability that she may have a child with ptosis? 4. Two highly inbred strains of laboratory mice were crossed. One parent has black fur, and the other parent has gray fur. All of their offspring possess black fur. What is the expected genotypic and phenotypic ratio if all of their offspring intercrossed? 5. An Arabidopsis plant heterozygous for three independently assorting genes C, D, and E is self-fertilized. Among the offspring of the cross, predict the frequency of (a) CCDDEE individuals, (b) ccddee individuals, (c) CcDdEe individuals, (d) an individual that is dominant for all genes. Photo Credit Wanitetlefthand by Cplbeaudoin at English Wikipedia is licensed under CC BY-SA 3.0 via Wikimedia Commons. 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 27 Unit 1: Laws of Inheritance Bibliography Brooker, J. Concepts of Genetics (1st ed.). New York, USA: McGraw-Hill Companies Inc., 2012. Klug, W.S, and Cummings, M.R. Concepts of genetics (6th ed). Upper Saddle River, N.J: Prentice-Hall. 2003. Pierce, B. Genetics: a conceptual approach (8th ed). New York: W.H. Freeman. 2012. Reece J., Taylor M., Simon E., and Dickey J. Campbell Biology: Concepts and Connections (7th ed.). Boston: Benjamin Cummings/Pearson. 2011. Snustad, D.P., and Simmons, M.J. Principles of Genetics (6th ed.). Hoboken, NJ: Wiley. 2012. Key to Try It! 1. The genotypic ratio of the offspring of the cross AaBb × aaBb (given that A–round seed, a–wrinkled seed, B–yellow seeds, and b–green seeds) is 1/8 AaBB: 1/8 Aabb: 2/8 AaBb: 2/8 aaBb: 1/8 aaBB: 1/8 aabb. The phenotypic ratio is 3/8 round, yellow, 3/8 wrinkled, yellow, 1/8 round, green: 1/8 wrinkled, green. AB Ab aB ab AaBB AaBb aaBB aaBb aB round, round, wrinkled, wrinkled, yellow yellow yellow yellow AaBb Aabb aaBb aabb ab round, round, wrinkled, wrinkled, yellow green yellow green 2. Given the testcross of MmNnOo (thus MmNnOo × mmnnoo), the genotypes are given through the forked-line method below. The genotypic ratio is 1:1:1:1:1:1:1:1 (genotypes provided in the last column below). 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 28 Unit 1: Laws of Inheritance 3. The genotypes for the provided individuals in the cross are as follows. Assume that gene A codes for seed shape, B for flower color, and C for seed color. Cross: smooth pods, white flowers, and yellow seeds × smooth pods, violet flowers, and yellow seeds A__bbC__ × A__B__C__, then to complete, AabbCc × AaBBCc F1 generation: smooth, violet, yellow AaBbCC or AaBbCc wrinkled, violet, green aaBbcc smooth, violet, green AaBbcc wrinkled, violet, yellow AABbCC or AaBbCc 4. Given the cross MmNnooPp × mmNnOoPp, the frequency of the following are as follows: a. mmNnOoPp = (1/2) × (1/2) × (1/2) × (1/2) = 1/16 b. MmnnooPP = (1/2) × (1/4) × (1/2) × (1/4) = 1/64 c. M__N__O__P__= (1/2) × (3/4) × (1/2) × (3/4) = 9/64 d. mmnnoopp = (1/2) × (1/4) × (1/2) × (1/4) = 1/64 1.2. Using Laws of Inheritance to Predict Genotypes and Phenotypes 29