Additional STR allele frequency information can be added to improve the analysis in Problem
E.8 Figure
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Genetic Analysis: An Integrated Approach (3rd Edition)
- 1. DO NOT ROUND OFF YOUR ANSWERS PREMATURELY. Use the unrounded and complete values when computing for the genetic gain; 2. Upon encoding, all final answers should be rounded-off to TWO decimal places. Examples: 3.5798 = 3.58, 0.0079 = 0.01, 0.00000008 = 0.00; 2. Report your answers as NUMERIC VALUES only. Please exclude the units; 3. Write ONLY one answer for each question. Avoid using "or", "/", "&" and "and". Provide only what is asked to you; 4. AVOID using unnecessary PUNCTUATIONS (. , ? 1; ; etc.), SYMBOLS ( []{ } etc.) and SPACES (before or after your answer). Please follow these instructions carefully. Prediction of Genetic Gain. If the milk yield of a herd of cow is 6,479 kg while those kept for breeding is 6,890 kg, and the h2 estimate of milk yield is 0.85. And the generation interval of dairy cattle is 4 years. What will be the expected genetic gain in the milk yield (kg/year) of the next generation?arrow_forwardPlease answer all parts along with the reason. I'll definitely give a like. Thank you in advance! 1A) From the cross Ab/aB x ab/ab, what is the recombination frequency if the progeny numbers are 17 AB/ab, 72 Ab/ab, 68 aB/ab, and 21 ab/ab? 1B)In human gene mapping, a LOD score is calculated to see if a gene causing a rare disease is linked to a known SNP. The LOD score is -4. This means that 1C) A three-point testcross is used to determine the order of three linked genes. The following crossover progeny result: single crossovers, double crossovers, and no crossovers. To determine the order, the no-crossover progeny must be compared to what other class of progeny?arrow_forward33arrow_forward
- PLease help, double and triple check your answers, im using this to study, these questions are NOT graded they are PRACTICE problemsarrow_forwardPLease help, double and triple check your answers, im using this to study, these questions are NOT graded they are PRACTICE problems. PLease help me with all parts of this questionarrow_forwardUse keyboard only to enter your answer below. ALL WORKING MUST BE SHOWN Problem 1) mutation in a gene on chromosome 15 that codes for an enzyme. The disease is an inherited autosomal recessive condition which is Tay-Sachs disease is caused by loss of function found amongst Ashkenazi Jews of Central European origin. In this population, 3 in 5,200 children are born with the disease. What proportion of the population are carriers (heterozygotes) for this disease?arrow_forward
- Can I get an explanation for 1a, 1d, 1e, 2c, 2d and the calculations for the level of interference please.arrow_forwardShort hair (S) in rabbits is dominant over long hair (s). The following crosses are carried out, producing the progeny shown. Give all possible genotypes of the parents in each cross.arrow_forwardDraw if needed .. show workarrow_forward
- a. What are the linkage distances between m and r, between r and t, and between m and t?b. Determine the linkage order for the three genes.?c. What is the coefficient of coincidence (see Chapter 4) in this cross? What does it signify?arrow_forwardThe DNA of every individual in the pedigree shown in image B (below) has been sequenced at the causative locus, all the non-shaded individuals are wild type apart from III.1 and III.6. III.1 and III.6 have both been proven to have the causative allele for the condition but they do not exhibit any of the phenotypic signs or symptoms. Based on this pedigree, what is the level of penetrance for the condition? Please give your answer as a percentage to one decimal place, give the number only, no percentage symbol. Given the information above I calculate the level of penetrance seen in image B to be "Blank" 1 percent.arrow_forwardThe DNA of every individual in the pedigree shown in image B (below) has been sequenced at the causative locus, all the non- shaded individuals are wild type apart from III.1 and III.6. III.1 and III.6 have both been proven to have the causative allele for the condition but they do not exhibit any of the phenotypic signs or symptoms. Based on this pedigree, what is the level of penetrance for the condition? Please give your answer as a percentage to one decimal place, give the number only, no percentage symbol. ANSWER: Given the information above I calculate the level of penetrance seen in image B to be Blank 1 percent. A KEY Homozygous Homozygous Heterozygous Heterozygous Wild Type Male Female Male Female Male Note: Completely red symbol denotes an individual exhibiting the phenotype of interest CI || III IV V 1/4 1/2 1/2 1/2 1/2 Wild Type Female 1/4 1/2 Affected Known carrier Affected female Normal female Affected male Normal male D ●●●arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning