Concept explainers
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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Chapter 15 Solutions
Human Heredity: Principles and Issues (MindTap Course List)
- 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
- Genes A and B are on two different chromosomes. You construct a Punnett Square to to show the expected genotypes in the offspring of a cross between these two genotypes: AaBB x AaBb. What are the dimensions of the smallest Punnett square you can make to show the expected results? (e.g., 2x2, 4x1. Don’t worry about the order of the two numbers if they differ. That is, 4x8 is the same as 8x4)arrow_forwardShown 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_forwardThe following is a linkage map of chromosome 5 for three genes in tomato:Lf (normal) vs. lf (leafy)J (jointed) vs. j (jointless)W (non-wilty) vs. w (wilty)The cross between the triple heterozygote (Lf J W/ lf j w) and a triple homozygous recessive produced500 progeny. Assume that there is no interference in the Lf-W region. Give the expected number ofindividuals for each of the following progeny types and show complete solutions.a. with crossover in the Lf-J and J-W regionsb. with crossover in the Lf-J regionc. with crossover in the J-W regiond. without crossover in the Lf-W regionarrow_forward
- In 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_forward1-1 I-1I II II-1 II-2 П-3 П-4 II III-1 III-2 III-3 a) In humans, brown eyes are dominant and blue eyes are recessive. Determine which individuals in the pedigree above would have blue eyes. Justify your decision. b) Predict the phenotype of the possible offspring between III-1 and someone who has blue eyes.arrow_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
- In corn, the genes v (virescent seedlings), pr (red aleurone), and bm (brown midrib) are all on chromosome 5, but not necessarily in the order given. The cross: v+ pr bm/ v pr+ bm+ with v pr bm/ v pr bm produces 1000 progeny with the following phenotypes: v+ pr bm 226 v pr+ bm+ 229 v+ pr bm+ 153 v pr+ bm 185 v+ pr+ bm 59 v pr bm+ 71 v+ pr+ bm+ 36 v pr bm 41 What is the gene order, and the (b) genetic map of these three genes?arrow_forwardWith 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_forwardIn Drosophila flies, the allele b gives a black body, and the allele b+ gives brown, the wild-type phenotype. The allele wx of a separate gene gives waxy wings, and wx+ gives non-waxy. The allele cn of a third gene gives cinnabar eyes, and cn+ gives red. A female heterozygous for these three genes is testcrossed, and 745 progenies are produced which are phenotypically classified as follows: (see image) Make a linkage map of the three genes. Compute for interference and explain what the derived value means. Show complete solutions to support your answersarrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning