BICD 158 Problem Set 1

Provide a basic definition for the term active genetics

Copying of a genetic element from one chromosome to its homolog in response to a double-strand DNA break being generated in the homolog at the same genomic site that the active genetic element is inserted.

Two individuals A and B are heterozygous for a recessive loss-of-function pigmentation allele that is viable when homozygous. If one of them (A) carries an active genetic element active in all cells of the organism, how does its phenotype compare to that of the other (B), which carries a loss-of function allele in the gene that is inherited in a standard Mendelian fashion

Individual A will express the phenotype of the recessive loss-of-function pigmentation allele since all of the cells contain an active genetic element. This element will cause all of the wildtype alleles to be converted to the mutant recessive loss-of-function pigmentation allele, resulting in 100% mutant allele inheritance. However, individual B will only express 50% inheritance of allele for the mutant phenotype (recessive loss-of-function pigmentation allele) due to the mendelian inheritance. This means that individual B will express the wildtype phenotype.

A pigmentation gene W is responsible for a dominant wild-type red flower phenotype (+) in peas and a recessive mutant allele (w) results in an unpigmented white flower phenotype.

What are the phenotypes of the following genotypes:

+/+:

+/w:

w/+:

w/w:

Draw a Punnett square that depicts the genotypes and phenotypes of progeny from a cross of two heterozygous parents (e.g., +/w X +/w). What are the ratios of the various phenotypes? [WHITEBOARD]

+/+: red flower

+/w: red flower

w/+: red flower

w/w: unpigmented white flower

A cross of two heterozygous parents would result in phenotype ratios of 75% red flower progeny and 25% unpigmented white flower progeny (Ratio 3:1). (see Punnett square below)

A second pigmentation gene P in peas is also responsible for a dominant wild-type red flower phenotype (+) and a semi-dominant mutant allele (p) results in a pink flower phenotype. Homozygotes for the p allele have unpigmented white flowers.

What are the phenotypes of the following genotypes:

+/+:

+/p:

p/+:

p/p:

Draw a Punnett square that depicts the progeny from a cross of two heterozygous parents

(e.g., +/p X +/p). What are the ratios of the various phenotypes? [WHITEBOARD]

+/+: red flower

+/p: pink flower

p/+: pink flower

p/p: unpigmented white flower

A cross of two heterozygous parents will result in phenotypic ratios of 25% red flowers, 50% pink flowers, and 25% unpigmented white flowers (Ratio 1:2:1). (see Punnett square below).

The Curly gene (C) in fruit flies is essential for viability (i.e., homozygous loss-of- function results in death of the individual during early developmental stages). Heterozygous loss-of-function C mutants have a semi-dominant phenotype (curled-up wings).

What are the phenotypes of the following genotypes:

+/+:

+/C:

C/+:

C/C:

Draw a Punnett square that depicts the progeny from a cross of two heterozygous parents

(e.g., +/C X +/C). What are the ratios of the various phenotypes? [WHITEBOARD]

+/+: normal, non-curly wings phenotype

+/C: curly wings phenotype

C/+: curly wings phenotype

C/C: death of individual during early developmental stages

A cross of two heterozygous parents will result in phenotypic ratios of 25% normal wing phenotype (non-curly), 50% curly wing phenotype (semi-dominant phenotype), and 25% of progeny will die in early developmental stages (Ratio 1:2:1). (see Punnett square below).

The E gene in peas is involved in determining the wild-type (+) round shape of leaves. Recessive e mutants have elongated leaves. The E gene resides on a different chromosome (i.e., is unlinked) from the W gene involved in flower pigmentation in question 1. Draw a Punnett square depicting the various genotype and phenotype classes of progeny that would be generated from a cross of two doubly heterozygous parents carrying mutations in the W and E genes: (+/w; +/e X +/w; +/e). What are the ratios of the various phenotypes? [WHITEBOARD]

Crossing two doubly heterozygous parents carrying mutations in the W and E genes would result in a 9:3:3:1 ratio of the various phenotypes. From 16 total progeny, it is expected that 9 will have round leaves and red flowers, 3 will have round leaves and white flowers, 3 will have elongated leaves and red flowers, and 1 will have elongated leaves and white flowers. (see Punnett square below).

The H gene is involved in determining the wild-type (+) height of peas. Heterozygous semi-dominant h mutants are shorter that wild-type plants (short phenotype) while homozygous mutants have a tiny phenotype. The H gene resides on a different chromosome (i.e., is unlinked) from the p gene involved in flower pigmentation in question 2. Draw a Punnett square depicting the various genotype and phenotype classes of progeny that would be generated from a cross of two doubly heterozygous parents carrying mutations in the P and H genes: (+/p; +/h X +/p; +/h). What are the ratios of the various phenotypes? [WHITEBOARD]

The ratio of the various phenotypes generated from a cross of two doubly heterozygous parents carrying mutations in the P and H genes would be 1:2:1:2:4:2:1:2:1. From 16 total progeny, it is expected that 1 will have wildtype height and red flowers, 2 will have wildtype height and pink flowers, 1 will have wildtype height and unpigmented white flowers, 2 will be short with red flowers, 4 will be short with pink flowers, 2 will be short with unpigmented white flowers, 1 will be tiny with red flowers, 2 will be tiny with pink flowers, and 1 will be tiny with unpigmented white flowers. (see Punnett square below).

The B gene is involved in determining the flavor of wild-type (+) peas. Recessive loss- of-function b mutants have a bitter taste. Two doubly heterozygous parents carrying the b mutation and the w mutation (analyzed in question 3) are crossed to each other. 25% of the resulting progeny have white flowers and bitter leaves while 75% have wild- type red flowers and normal sweet taste. What genetic phenomenon can account for this inheritance pattern?

A genetic phenomenon that can account for this inheritance pattern could be that there is genetic linkage between the B gene and the W gene. It is likely that the bitter taste allele is more likely to be inherited with the white flower pigmentation allele and that the wildtype taste allele is more likely to be inherited with the wildtype red flower allele.

The V gene is involved in determining the wild-type (+) solid green color of leaves in peas. Recessive v mutants have a so-called variegated pattern in which leaves have random small spots of green on an otherwise white background. Two doubly heterozygous parents carrying both the v and the w alleles (w analyzed in question 3) are crossed to each other. 20% of the resulting progeny have white flowers and variegated leaves, just under 5% have white flowers but normal full green leaves, and 75% have wild-type red flowers and normal full green leaves. What two separate genetic phenomena could account for this unexpected inheritance pattern?

One potential genetic phenomenon that could account for this unexpected inheritance pattern could be that the WT red flower phenotype is genetically linked to the solid green leaf phenotype, meaning that any plant that has red flowers will have solid green leaves. However, a plant that has white flowers could possibly have solid green or variegated leaves. This accounts for the 75% plants that have red flowers and normal full-green leaves, and the other phenotypes making up the other 25%. Another potential explanation of this unexpected inheritance pattern could be explained by Barbara McClintock's discovery of transposons. Since it is inferred that red flowers and solid green leaves are genetically linked, it would be expected that the other 25% would have white flowers and variegated leaves. However, there is 5% of the progeny that have white flowers and normal green leaves. This could be due to the fact that the transposons flip the allele for the leaf pigmentation allele for the V gene in the white flowered plants, making it such that a white flowered plant can have variegated or solid green leaves.

The v allele of the V gene described in question 7 exerts its effects both in somatic cells of the leaf as well as in the germline. What are the manifestations (phenotypes) associated with those different activities?

If the v allele of the V gene exerts its effects in the somatic cells of the leaf, it will be expected that the phenotype will only be expressed in the somatic cell that exhibit that change. Whereas, if the v allele of the V gene exerts its effects in the germline cells of the leaf, then it is expected that all the cells in the leaf will express the same phenotype caused by this change. In other words, if the v allele is expressed in somatic cells, then it is expected that the leaves will show a variegated phenotype, whereas if the v allele is expressed in the germline cells, all the cells will exhibit the solid green phenotype.