Concept explainers
Feather color in parakeets is produced by the blending of pigments produced from two biosynthetic pathways shown below. Four independently assorting genes (A, B, C, and D) produce enzymes that catalyze separate steps of the pathways. For the questions below, use an uppercase letter to indicate a dominant allele producing full enzymatic activity and a lowercase letter to indicate a recessive allele producing no functional enzyme. Feather colors produced by mixing pigments are green (yellow + blue) and purple (red + blue). Red, yellow, and blue feathers result from production of one colored pigment, and white results from absence of pigment production.
a. What is the genotype of a pure-breeding purple parakeet strain?
b. What is the genotype of a pure-breeding yellow strain of parakeet?
c. If a pure-breeding blue strain of parakeet (aa BB CC DD) is crossed to one that is pure-breeding purple, predict the genotype(s) and phenotype(s) of the
d. If
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- Scale color in dragons is determined by two genes. The W gene is epistatic to the G gene. The W allele is dominant and allows the production of pigment, while w is recessive and blocks pigment formation, resulting in white dragons. When pigment formation is possible, the dominant G allele results in green dragon scales, while the recessive g allele results in black dragon scales. The pigment formation pathway is shown below W (Enzyme 1) G (Enzyme 2) White precursor Black intermediate pigment Green pigment а. What color scales would you expect in each of the following dragons? WwGg Wwgg WWGG wwgg b. If you cross two dragons with WwGg genotype, what is the expected phenotypic ratios of the progeny? c. What kind of epistasis do we observe?arrow_forwardLeber Congenital Amaurosis (LCA) causes progressive vision loss due to defects in the gene that encodes RPE65 isomerase. Affected individuals are homozygous recessive for mutant alleles of the RPE65 gene. You are trying to determine the molecular nature of the mutations in three individuals with LCA. For ease of analysis, you may assume that each individual is homozygous for the same mutant allele (though the three individuals have different mutations than each other). You use the polymerase chain reaction to amplify DNA from each patient and you determine the sequence of the DNA and compare it to unaffected individuals. You identify the following differences. Note that the non-template strand of DNA is given and the changes are highlighted using red boldface. You can assume that the sequences are in the first reading frame (eg. the first three nucleotides of each sequence is a codon). The coding region of the gene is 1602 bp and the position of the sequences shown below is…arrow_forwardMany aspects of gene function can be nicely explained with the one-gene-one-enzyme hypothesis, which states that a gene controls the production of an enzyme. Which of the following findings about gene expression, though, requires an expansion of this simple concept? Choose an answer below: Non-enzyme proteins are made from genes too. Some genes code for RNA molecules only. Enzymes composed of different polypeptides are coded for by more than one gene. a and c, but not b a, b, and carrow_forward
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- A complex biochemical pathway is shown below, along with the alleles that either promote or inhibit each step of the pathway leading to a phenotype. Gene A has alleles A and a, B has alleles B and b, and so forth. Genes B and C are duplicate dominant epistatic lethal as heterozygotes (i.e. Bb Cc are lethal). Genes D and E are duplicate dominant epistatic (i.e. dd eg = desired phenotype). If I were to cross AA Bb cc Dd Ee with aa BB Cc Dd e, (i) (ii) What proportion of all offspring don't show the phenotype? What proportion of offspring survive? Gene A Gene B B Gene D a Gene C Gene Earrow_forwardIn roses, the synthesis of red pigment is produced by two steps in a pathway. gene O magenta intermediate - gene P colorless intermediate- red pigment What would the phenotype be of a plant homozygous for a null mutation of gene P? What would the phenotype be of a plant homozygous for a null mutation of gene Q? What would the phenotype be of a plant homozygous for null mutations of genes P and Q? magenta red Match a genotype to each strain. colorless Strain P locus Q locus homozygous null mutation of gene P homozygous null mutation of gene Q homozygous null mutations of genes P and Q Answer Bank plp PIP What F2 ratio is expected from crossing a plant that is homozygous for a null mutation of gene P with a plant that is homozygous for a null mutation of gene Q? Assume independent assortment. 9 colorless : 4 magenta : 3 red 9 red : 4 colorless : 3 magenta O 9 red : 4 magenta : 3 colorlessarrow_forwardMany aspects of gene function can be nicely explained with the one- gene-one-enzyme hypothesis, which states that a gene controls the production of an enzyme. Which of the following findings about gene expression, though, requires an expansion of this simple concept? Non-enzyme proteins are made from genes too. Some genes code for RNA molecules only. Enzymes composed of different polypeptides are coded for by more than one gene. a and c, but not b a, b, and carrow_forward
- Gene A, which encodes alcohol dehydrogenase, is labeled with a red fluorophore in a FISH experiment with one chromosome of a homologous pair, with results shown below. Gene B is labeled in green and encodes Notch2NL, a gene which may help to explain brain size differences between humans and chimpanzees, and which with increased dosage, may lead to a condition referred to as macrocephaly (enlarged head circumference sometimes associated with learning disabilities). What can you infer about these two genes given the results seen below?arrow_forwardSuppose that you are studying the role of Protein B, which you believe plays a role in regulating PCD/Apoptosis in mice. You create two lines of mutant mice. One (bb) is homozygous for a loss-of-function allele of gene B. The other (Bb) is heterozygous, with one wild-type allele and one loss-of function allele. Initially you pay particular attention to two phenotypes of the resulting mice:(i) The morphology of their paws (see picture) (ii) The size of their brains & shape of their skulls. The bb mice have unusually large brains and unusual protrusions from their skulls. Based on these data, does it appear that Protein B, when present and active, favors or inhibits PCD/Apoptosis?Briefly explain your reasoning. The answer should address both the paw and brain/skull data.arrow_forwardSandhoff disease is due to a mutation in a gene that encodes a proteincalled hexosaminidase B. This disease has symptoms that aresimilar to those of Tay-Sachs disease. Weakness begins in the first 6 months of life. Individuals exhibit early blindness and progressive mental and motor deterioration. The family in the pedigree shown below has three members with Sandhoff disease, indicated with black symbols.arrow_forward
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