Concept explainers
Alleles of the IGF-1 gene in dogs, encoding insulin-like growth factor, largely determine whether a domestic dog will be large or small. Dogs with an ancestral domi-nant allele are large, whereas dogs homozygous for the mutant recessive allele are small. Chondrodysplasia, a short-legged phenotype (as in dachshunds and basset hounds), is caused by a dominant gain-of-function allele of the FGF4 gene. The MSTN gene encodes myostatin, a regulator of muscle development. Dogs with a dominant ancestral allele of the MTSN gene have normal muscle development, while dogs homozygous for recessive mutants in the MTSN gene are “double muscled” and have trouble running quickly. However, dogs heterozy-gous for the mutant allele run faster than either of the homozygotes. You breed a pure-breeding small basset hound ofnormal musculature with a pure-breeding “bully” whip-pet, a double-muscled large dog with normal legs.
a. What are the genotypes and
b. If the
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- In the fish Species a, the expression of different combinations of four Hox genes results in structural differences in tissues and organs along the anterior-posterior axis. Figure 3 shows the expression pattern of four genes, i.e. Genes 1 – 4, and the tissue produced along the embryonic anterior - posterior plane of fish Species a wild type. The influence of Hox genes on the formation of specific types of tissue can be determined by introducing gain-of-function and loss-of-function mutations in specific Hox genes. Looking at Figure 3, predict what would happen to each one of the four types of tissues, if the expression of gene 3 was expanded anteriorly by a gain-of-function mutation so that the pattern of gene 3 expression would be the same as that of gene 2.arrow_forwardThere are two genes that determine the coat colour expression in some dogs: eumelanin and merle. These genes are located on two separate chromosomes. For the eumelanin gene, black coat colour (E) is dominant over red coat colour (e). The merle gene controls the degree to which these coat colours are expressed through incomplete dominance. The following table describes the merle gene expression. Genotype Phenotype MM White Mm Half colour (Grey or light red) mm Full colour (Black or Red) A grey dog that is heterozygous for the eumelanin gene mates with a light red dog. 1. What is the probability they would have a grey puppy? 2. What is the probability they would have a grey or light red puppy?arrow_forwardSuppose in a hospital in the United Kingdom, out of 98,543 of babies born in the last 20 years, 13 children were born with myotonic dystrophy. Myotonic dystrophy is caused by an autosomal dominant mutation in the dystrophia myotonica protein kinase gene, DMPK. Myotonic dystrophy causes progressive muscle weakness, cataracts, and difficulty relaxing clenched hands. Of the 13 affected children, only 3 were born to a parent with a DMPK mutation. Given this information, what is the mutation rate of the DMPK gene?arrow_forward
- people with osteogenesis imperfecta have a dominant mutation in one of the two genes that produce type 1 collagen. people with OI have weak bones, bkuish color in teh whites of eyes, and a variety of afflictions that cause weakness in their joint and teeth. However, some people can carry the mutation but have no symptoms. Thus, families can unknowingly transmit the mutation but does not express the OI phenotype. This is an example of which of the following? a. incomplete penetrance b. variable expressivity c. epistasis d. incomplete dominancearrow_forwardOne of the two genes known to be mutated in cases of Hypokalemic periodic paralysis (which is inherited in an autosomal dominant pattern but known to affect males more often than females) is the calcium voltage-gated channel subunit alpha1 S (CACNA1S). What is known about the gene is recorded here: https://www.ensembl.org/Homo_sapiens/Gene/Summary?db=core;g=ENSG00000081248;r=1:201039512-201112451 Please navigate to the link above and use the information and link-outs from the page to answer the following question. What is the NCBI accession number (including the version) of the RefSeq Match for the first transcript (CACNA1S-201)?arrow_forwardOne of the two genes known to be mutated in cases of Hypokalemic periodic paralysis (which is inherited in an autosomal dominant pattern but known to affect males more often than females) is the calcium voltage-gated channel subunit alpha1 S (CACNA1S). What is known about the gene is recorded here: https://www.ensembl.org/Homo_sapiens/Gene/Summary?db=core;g=ENSG00000081248;r=1:201039512-201112451 Please navigate to the link above and use the information and link-outs from the page to answer the following questions ANSWER ONLY IN UPPERCASE LETTERS, NO UNITS: Using the left-hand menu to view the sequence for CACNA1S, what are the last three nucleic acid bases of exon 1?arrow_forward
