QUESTION 4 In fruit flies, white eyes (we) and miniature wings (mw) are encoded by two mutant alleles that are recessive to those that produce wild- type traits (we+ and mw+); these two genes are on the same chromosome. A fly homozygous for white eyes and miniature wings is crossed with a fly homozygous for the wild-type traits. The F1 have normal eyes and normal wings. The F1 are crossed with flies that have white eyes and miniature wings in a testcross. The progeny of this testcross are (1000 total): wild-type eyes, wild-type wings 336 wild-type eyes, miniature wings 108 vhite eyes, wild-type wings 142 vhite eyes, miniature wings 414 Vhat is the genetic distance between these two loci? О 25сМ О 33.3сМ 44.4cM 10.8cM OThese loci are not linked so they are 50CM apart.
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- 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?ISlate edu/ d2l/le/content/5003190/viewContent/44248878/View Google Tranx 4 My Drive-G X 4. Suppose that a parent Drosophila is e ca* ca The gamete frequency is as follows: e'ca e ca 16% e'ca е са 31% 14% 29% a. Circle the recombinant gametes. b. What is the map distance between the ebony and claret genes?
- I. Male Drosophila from a true-breeding wild-typestock were irradiated with X-rays and then mated withfemales from a true-breeding stock carrying the following recessive mutations on the X chromosome:yellow body (y), crossveinless wings (cv), cut wings(ct), singed bristles (sn), and miniature wings (m).These markers are known to map in the order:y - cv - ct - sn - mMost of the female progeny of this cross were phenotypically wild type, but one female exhibited ct and snphenotypes. When this exceptional ct sn female wasmated with a male from the true-breeding wild-typestock, twice as many females as males appearedamong the progeny.a. What is the nature of the X-ray-induced mutationpresent in the exceptional female?b. Draw the X chromosomes present in the exceptional ct sn female as they would appear duringpairing in meiosis.c. What phenotypic classes would you expect to seeamong the progeny produced by mating the exceptional ct sn female with a normal male from a truebreeding wild-type…Topic: Penetrance. Petal number is controlled by a single gene in merigonias. The gene has a completely dominant wild type allele F that makes a plant have five petals and a mutant recessive six petal allele(f). However the six petal trait is only 50% penetrant. You do the cross Ff x Ff. What fraction of the progeny do you have the 6 petals? what is the meaning for 50% penetrant.Miniature wings, X in Drosophila melanogaster result from an X-linked allele that is recessive to the allele for long wings, 9 + X. Match the genotypes for each parent in the crosses. Male parent phenotye long miniature miniature 111 long long Female parent phenotype long long long miniature long m m X X Male offspring phenotypes 231 long, 250 miniature 610 long 410 long, 417 miniature 753 miniature 625 long m X Y Female offspring phenotypes 560 long 632 long 412 long, 415 miniature 761 long 630 long Answer Bank ++ X X Male parent genotype + X Y Female parent genotype + X X m 00
- Topic: Gene Locus Height in gryphons is determined by a single locus. You cross a two 9' tall gryphons and get 1/4 10'; 1/2 9' and 1/4 8' tall gryphons. If you across two 7' tall gryphons you get 1/4 8', 1/2 7'; 1/4 6' tall gryphons. pls kindly help to explanation: why In gryphons, alleles of the height lous are semidominant and form an allelic series. A.Drosohpila Punnet Square of Crosses. I need results of F1 & F2 generation using Punnett Squares for: Make Punnet Squares of the following crosses •Drosophila Female wildtype cross Male White-eye •Drosophila Male wildtype cross Female White-eye •Drosophila Female Wild Type cross Male Scarlet Eye •Drosophila Male Wild Type cross Female Scarlet Eye Also, Which allele is heterozygous and which is homozygous, & which is dominant and which is recessive?Now cross two of the F₁ offspring. Parent 1 Gametes F2 Offspring Parent 2 Gametes 24. What is the phenotypic ratio in the F2 generation? 25. In the dihybrid cross you have considered two traits at a time. Although the number of traits has increased by one, what has happened to the number of possible phenotypes of offspring produced in the F2 generation? ni llit asa2013 bhidydenom sih at Simons 1:11 sabrax gatame
- (X^X^) and one male (X"). 12. The following are the gene order on the chromosomes of an individual who is heterozygous for this translocation. translocations (a.unbalanced breciprocal C. Robertsonian MN OP QR N067 89 34 5P QR 3 4 5 6 7 8 9 a. 0 LMNOP QR LMN067 89 345P QR 3456789 hmm Draw the synapsis (Prophase/Metaphase I) configuration dark spots b. What type of translocation is depicted by these chromosomes and what are the consequences of this chromosome rearrangement to the individual? this type of translocation is unbalanced and the major consequence here is that the chromosome number is not reciprocated leading to all sorts of problems with non-disjunction.Pedigree 2: A. What is the most likely mode of inheritance of this disease? Choose from: autosomal dominant, autosomal recessive, X-linked dominant, X-linked recessive. B State the genotypes of individuals # 1 #4. C If individual #3 has another daughter with the same partner, what is the probability that this daughter will be affected (show the disease)?dd-ons Help в I U A Calibri 12 三 三1 |:三 6. Consider a guinea pig with a homozygous genotype and a white fur color phenotype. a. What is the probability this parent will produce a gamete with the dominant allele? b. What is the probability this parent will produce a gamete with the recessive allele? C. If 31 sperm cells are collected from this guinea pig, how many would you expect to have the recessive allele (as determined by sequencing the gene)? !!!