Concept explainers
Canavan disease, caused by homozygosity for a recessive allele, is a severe neurodegenerative syndrome usually resulting in death by the age of 18 months. The frequency of Canavan disease is particularly high in Jewish populations. In an effort to map the gene causing this condition, researchers looked at 10 SNPs (1–10) spaced at roughly 100 kb distances along chromosome 17 in five affected Jewish patients (Cases) and four unaffected Jewish individuals (Controls). In the accompanying table, each row depicts a single haplotype. (Every individual is diploid and therefore has two haplotypes, although only one is shown in the table.) G, C, A, and T represent the actual
Case | SNP1 | SNP2 | SNP3 | SNP4 | SNP5 | SNP6 | SNP7 | SNP8 | SNP9 | SNP10 |
1 | G | T | G | T | T | T | C | A | G | T |
2 | A | T | G | T | T | T | C | A | G | T |
3 | G | T | G | T | T | T | C | A | G | C |
4 | A | A | G | T | T | T | C | T | C | C |
5 | G | A | G | C | C | T | G | A | C | C |
Control | ||||||||||
6 | A | A | G | T | T | T | C | A | G | T |
7 | G | T | G | G | C | T | G | A | G | T |
8 | A | T | C | T | C | G | C | T | C | C |
9 | G | T | C | G | T | G | G | A | C | T |
a. | Does the disease-causing mutation appear to be in linkage disequilibrium with any of the SNP alleles? If so, which ones? |
b. | Where is the most likely location for the Canavan disease gene? About how long is the region to which you can ascribe the gene? |
c. | How many independent mutations of the Canavan gene are suggested by these data? |
d. | Suppose that individuals 2–9 are Ashkenazic (whose ancestors lived in the Rhine river basin of Germany and France after the Jews were expelled from Judea in 70 A.D.) while individual 1 is Sephardic (a non-Ashkenazic Jew). Would these facts provide any information about the history of the mutations causing Canavan disease? |
e. | For mapping genes by haplotype association, why is it often helpful to focus on certain subpopulations? Does this strategy have any disadvantages? |
f. | Human chromosome 17 is an autosome, so each person has two copies of each region along the chromosome. With this in mind, explain the practical difficulty in determining haplotypes. (Hint: Consider heterozygosity.) In light of this difficulty, how could the researchers determine any individual haplotype, such as any of those shown in the table? |
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ND STONY BROOK UNIVERSITY LOOSELEAF GENETICS: FROM GENES TO GENOMES
- Using figure 1 and the following background information answer the following questions. Identification of the genetic cause of hornlessness in cattle has been the subject of intensive genetic and genomic research, culminating in the nomination of two different candidate neomutations on cattle chromosome 1 that are predicted to have arisen 500-1,000 years ago: a complex allele of Friesian origin (PF), an 80,128 base pair (bp) duplication (1909352–1989480 bp), and a second, simple allele of Celtic origin (PC) corresponding to a duplication of 212 bp (chromosome 1 positions 1705834–1706045) in place of a 10-bp deletion (1706051–1706060)We report the use of genome editing using transcription activator-like effector nucleases (TALENs) to introgress the putative PC POLLED allele into the genome of bovine embryo fibroblasts to try and produce a genotype identical to what is achievable using natural mating, but without the attendant genetic drag and admixture. In our previous studies, we…arrow_forwardThe wild-type (normal) fruit fly, Drosophila melanogaster, has straight wings and long bristles. Mutant strains have been isolated that have either curled wings or short bristles. The genes representing these two mutant traits are located on separate chromosomes. Carefully examine the data from the following five crosses shown below (running across both columns). (a) Identify each mutation as either dominant or recessive. In each case, indicate which crosses support your answer. (b) Assign gene symbols and, for each cross, determine the genotypes of the parents.arrow_forwardOrangutans are an endangered species in their natural environment (the islands of Borneo and Sumatra), and so a captive-breeding program has been established using orangutans currently held in zoos throughout the world. One component of this program is research into orangutan cytogenetics. This research has shown that all orangutans from Borneo carry one form of chromosome 2, as shown in the accompanying diagram, and all orangutans from Sumatra carry the other form. Before this cytogenetic difference became known, some matings were carried out between animals from different islands, and 14 hybrid progeny are now being raised in captivitya. What term or terms describe the differences between these chromosomes? b. Draw the chromosomes 2, paired in the first meiotic prophase, of such a hybrid orangutan. Be sure to show all the landmarks indicated in the accompanying diagram, and label all parts of your drawing. c. In 30 percent of meioses, there will be a crossover somewhere in the region…arrow_forward
- Orangutans are an endangered species in their natural environment (the islands of Borneo and Sumatra), and so a captive-breeding program has been established using orangutans currently held in zoos throughout the world. One component of this program is research into orangutan cytogenetics. This research has shown that all orangutans from Borneo carry one form of chromosome 2, as shown in the accompanying diagram, and all orangutans from Sumatra carry the other form. Before this cytogenetic difference became known, some matings were carried out between animals from different islands, and 14 hybrid progeny are now being raised in captivity.q3.1 q2.2 q2.1 q1.2 q1.1Borneop1.1 p1.2cenq3.1 q2.2 q2.1 p1.1Sumatrap1.2cenq1.1 q1.2a. What term or terms describe the differences between these chromosomes?b. Draw the chromosomes 2, paired in the first meiotic prophase, of such a hybrid orangutan. Be sure to show all the landmarks indicated in the accompanying diagram, and label all parts of