Chapter 11 Genetics 1 Assignment

Mendel studied pea plants dihybrid for seed shape (round versus wrinkled) and seed color (yellow versus green). Recall that the round allele (R) is dominant to the wrinkled allele (r) and the yellow allele (Y) is dominant to the green allele (y). The table below shows the F1 progeny that result from selfing four different parent pea plants. Use the phenotypes of the F1 progeny to deduce the genotype and phenotype of each parent plant. Complete the table by dragging the correct label to the appropriate location. Labels can be used once, more than once, or not at all.

Parent pheno: g,y,y,wg Parent geno:Rryy,RrYy,RRYy,rry F1 progeny phenotypes from selfed parent: The ability to deduce an organism's genotype from the phenotype(s) of its progeny is an important skill in solving genetics problems. In this example, the logic was simplified because the parent plants were selfed, and therefore only one parental genotype was involved.

A plant grown from a [round, yellow] seed is crossed with a plant grown from a [wrinkled, yellow] seed. This cross produces four progeny types in the F1: [round, yellow], [wrinkled, yellow], [round, green], and [wrinkled, green]. Use this information to deduce the genotypes of the parent plants. Indicate the genotypes by dragging the correct label to the appropriate location.

RrYy, rrYy Answering this question requires two logical steps: First, eliminate those genotypes that are inconsistent with the phenotypes of the parents. Then examine the remaining possibilities: Which are consistent with the phenotypes of the progeny? Because the cross produces green progeny (and both parents are yellow), both parents must be Yy: If either parent were YY, only yellow progeny would result. The cross also produces wrinkled progeny, so the [round, yellow] parent must be heterozygous (Rr): If it were RR, all of the progeny would be Rr and have a round phenotype

Diagramming a cross using a Punnett square Punnett squares can be used to predict the two possible outcomes of the botanist's test cross. The Punnett square on the left shows the predicted result if the unknown plant is homozygous (GG); the Punnett square on the right shows the predicted result if the unknown plant is heterozygous (Gg).

The genotypes in a Punnett square show all the possible combinations of alleles in offspring that could result from the particular cross. A Punnett square reveals the expected probabilities of each genotype among the offspring. For example, the Punnett square on the right reveals that there is a 50% chance that each offspring will have green pods and a 50% chance that each offspring will have yellow pods.

Based on the inheritance pattern, which mode of inheritance must be the cause of galactosemia? -autosomal dominant -autosomal recessive -sex-linked dominant -sex-linked recessive

autosomal recessive Because neither Jane nor John has the same condition as their daughter, and there is no evidence of sex-linkage, galactosemia must be an autosomal recessive trait.

Suppose that the botanist carried out the test cross described in Parts A and B and determined that the original green-pod plant was heterozygous (Gg). Which of Mendel's findings does her test cross illustrate? Suppose that the botanist carried out the test cross described in Parts A and B and determined that the original green-pod plant was heterozygous (Gg). Which of Mendel's findings does her test cross illustrate? -chromosome theory of inheritance -law of segregation -linkage -law of independent assortment

law of segregation The law of segregation states that the two alleles for a gene separate during gamete formation, and end up in different gametes. In the case of the heterozygous green-pod plant (Gg), one gamete will receive the dominant allele (G), and the other gamete will receive the recessive allele (g). The law of segregation accounts for the prediction that 50% of the offspring of the test cross will have green pods and 50% will have yellow pods

During which part of meiosis (meiosis I or meiosis II) do the two alleles of a gene separate? During which phase does the separation occur?

meiosis I, anaphase Alleles separate from one another during anaphase of meiosis I, when the homologous pairs of chromosomes separate.

How could the botanist best determine whether the genotype of the green-pod plant is homozygous or heterozygous? -Self-pollinate the green-pod plant. -Cross the green-pod plant with another green-pod plant. -Cross the green-pod plant with a yellow-pod plant.

-Cross the green-pod plant with a yellow-pod plant. A cross between a plant of unknown genotype and one that is known to be homozygous recessive is called a test cross because the recessive homozygote tests whether there are any recessive alleles in the unknown. Because the recessive homozygote will contribute an allele for the recessive characteristic to each offspring, the second allele (from the unknown genotype) will determine the offspring's phenotype.

For the cross in Part B, predict the frequencies of each of the phenotypes in the F1 progeny, and determine the genotype(s) present in each phenotypic class. Complete the diagram by dragging the correct label to the appropriate location. Labels can be used once, more than once, or not at all.

3/8 - RrYY, RrYy (x2) 3/8 - rrYY, rrYy (x2) 1/8 - Rryy 1/8 - rryy The frequencies of the four phenotypic classes are determined by applying the multiplication rule. The green progeny, for example, make up ¼ of the total progeny because the probability of their formation depends on receiving a y allele from both parents (probability = ½ x ½ = ¼). Similarly, the wrinkled progeny make up ½ of the total progeny because the probability of their formation depends on receiving an r allele from both parents (probability = 1 x ½ = ½). Because the [wrinkled, green] progeny are simultaneously green and wrinkled, and these events are independent, apply the multiplication rule to obtain their frequency: ¼ x ½ = 1/8.