- One of the two genes known to be mutated in cases of Hypokalemic periodic paralysis (which is inherited in an autosomal dominant pattern but known to affect males more often than females) is the calcium voltage-gated channel subunit alpha1 S (CACNA1S). What is known about the gene is recorded here: https://www.ensembl.org/Homo_sapiens/Gene/Summary?db=core;g=ENSG00000081248;r=1:201039512-201112451 Please navigate to the link above and use the information and link-outs from the page to answer the following question. GIVE YOUR ANSWER AS A NUMBER ONLY, NO UNITS: What is the size in amino acid residues of the CACNA1S transcript named CACNA1S-202? Answer: The size of the CACNA1S transcript named CACNA1S-202 is how many amino acid residues.arrow_forwardOne of the two genes known to be mutated in cases of Hypokalemic periodic paralysis (which is inherited in an autosomal dominant pattern but known to affect males more often than females) is the calcium voltage-gated channel subunit alpha1 S (CACNA1S). What is known about the gene is recorded here: https://www.ensembl.org/Homo_sapiens/Gene/Summary?db=core;g=ENSG00000081248;r=1:201039512-201112451 Please navigate to the link above and ensure that you click to reveal the transcript table. Then use the information in the table to answer the following question. PLEASE GIVE YOUR ANSWER AS A NUMBER ONLY, NO UNITS What is the size in base pairs of the CACNA1S transcript named CACNA1S-202? Answer: The size of the CACNA1S transcript named CACNA1S-202 isarrow_forwardIn squirrels, individuals that are heterozygous for the mutant LDL receptor gene( Fa) begin to experience heart attacks at the squirrel equivalent of the 30’s and 40’s in humans, while individuals that are homozygous for the mutant LDL receptor allele experience heart attacks much earlier. Closer examination reveals that the hepatocytes of the homozygous normal squirrels contain ONLY normal receptors. In the hepatocytes of the heterozygous squirrels, 50% of the receptors are of the mutant type, and fail to bind the LDL cholesterol, while the other 50% are normal. In individuals homozygous for the mutant LDL receptor allele, only mutant receptors are present. NAME and DEFINE the genetic phenomenon observed here.arrow_forward
- Human females who are heterozygous for an X-linked recessive allele sometimes exhibit mild expression of the trait. However, such mild expression of X-linked traits in females who are heterozygous for Xlinked alleles is not seen in Drosophila. What might cause this difference in the expression of X-linked genes between human females and female Drosophila? (Hint: In Drosophila, dosage compensation is accomplished by doubling the activity of genes on the X chromosome of males.)arrow_forwardDescribe modifying epistasis. Modifying epistasis can be observed in the coat color of Dobermans. The doberman has two color genes: a black(B) and a color-dilution(D) gene. These two genes can create four different colors combinations in coat. Among the two color genes in the Doberman dogs, which one is the epistatic gene, and which one is the hypostatic gene. Why?arrow_forwardAt least two pairs of genes control eye color. Both pairs influence the production of the pigment, melanin, but act independently. One pair of alleles is B (Brown color; dominant) and b (blue color); the other pair is A (pigment production; dominant) and a (no pigment production; albino). The gene pair aais epistatic to (masks) B and b and produces the nonpigmented eyes of the albino. What is the typeof gene interaction that exists between the two gene pairs? Give the genotypes and phenotypes of the possible offspring of the mating bbAa x Bbaa andi ndicate the parental phenotypes. (any method)arrow_forward
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