your…arrow_forwardUsing the figure and the following background information answer the following questions about the figure Background: Identification of the genetic cause of hornlessness in cattle has been the subject of intensive genetic and genomic research, culminating in the nomination of two different candidate neomutations on cattle chromosome 1 that are predicted to have arisen 500-1,000 years ago: a complex allele of Friesian origin (PF), an 80,128 base pair (bp) duplication (1909352-1989480 bp), and a second, simple allele of Celtic origin (PC) corresponding to a duplication of 212 bp (chromosome 1 positions 1705834-1706045) in place of a 10-bp deletion (1706051-1706060)We report the use of genome editing using transcription activator-like effector nucleases (TALENs) to introgress the putative PC POLLED allele into the genome of bovine embryo fibroblasts to try and produce a genotype identical to what is achievable using natural mating, but without the attendant genetic drag and admixture. In…arrow_forwardSusan’s grandfather was deaf, and passed down a hereditary form of deafness within Susan’s family as shown in Figure Q19–12.A. Is this mutation most likely to be dominant or recessive?B. Is it carried on an autosome or a sex chromosome? Why?C. A complete SNP analysis has been done for all of the 11 grandchildren (4 affected, and 7 unaffected). In comparing these 11 SNP results, how long a haplotype block would you expect to find around the critical gene? How might you detect it?arrow_forward
- An STR on chromosome 5 is very closely linked to a gene involved in a dominant metabolism disorder (one one copy of the mutation causes the disease). There are four different alleles of the STR (Q, B, S, and L), each with different numbers of GATCTCG repeats. The S allele has 2 repeats, the B allele has 7 repeats, the Q allele has 10 repeats, and the L allele has 5 repeats. Part 1 Lane 1 of the gel shown below indicates the locations of PCR products corresponding to all four STR alleles when run together in one lane. DNA was loaded into the wells at the top of the gel. Move the letter of each allele to its correct position at the left of Lane 1 to indicate where each allele runs on the gel. 0000 Gabriel Andre Lane 1 O Maria Andre Maria Juliana B L Part 2 Andre and Maria are the parents in this family; all others shown on the gel are children. Daughter Juliana is born with the metabolism disorder, but daughter Camila does not have the disorder. Assuming no recombination between the STR…arrow_forwardIn 1983, a sheep farmer in Oklahoma noticed in his flock a ram that possessed increased muscle mass in his hindquarters. Many of the offspring of this ram possessed the same trait, which became known as the callipyge phenotype (callipyge is Greek for “beautiful buttocks”). The mutation that caused the callipyge phenotype was eventually mapped to a position on the sheep chromosome 18. When the male callipyge offspring of the original mutant ram were crossed with normal females, they produced the following progeny: 14 male callipyge, 14 female callipyge, 14 male normal, and 14 female normal. When the female callipyge offspring of the original mutant ram were crossed with normal males, all of the offspring were normal. Analysis of the chromosomes of these offspring of callipyge females showed that half of them received a chromosome 18 with the allele encoding callipyge from their mother. Propose an explanation for the inheritance of the allele for callipyge. How might you test your…arrow_forwardConcordance studies of twins for a neurodegenerative disorder show MZ= 46% and DZ= 15%. Further studies have shown a possible link to a gene on chromosome 9, however, there are some individuals in the study who have the allele but do not develop the disorder (group 1), and there are other individuals who do not have the allele yet develop the disorder (group 2). Amita's older sister and maternal uncle have this disorder. Currently, Amita & her 2 younger brothers do not show symptoms. Amita's paternal grandfather was rumored to have this disorder. 1. Draw the pedigree for Amita's family and determine the mode of inheritance if any. 2. Explain how the 2 groups in the study could be possible? 3. What would you tell Amita about the heritability of this disorder?arrow_forward
- A small-scale pedigree study involving 15 families with a total of 50 children was performed to determine whether a SNP in a candidate gene is linked to a disease. The result showed that among 22 children who developed the disease, 19 belonged to non-recombinants while 3 recombinants. For the remaining 28 children, 21 were non-recombinants and 7 belonged to recombinants. A Lod (Z) score can be calculated and used to determine whether this SNP is linked to the disease. By using a very stringent threshold, which of the following conclusions is correct? O A.Z is approximately 4.03, indicating a linkage O B. Z=0.25, indicating no linkage OCZ=0.2, indicating no linkage O D.Z is approximately 4.19, indicating a linkagearrow_forwardA woman has her personal genome analyzed for the BRCA1 mutation after learning that her father is heterozygous and carries one mutant allele. What is her chance of inheriting the mutant allele from her father? 0: men cannot transmit genes affecting breast cancer. 25% 50% 75% 100% Among the progeny of a heterozygous round (Aa) x homozygous wrinkled ( aa) cross, three seeds are chosen at random. What is the probability that all three seeds are round? (1/4)3 1/4 (1/2) 3 1/2 A single gene can produce different proteins. True Falsearrow_forwardWhen a female melanotic fly is crossed with a normal male, the progeny are produced: 123 normal females, 125 melanotic females, and 124 normal males. In subsequent crosses between melanotic females and normal males, melanotic females are frequently obtained, but never any melanotic males. Provide a possible explanation for the inhertiacne of the melanotic mutation (Hint: The cross produces twice as many female progeny as male progeny)arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning