Human Heredity: Principles and Issues (MindTap Course List)
11th Edition
ISBN: 9781305251052
Author: Michael Cummings
Publisher: Cengage Learning
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Chapter 15, Problem 2QP
Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters’ sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?
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Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome . To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness , their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters’ sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia 24 have neither of the traits , 1 has color blindness only and 1 has hemophilia only . how many centimorgans seperate the hemophilia locus from the locus of the color blindness.
Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromo-some with two mutant alleles), and the daughters’ sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?
If the genes are not assorting independently, what is the recombinationfrequency between them?
Chapter 15 Solutions
Human Heredity: Principles and Issues (MindTap Course List)
Ch. 15.1 - Who Owns Your Genome? John Moore, an engineer...Ch. 15.1 - Who Owns Your Genome? John Moore, an engineer...Ch. 15 - James sees an online ad for an at-home genetic...Ch. 15 - James sees an online ad for an at-home genetic...Ch. 15 - James sees an online ad for an at-home genetic...Ch. 15 - James sees an online ad for an at-home genetic...Ch. 15 - The gene controlling ABO blood type and the gene...Ch. 15 - Hemophilia and color blindness are both recessive...Ch. 15 - Prob. 3QPCh. 15 - Prob. 4QP
Ch. 15 - How many nucleotides does the human genome...Ch. 15 - Which of the following best describes the process...Ch. 15 - Which of the following is NOT an activity carried...Ch. 15 - Prob. 8QPCh. 15 - Prob. 9QPCh. 15 - What percentage of the DNA in the genome actually...Ch. 15 - When the human genome sequence was finally...Ch. 15 - One unexpected result of the sequencing of the...Ch. 15 - Prob. 13QPCh. 15 - Prob. 14QPCh. 15 - Prob. 15QPCh. 15 - Prob. 16QPCh. 15 - Prob. 17QP
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- In the mapping example in Fig 2, the dominant alleles were on one chromosome and the recessive alleles were on the homolog. Let’s consider a twofactor cross in which the dominant allele for one gene is on onechromosome, but the dominant allele for a second gene is on thehomolog. A cross is made between AAbb and aaBB parents. The F1offspring are AaBb. The F1 heterozygotes are then testcrossed to aabbindividuals. What topic in genetics does this question address?arrow_forwardIn the mapping example in Fig 2, the dominant alleles were on one chromosome and the recessive alleles were on the homolog. Let’s consider a twofactor cross in which the dominant allele for one gene is on onechromosome, but the dominant allele for a second gene is on thehomolog. A cross is made between AAbb and aaBB parents. The F1offspring are AaBb. The F1 heterozygotes are then testcrossed to aabbindividuals. Which F2 offspring are recombinant?arrow_forwardthree recessive genes a, b, and c in the model plant Arabidopsis are found to be linked on chromosome 4. A three point test cross is done with a homozygous recessive plant with a heterozygous for all three genes. Following is the number of progenies a b C 65 A B c 56 A B C 1267 a b c 1310 A b C 550 a B c 515 a B C 470 A b c 489 Total = 4,722 Determine the middle locus by your choice of method and after that calculate the map distance between the genes in map unit (m.u.).arrow_forward
- Shown below is a pedigree for a completely penetrant trait called Adams syndrome in which babies are born blind. This trait occurs when an allele of the adams gene is associated with ≥200 tandem trinucleotide repeats (the normal number is 10). First cousins, III-1 and III-2 married and their first child (IV-1) was blind. For their next child, they decided to do in vitro fertilization with III-1's sperm and III-2's eggs to generate six embryos (labeled E1-6). When each embryo contained eight cells, a single cell was removed and genomic DNA was isolated. PCR reactions using primers that flank the trinucleotide repeat region were then performed and the resulting fragments were fractionated on an agarose gel. PCR reactions using genomic DNA from III-1, III-2 and IV-1 were included as controls. The DNA was visualized using a fluorescent dye and the gel is shown below. Based on this information, select the best answer from the list to the questions below. || E1 = embryo 1 IV E1 E2 E3 E4 E5…arrow_forwardIn the pedigree below, male II-1 has Klinefelter syndrome, which is the result of an XXY karyotype. On the X chromosome, a gene called G6PD has two codominant alleles, G6PDA and G6PDB. In this pedigree, A, B, and AB refer to the phenotypes associated with the alleles of this gene. (Note: In this family, no individuals have the AB version of the phenotype.) A A B Based on the information in the pedigree, when could nondisjunction have occurred? Select all correct answers. In Il-1's father, during meiosis I In II-1's mother, during meiosis I In II-1's mother, during meiosis II In Il-1's father, during meiosis IIarrow_forwardThe pedigree below shows three generations of a family that carries albinism, an autosomal recessive genetic disease. In the third generation, a child was born with albinism but the genotypes of the rest of the family are unknown. No other family members have the disease. Assume normal, Mendelian genetics with no new mutations. What are the genotypes of the parents of the affected child? A) There is not enough information to determine their genotypes B) Both are homozygous for albinism C) One is hemizygous and one is heterozygous for albinism D) Both are heterozygous for albinism E) One is homozygous and one is heterozygous for abinismarrow_forward
- With three-point genetic mapping, we can look at the inheritance of three linked genes and determine their order and map distance relative to each other on the chromosome. Suppose Gene G, Gene J, and Gene M are linked. An organism with the genotype GgJjMm was mated to an organism with the genotype ggjmm. The following phenotypes were seen in the offspring: Dominant for all three 13 Dominant for G and J Dominant for G and M 84 Dominant for J and M 389 Dominant for G only 401 Dominant for J only 96 Dominant for M only Recessive for all three 12 a. What the alleles in the parental gametes? b. What are the alleles in the double crossover gametes? c. What gene is in the middle of the three? d. What is the map distance between Gene G and Gene J? (Show all your work.) е. What is the map distance between Gene J and Gene M? (Show all your work.)arrow_forwardTwo pure-breeding parents produced a heterozygous female offspring (AaBb) that was then testcrossed with an aabb male. The offspring produced from the testcross included 50 AaBb, 450 Aabb, 450 aaBb, 50 aabb individuals. Describe how you can tell if these two genes are linked or unlinked (What ratio would you expect to see from the testcross if they were not linked?). What were the genotypes of the original parents that produced the heterozygous female? What is the genetic map distance between the two genes?arrow_forwardIn D. melanogaster, scarlet eyes (st), curled wings (cu), and spineless bristles (ss) are recessive traits located on chromosome III in that order. The distance between st and cu is 10 cM and the distance between cu and ss is 12 cM. The CoC for this chromosome is 1. A true-breeding female expressing scarlet eyes and spineless bristles is crossed to a true-breeding male expressing curled wings. The F1 are wild type. An F1 female is then test crossed to a male expressing all three traits. Among the F2 progeny, how often would you expect to observe flies expressing only scarlet eyes? Using the same information from above, among the F2 progeny, how often would you expect to observe flies expressing only spineless bristles? Using the same information from above, among the F2 progeny, how often would you expect to observe flies expressing only curled wings?arrow_forward
- A researcher hypothesizes that, in mice, two autosomal dominant traits, trait Q and trait R, are determined by separate genes found on the same chromosome. The researcher crosses mice that are heterozygous for both traits and counts the number of offspring with each combination of phenotypes. The total number of offspring produced was 64. The researcher plans to do a chi-square analysis of the data and calculates the expected number of mice with each combination of phenotypes. Which of the following is the expected number of offspring that will display both trait Q and trait R? a.4 b.12 c.36 d.48arrow_forwardSemi-sterility in corn, as seen by unfilled ears with gaps due to abortion of approximately half the ovules, is an indication that the strain is a translocation heterozygote. The chromosomes involved in the translocation can be identified by crossing the translocation heterozygote to a strain homozygous recessive for a gene on the chromosome being tested. The ratio of phenotypic classes produced from crossing semi-sterile F1 progeny back to a homozygous recessive plant indicates whether the gene is on one of chromosomes involved in the translocation. For example, a semi-sterile strain could be crossed to a strain homozygous for the yg mutation on chromosome 9. (The mutant has yellow-green leaves instead of the wild-type green leaves.) The semi-sterile F1 progeny would then be backcrossed to the homozygous yg mutant. (a) What types of progeny (fertile or semi-sterile, green or yellow-green) would you predict from the backcross of the F1 to the homozygous yg mutant if the gene was not on…arrow_forwardYou are studying a gene that controls ossicone (horn) length in giraffes. The wild type long-ossicone allele (L) is dominant to the mutant short-ossicone (I) allele. However the L allele is only 60% penetrant. You cross a long-horn heterozygous giraffe to a short- horn giraffe. What phenotypic progeny classes do you expect and at what frequencies?arrow_forward
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