4.6 Dihybrid Crosses. offspring produced from such a cross are heterozygous for both the yellow and round genotypes. YYRR. YR YR yr.

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1 (a) Indicate the genotypes and phenotypes of the F generation from the mating of a heterozygous Himalayan rabbit with an albino rabbit. (b) The mating of a full-coloured rabbit with a light-grey rabbit produces two full-coloured offspring, one light-grey offspring, and one albino offspring. Indicate the genotypes (c) of the parents. A chinchilla rabbit is mated with a light-grey rabbit. The breeder knows that the light-grey rabbit had an albino mother. Indicate the genotypes and phenotypes of the F generation from this mating. (d) A test cross is performed with a light-grey rabbit, and the following offspring are noted: five Himalayan rabbits and five light-grey rabbits. Indicate the genotype of the lightgrey rabbit. 6. Palomino horses are known to result from the interaction of two different alleles. The allele C r in the homozygous condition produces a chestnut, or reddish, coat. The allele C m in the homozygous condition produces a very pale cream coat, called cremello. The palomino colour is caused by the interaction of both the chestnut and cremello alleles. Indicate the expected ratios in the F generation from mating a palomino with a cremello. 7. There are four different ABO blood types (Table 5). The alleles for blood types A and B are codominant but are dominant over the allele for blood type O. Table 5 yellow, round YYRR yyrr green, wrinkled Phenotypes type A type B type AB type O Genotypes I A I A, I A I O I B I B, I B I O I A I B I O I O formed by meiosis YR YR yr fertilization of yr Determine the possible phenotypes and genotypes of the F generation offspring of a parent with type A blood and a parent with type B blood. Use Punnett squares to show your work. All members of the F generation have the same genotype and phenotype. Figure A dihybrid cross between a pure-breeding pea plant with yellow, round seeds and a purebreeding pea plant with green, wrinkled seeds dihybrid cross: a type of cross that involves two genes, each consisting of nonidentical alleles.6 Dihybrid Crosses Mendel also studied the inheritance of two separate traits in crossbreeding, following the same procedure he had used for studying single traits. He crosspollinated pure-breeding plants that produced yellow, round seeds with purebreeding plants that produced green, wrinkled seeds to study the inheritance of two traits. The laws of genetics that apply for a single-trait inheritance (monohybrid cross) also apply for a two-trait inheritance (dihybrid cross). Figure shows a dihybrid cross. The pure-breeding round seed is indicated by the symbol RR, and the pure-breeding wrinkled seed by the recessive alleles rr. The pure-breeding yellow seed is indicated by the alleles YY and the green by yy. The genotype for the yellow, round parent is YYRR, while the genotype for the green, wrinkled parent is yyrr. The F offspring produced from such a cross are heterozygous for both the yellow and round genotypes. 50 Chapter

2 .6 Now consider a cross between a pure-breeding green, round pea plant and a pure-breeding yellow, wrinkled pea plant. Figure shows the resulting offspring. Inheritance of the gene for colour is not affected by either the wrinkled or round alleles. By doing other crosses, Mendel soon discovered that the alleles assort independently, even though he did not know about the existence of chromosomes or the process of meiosis. Today, this phenomenon is referred to as the law of independent assortment. The genes that govern pea shape are inherited independently of the ones that control pea colour. Mendel allowed plants of the F generation to self-fertilize in order to produce an F generation. Each heterozygous, yellow, round plant can produce four different phenotypes. As the homologous chromosomes move to opposite poles during meiosis, the yellow allele will segregate with the round and wrinkled alleles in equal frequency (see Figure ). This means that the sex cells containing YR will equal the number of sex cells containing. Similarly, the green allele will segregate with round and wrinkled alleles in equal frequency. The number of containing Yr will be equal to the number of containing yr. law of independent assortment: If genes are located on separate chromosomes, they are inherited independently of each other. green, round yr yellow, wrinkled YYrr Yr Yr F generation yellow, round for yellow, wrinkled Y R y r parent cell Figure The inheritance of the gene for pea colour is not affected by the gene for pea shape. Y R y r Y r y R Figure Four possible combinations for the of the genotype A Punnett square can be used to predict the genotypes and phenotypes of the F generation. The phenotype ratios of the F generation shown in the Punnett square in Figure (page 5) are 9 yellow, round green, round 6 6 yellow, wrinkled green, wrinkled 6 6 Probability Genotypic and phenotypic ratios are determined by the probability of inheriting a certain trait. The probability of an event is the likelihood that the event will occur. For example, you can calculate the probability of getting heads when you toss a coin. Probability can be expressed by the following formula: Probability = number of ways that a given event could occur total number of possible events In the coin-toss example, there is only one way of tossing heads, so the numerator is. The denominator is because there are two possible events in total, heads or tails. Therefore, the probability of tossing heads is. Two important rules will help you understand probability: Chance has no memory. For example, if you tossed two heads in a row, the probability of tossing heads once again would still be. The probability of Genes and Heredity 5

3 YR Yr yr YYRR YR YYRr YR yr yr YYRr YYrr Yyrr yr Yyrr yyrr YR Yr Figure A Punnett square used to find the F generation of a dihybrid cross. Each heterozygous, yellow, round plant can produce four different phenotypes. yr each event is not affected by the results of the others, so the events are said to be independent. The probability of independent events occurring together is equal to the product of those events occurring separately. For example, the chances of tossing heads once is ; the probability of tossing heads twice is =, and the probability of tossing heads three times in a row is = 8. (a) (b) Sample Problem In humans, free earlobes are controlled by the dominant allele E, and attached earlobes by the recessive allele e (Figure 5 (a) and (b)). The widow s peak hairline is regulated by the dominant allele H, while the straight hairline is controlled by the recessive allele h (Figure 5 (c) and (d)). Consider the mating of the following genotypes: EeHh EeHh (c) What are the probabilities of obtaining F offspring with the following characteristics? widow s peak and free earlobes straight hairline and free earlobes widow s peak and attached earlobes straight hairline and attached earlobes (d) Figure 5 The shape of both earlobes and hairline are inherited characteristics in humans. The free earlobe (a) is dominant over the attached earlobe (b), and the widow s peak (c) is dominant over the straight hairline (d). Solution Dihybrids can be treated as two monohybrids, as shown in Figure 6. Isolate the gene for earlobes and for hairline and work with each as a monohybrid. The F generation resulting from a cross between the heterozygous parents can be determined. The following are the phenotype probabilities for earlobes: free earlobes 5 Chapter attached earlobes

4 .6 EeHh EeHh E e H h E EE Ee H HH Hh e Ee ee h Hh hh Figure 6 Punnett squares showing monohybrid crosses between heterozygous parents for free earlobes and for a widow s peak. The following are the phenotype probabilities for hairlines: widow s peak straight hairline The monohybrids can now be combined to calculate the probabilities of the dihybrid crosses. For example, the chances of producing an F offspring from the mating of EeHh EeHh who has a widow s peak and free earlobes is,or 9 ; 6 a straight hairline and free earlobes is,or ; 6 a widow s peak and attached earlobes is,or 6 ; a straight hairline and attached earlobes is,or. 6 Figure 7 shows the results of the dihybrid cross. EH eh Eh eh EH EEHH EeHH EEHh EeHh eh Eh eh EeHH EEHh EeHh eehh EeHh eehh EeHh EEhh Eehh eehh Eehh eehh Figure 7 Punnett square showing the results of the dihybrid cross of EeHh EeHh. If 6 offspring were generated, it is expected that 9 would have a widow s peak and free earlobes (orange squares), would have a straight hairline and free earlobes (yellow squares), would have a widow s peak and attached earlobes (green squares), and would have a straight hairline and attached earlobes. Sample Problem What is the probability that a child from the mating of the EeHh EeHh parents would be a male with a widow s peak and have attached earlobes? Genes and Heredity 5

5 Solution The calculation indicates that the dihybrid cross is equivalent to two separate monohybrid crosses. Consider each of the separate probabilities: The probability of producing a male is. The probability of having a widow s peak is. The probability of having attached earlobes is. Therefore, the probability of producing a male with a widow s peak and attached earlobes is,or. Practice Applying Inquiry Skills. In guinea pigs, black coat colour (B) is dominant over white (b), and short hair length (H) is dominant over long (h). Indicate the genotypes and phenotypes from the following crosses: (a) A guinea pig that is homozygous for black and heterozygous for short hair is crossed with a white, long-haired guinea pig. (b) A guinea pig that is heterozygous for black and for short hair is crossed with a white, long-haired guinea pig. (c) A guinea pig that is homozygous for black and for long hair is crossed with a guinea pig that is heterozygous for black and for short hair.. Black coat colour (B) in cocker spaniels is dominant over white coat colour (b). Solid coat pattern (H) is dominant over spotted pattern (h). The gene for pattern arrangement is located on a different chromosome than the one for colour, and the pattern gene segregates independently of the colour gene. A male that is black with a solid pattern mates with three females. The mating with female A, which is white and solid, produces four pups: two black, solid; and two white, solid. The mating with female B, which is black and solid, produces a single pup, which is white, spotted. The mating with female C, which is white and spotted, produces four pups: one white, solid; one white, spotted; one black, solid; one black, spotted. Indicate the genotypes of the parents.. The alleles for human blood types A and B are codominant, but both are dominant over the type O allele. The Rh factor is separate from the ABO blood group and is located on a separate chromosome. The Rh-positive allele is dominant over Rh-negative. Indicate the possible phenotypes from the mating of a woman with type O, Rh-negative, with a man with type A, Rh-positive. Making Connections. Two pea plants are crossbred. Using a Punnett square and probability analysis, you predict that of the offspring will be tall. However, less than grow to be tall. What other factors can affect phenotype? How much trust should be put on probability calculations? Reflecting 5. When solving genetic problems, do you check over your work before giving your final answer? How do you check by redoing the questions, working backwards, or some other method? 5 Chapter

6 .6 Activity.6. Genetics of Corn Corn is one of the world s most important food crops. It has been subject to selective breeding techniques and hybridization for many years, which have resulted in vigorous, high-yielding varieties. Nearly all corn grown today is hybrid corn. Some varieties of corn are chosen for their sweet flavour while the mixed coloration of the Indian corn varieties makes them popular decorations during the autumn months. In this activity, you will determine the probable genotypes of parents by examining the phenotypes of corn for two different and independently assorted traits. Materials dihybrid corn ears (sample A, sample B) Procedure. Obtain a sample A corn ear from your teacher (Figure 8). The kernels display two different traits whose genes are located on different chromosomes.. Describe the two different traits: colour and shape. Predict which phenotypes are dominant and which are recessive.. Assume that the ear of corn is from the F generation. The parents of the F corn were pure-breeding homozygous for each of the characteristics. Assign the letters P and p to the alleles for colour, and A and a to the alleles for shape. Use the symbols PPaa and ppaa for the parents of the F generation. Describe the phenotype of the PPaa parent and ppaa parent. The cross for the F generation is PPaa ppaa.. Count 00 of the kernels in sequence; describe the phenotypes and record the number of each in a table similar to Table. Figure 8 Sample A Table Phenotype Number Ratio dominant alleles for colour and shape dominant allele for colour, but recessive allele for shape recessive allele for colour, but dominant allele for shape recessive alleles for colour and shape 5. Obtain sample B. Assume that this ear was produced from a test cross. Count 00 kernels in sequence and record your results as in step. Analysis and Evaluation (a) What are the expected genotypes and phenotypes of the F generation resulting from a cross between the parents PPaa and ppaa? (b) Use a Punnett square to show the expected genotypes and the phenotypic ratio of the F generation. Compare your results with what you obtained in step. What factors might account for discrepancies? Would your results be any different if you took larger samples or took multiple samples and averaged the results? Genes and Heredity 55

7 (c) Assuming that sample B was produced from a test cross (i.e., B is from the F generation), what is the phenotypic ratio of the F generation? (d) What is the phenotype of the unknown parent? Try This Activity Virtual Fruit Fly Simulation GO TO Follow the links for Nelson Biology,.6 to begin the activity. Test your understanding of Mendelian genetics by performing crosses in the Virtual Fruit Fly lab. You will be able to select traits that demonstrate the concepts of dominant and recessive alleles, segregation, and independent assortment. Create a classroom list of all possible crosses. Working in groups of or, select at least crosses to examine. Write a report of the crosses you tested and the results. State the genotypes and phenotypes of the parents and F and F offspring. Indicate the number of each type of offspring produced in each generation. Record the information on the classroom list. As a class, analyze the data. Compare the number of flies for eye colour and for the other traits that were examined. Section.6 Questions Understanding Concepts. Why are test crosses important to plant breeders?. A dihybrid cross can produce 6 combinations of alleles, 9 of which are different. Explain why 00 seeds were counted rather than only 6 or 9. Applying Inquiry Skills. A dominant allele Su, called starchy, produces smooth kernels of corn. The recessive allele su, called sweet, produces wrinkled kernels of corn. The dominant allele P produces purple kernels, while the recessive allele p produces yellow kernels. A corn plant with starchy, yellow kernels is cross-pollinated with a corn plant with sweet, purple kernels. One hundred kernels from the hybrid are counted, and the following results are obtained: 5 starchy, yellow kernels and 8 starchy, purple kernels. What are the genotypes of the parents and of the F generation? Making Connections. Thousands of years ago, the ancestor of corn grew only in Mexico. Scientists have used technology and selective breeding methods to develop varieties of corn that can grow in a wide range of environmental conditions. As a result, corn is now grown in many places where it would not occur by nature. What are some risks associated with growing a species in a foreign environment? Start your research on the Internet. Follow the links for Nelson Biology,.6. GO TO 56 Chapter