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A mutation occurs in the bicoid gene in a female fly. What will result if her eggs are fertilized?
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- 1. What role does a patient's genetics have in their symptoms of muscular dystrophy? What causes various mutations to cause distinct symptoms? (two to three sentences) (Think about why various mutations in different genes cause different illnesses.)2. What can family history teach you about a patient's muscular dystrophy inheritance and, hence, genetic basis? (two to three sentences)3. The X-chromosome contains the mutation that causes Duchenne Muscular Dystrophy (DMD). Explains how this impacts DMD inheritance and why DMD patients are disproportionately male. (two to three sentences)Almost all calico cats (one is pictured in FIGURE 10.7B) are female. Why? B When this calico cat was an embryo, one of the two X chromosomes was inactivated in each of her cells. The descendants of the cells formed her adult body, which is a mosaic for expression of X chromosome genes. Black fur arises in patches where genes on the X chromosome inherited from one parent are expressed; orange fur arises in patches where genes on the X chromosome inherited from the other parent are expressed. FIGURE 10.7 Animated X chromosome inactivation.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 if the couple wanted prenatal testing so that a normal fetus could be aborted?
- 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?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. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?Another gene in Drosophila determines wing length. The dominant wild-type allele of this gene produces long wings; a recessive allele produces vestigial (short) wings. A female that is true- breeding for red eyes and long wings is mated with a male that has purple eyes and vestigial wings. F1 females are then crossed with purple-eyed, vestigial-winged males. From this second cross, a total of 600 offspring are obtained with the following combinations of traits: 252 with red eyes and long wings 276 with purple eyes and vestigial wings 42 with red eyes and vestigial wings 30 with purple eyes and long wings Are the genes linked, unlinked, or sex-linked? If they are linked, how many map units separate them on the chromosome?
- 16 - The phenotype of vestigial (short) wings (vg) in Drosophila melanogaster is caused by a recessive mutant gene that independently assorts with a recessive gene for hairy (h) body. Assume that a cross is made between a fly the is homozygous for normal wings with a hairy body and a fly with vestigial wings that is homozygous for normal body hair. The wild-type F1 flies were crossed among each other to produce 1024 offspring. Which phenotypes would you expect among the 1024 offspring, and how many of each phenotype would you expect? a) Phenotypes: wild, vestigial, hairy, and vestigial hairy; Numbers expected: wild (256), vestigial (256), hairy (256), and vestigial hairy (256). b) O Phenotypes: wild, vestigial, hairy, and vestigial hairy; Numbers expected: wild (576), vestigial (192), hairy (192), and vestigial hairy (64). C) O Phenotypes: wild, vestigial, hairy, and vestigial hairy; Numbers expected: wild (192), vestigial (256), hairy (64), and vestigial hairy (192). d) All vestigial…Part 1. Norrie disease is caused by a recessive mutation on the X chromosome in a gene called norrin. The disease affects the eyes and can cause blindness. In the picture below, the left (A) and right (B) retinas from the same female mouse are shown. The X chromosome inherited from the father is labeled magenta and the X chromosome inherited from the mother is labeled yellow. А. В. From this picture, and the information you know about X chromosomes, what can you conclude? Imprinting randomly renders one of these X chromosomes inactive per cell, so this is what you would expect from that process. The maternal chromosome is likely imprinted in the left eye while the paternal chromosome is likely imprinted in the right eye. In the left eye, more of the cells have the paternal chromosome inactivated than in the right eye. In the left eye, more of the cells have the maternal chromosome inactivated than in the right eye. Part 2. Given the information in the previous question, if the norrin…40 Below is a pedigree of a family, some of whom have the autosomal dominant condition Huntington's disease. Affected individuals are indicated by a dark square or circle. The male in generation I (indicated by the arrow) is heterozygous for the Huntington's disease mutation. The following two questions relate to this pedigree. Generation II II If H represents the disease allele, and h the wild type allele, what is the genotype of the individual indicated by the *? Select one alternative: O hh O Hh O hH O HH
- 17. In rats, the following genotypes of two independently assorting autosomal genes determine coat color.A_B_ (gray)A_bb (yellow)aaB_ (black)aabb (cream)A third gene pair on a separate autosome determines whether or not any color will be produced.The CC and Cc genotypes allow color according to the expression of the A and B alleles.However, the cc genotype results in albino rats regardless of the A and B alleles present. Given the inheritance pattern of coat color in rats described in Problem 17, predict the genotype and phenotype of the parents who produced the following offspring: (c) 27/64 gray: 16/64 albino: 9/64 yellow: 9/64 black: 3/64 cream8. While vermilion is X-linked and brightens the eye color, brown is an autosomal recessive mutation that darkens the eye. Flies carrying both mutations lose all pigmentation and are white-eyed. Predict the Fl and F2 results of the following crosses: a) vermilion females X brown males b) brown females X vermilion malesAav AaBbCc Normal No Spacing Heading 1 Paragraph Styles In man, two abnormal conditions, cataracts (C) in the eyes and excessive fragility (F) in the bones, seem to depend on separate dominant genes located on different chromosomes. Normal vision and normal bones are recessive traits. A man with cataracts and nomal bones, whose father had normal eyes, married a woman free from cataracts but with fragile bones. Her father had normal bones. 11. What is the genotype of the man with cataracts and nomal bones? What is the genotype of the woman with normal vision and fragile bones? What type of offspring might this couple expect? Genotypes Phenotypes What is the probability that their first child will, (a) be free from both abnormalities (b) have cataracts but not fragile bones (c) have fragile bones but not cataracts (d) have both cataracts and fragile bones? lili
