Genetics Problems – Following Two Traits at Once
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Genetics Problems – Following Two Traits at Once 1. In peas round seeds (R) are dominant to wrinkled seeds (r) and yellow seeds (Y) are dominant over green seeds (y). Predict the possible phenotypes of the offspring and the frequency of each when 2 RrYy parents are crossed. Variables have been assigned for you already and you know the genotypes of the parents, but before you can do a Punnett Square, you need to figure out the gametes that each parent can make. Each gamete must have ONE allele for each trait. This means that each gamete must have one letter for shape (it may be either a R or a r) and one letter for color (it may be either a Y or a y). This means there are 4 possible gametes for each parent: RY Ry Since this is the same for both parents, you will 4 possible RrYy rY need a 16 square Punnett Square. gametes ry RY Ry rY ry RY RRYY Round, yellow RRYy Round, yellow RrYY Round, yellow RrYy Round, yellow Ry RRYy Round, yellow RRyy Round, green RrYy Round, yellow Rryy Round, green rY RrYY Round, yellow RrYy Round, yellow rrYY rrYy Wrinkled, yellow Wrinkled, yellow ry RrYy Round, yellow Rryy Round, green rrYy Wrinkled, yellow Wrinkled, green Counting boxes for each phenotype gives: 9/16 Round, yellow 3/16 Round, green 3/16 Wrinkled, yellow 1/16 Wrinkled, green rryy 2. In rabbits, spotted coat (S) is dominant over solid coat (s) and black coat (B) is dominant over brown coat (b). Predict the phenotypes and their frequencies for the offspring of the following crosses: A) SSBb X Ssbb In this case, each parent can make only 2 different kinds of gametes: (Meiosis) SB Sb SB Sb Sb Sb Sb sb sb (cross out duplicates) SB sb SSBb SsBb Spotted, black Spotted, black This means you only need a 4 square Punnett Square: Sb The expected offspring would be: ¼ SSBb ½ Spotted black ¼ SsBb ¼ SSbb ½ Spotted brown ¼ Ssbb SSbb Ssbb Spotted, brown Spotted, brown B) SsBb X ssBb The first parent can make 4 different gametes, while the second can only make 2. (Meiosis) SB Sb sB sB sb sB sb sb sB sb Since the second parent can only make 2 different gametes, you can make an 8 box “Punnett Rectangle”: (cross out duplicates) SB SsBB Spotted, black SsBb Spotted, black Sb SsBb Spotted, black Ssbb Spotted, brown sB ssBB sb ssBb Solid, black ssBb Solid, black Solid, black ssbb Solid, brown The expected offspring would be: 3/8 Spotted black 3/8 Solid black 1/8 Spotted brown 1/8 Solid brown C) A brown solid colored female X a heterozygous black spotted male (heterozygous for both) First write the genotypes for both parents: Brown solid = bbss since both traits are recessive Heterozygous black spotted = BbSs Now you can determine the gametes each parent can make: bbss X BbSs (Meiosis) (cross out duplicates) SB Sb sB sb sb sb sb sb Since the second parent can only make 1 different gamete, you can make an 4 box “Punnett Rectangle”: sb Sb SB SsBb Spotted, black Ssbb Spotted, brown sB ssBb sb ssbb Solid, black Solid, brown The expected offspring would be: 1/4 Spotted black 1/4 Spotted brown 1/4 Solid black 1/4 Solid brown 3. In peas, green pods is dominant over yellow pod and tall plants are dominant over short plants. Predict the phenotypes and their frequencies for the offspring of the following crosses: A) 2 parents that are heterozygous for both traits (This is the classic dihybrid cross.) First assign variable: Tt Gg T = tall t = short 4 possible gametes TG TG Tg tG tg G = green pods g = yellow pods Since this is the same for both parents, you will need a 16 square Punnett Square. Tg tG tg TG TTGG Tall, green pod TTGg Tall, green pod Tt GG Tall, green pod Tt Gg Tall, green pod Tg RRYy Tall, green pod TTgg Tall, yellow pod Tt Gg Tall, green pod Tt gg Tall, yellow pod tG Tt GG Tall, green pod Tt Gg Tall, green pod GG Short, green pod Short, green pod tg Tt Gg Tall, green pod Tt gg Tall, yellow pod tt Gg Short, green pod Short, yellow pod tt Gg tt ttgg Counting boxes for each phenotype gives: 9/16 Tall, green pod 3/16 Tall, yellow pod 3/16 Short, green pod 1/16 Short, yellow pod B) A plant that is heterozygous for both traits X a heterozygous tall plant that produces yellow pods Using the variables assigned above, this cross can be summarized as: Tt Gg X Tt gg (Meiosis) TG Tg tG Tg Tg tg tg tg Tg tg (cross out duplicates) TG TTGg Tall, green pod Tt Gg Tall, green pod Since the second parent can only make 2 different gametes, you can make an 8 box “Punnett Rectangle”: Tg TTgg Tall, yellow pod Ttgg Tall, yellow pod tG tg Tt Gg Tall, green pod tt Gg Short, green pod Ttgg Tall, yellow pod ttgg Short, yellow pod Counting boxes for each phenotype gives: 3/8 Tall, green pod 3/8 Tall, yellow pod 1/8 Short, green pod 1/8 Short, yellow pod C) A plant that is homozygous dominant for both traits X a plant that is homozygous recessive for both traits Using the variables assigned above, this cross can be summarized as: TTGG X ttgg (Meiosis) TG TG TG TG tg tg tg tg (cross out duplicates) Since each parent can only make one kind of gamete, all of the offspring will be the same. All of the of spring will be Tt Gg. 4. List all the possible gametes for an individual whose genotype for 3 traits is JjKkLl. JKL JKl J kL Jkl jKL jKl jkL jkl (There will always be 2n possible gametes, where n is the number of traits.)
