The Oregon Wolfe Barley mapping population is unique in having 12 easily-scored morphological markers, each showing monogenic inheritance. Do you consider these markers useful? Briefly defend your answer, pointing out advantages and disadvantages of morphological vs. molecular markers.
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- The Oregon Wolfe Barley mapping population is unique in having 12 easily-scored morphological markers, each showing monogenic inheritance. Do you consider these markers useful? Briefly defend your answer, pointing out advantages and disadvantages of morphological vs. molecular markers.
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- Given the karyotype shown at right, is this a male or a female? Normal or abnormal? What would the phenotype of this individual be?Why dont genetic markers on the Y chromosome undergo recombination? Why is this lack of recombination a necessity for these markers to be used in tracing migrations?The gene controlling ABO blood type and the gene underlying nail-patella syndrome are said to show linkage. What does that mean in terms of their relative locations in the genome? What does it mean in terms of how the two traits are inherited with respect to each other?
- Individuals of genotype AaBb were mated to individuals of genotype aabb. One thousand offspring were counted, with the following results: 474 Aabb, 480 aaBb, 20 AaBb, and 26 aabb. What type of cross is it? Are these loci linked? What are the two parental classes and the two recombinant classes of offspring? What is the percentage of recombination between these two loci? How many map units apart are they?Still referring to Problem 1, what will be the possible genotypes of offspring from the following matings? With what frequency will each genotype show up? a. AABB aaBB b. AaBB AABb c. AaBb aabb d. AaBb AaBbPedigree Analysis Is a Basic Method in Human Genetics Using the pedigree provided, answer the following questions. a. Is the proband male or female? b. Is the grandfather of the proband affected? c. How many siblings does the proband have, and where is he or she in the birth order?
- Figure 8.10 In pea plants, purple flowers (P) are dominant to white (p), and yellow peas (Y) are dominant to green (y). What are the possible genotypes and phenotypes for a cross between PpYY and ppYy pea plants? How many squares would you need to complete a Punnett square analysis of this cross?The young woman shown at right has albinismvery pale skin, white hair, and pale blue eyes. This phenotype is due to the absence of melanin, which imparts color to the skin, hair, and eyes. It typically is caused by a recessive allele. In the following situations, what are the probable genotypes of the father, the mother, and their children? a. Both parents have normal phenotypes; some of their children are albino and others are not. b. Both parents and all their children are albino. c. The mother is not albino, the father is albino, and one of their four children is albino.As it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women. During his tenure as guard, he had numerous clandestine affairs. In each case, children resulted. Subsequently, some of the childrenwho had no way of knowing that they were relatedmarried and had children of their own. Assume that two pairs of genes determine height. The genotype of the 7-foot-tall Potsdam Guard was A9A9B9B9, and the genotype of all of his 5-foot clandestine lovers was AABB. An A9 or B9 allele in the offspring each adds 6 inches to the base height of 5 feet conferred by the AABB genotype. a. What were the genotypes and phenotypes of all the F1 children? b. Diagram the cross between the F1 offspring, and give all possible genotypes and phenotypes of the F2 progeny
- What is probability, and how is it applied in genetic analysis?Half of the worlds population eats rice at least twice a day. Much of this rice is grown in flooded conditions, and different strains of rice are tolerant (survive) or intolerant (die) under these conditions. Rice breeders used genetic crosses to test whether tolerance to flooding is a dominant trait. Researchers used three true-breeding flood-tolerant strains, FR143, BKNFR, and Kurk, and two true-breeding flood- intolerant strains, IR42 and NB, in the crosses. Results were obtained from three sets of crosses and are reported in the Table below: Results of cross of F1 to tolerant parent: F1 plants were crossed with the tolerant parent of the cross. Number of Plants Progeny Analyzed from Intolerant Tolerant Cross Alive Dead Total 1. F2 results of cross: IR42 FR13A 187 77 264 IR42 BKNFR 192 73 265 NB Kurk 142 52 195 2. Results of cross of F1 to intolerant parent: (F1 of IR42 FR13A) IR42 14 17 31 (F1 of IR42 BKNFR) IR42 15 10 25 (F1 of NB Kurk) NB 21 35 56 3. Results of cross of F1 to tolerant parent: (F1 of IR42 FR13A) FR13A 31 0 31 (F1 of IR42 BKNFR) BKNFR 28 0 28 (F1 of NB Kurk) Kurk 40 0 40 Do the data support the hypothesis that the tolerance trait is dominant? Justify your conclusion by explaining the results from each of the three sets of crosses in terms of genotypes and phenotypic ratios. Source: T. Setter et al. 1997. Physiology and genetics of submergence tolerance in rice. Annals of Botany 79:6777.A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What is the chance that this couple will have a child with two copies of the dominant mutant gene? What is the chance that the child will have normal height?
