Answered: Parents Progeny 4 short and 2 long a.…

Name: Short hair (S) in rabbits is dominant over long hair (s). The followin
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Description: Short 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. 3:50 ullla 0.41KB/SA @ 763-Genetics_G..View only5. Solve and show

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Biology

Parents Progeny 4 short and 2 long a. short X short b. short X short c. short X long d. short X long 8 short 12 short 3 short and 1 long e. long x long 2 long

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter10: From Proteins To Phenotypes

Section: Chapter Questions

Problem 3CS: A couple was referred for genetic counseling because they wanted to know the chances of having a...

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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?
While sitting at home during Movement Control Order (MCO) because of pandemic covid19, observe two different traits of a couple in your family (eg. your mom & dad or your sister & her husband or your brother & his wife, etc). Draw a genetic cross that involves cross of the parents with the chosen 2 pairs of their contracting traits. Imagine that the cross obeys the Mendelian Laws, show the cross and gametes production for each generation (P, F1 and F2). By Using a Punnet square as symbolic representation of the results for the cross, determine the phenotypes, genotypes, phenotypic ratio and genotypic ratio of F2 generation in the family.
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?
Name: Genetic Crosses that Involve 2 Traits In rabbits, black hair is dominant to brown hair. Also in rabbits, long straight ears are dominant to floppy ears. These letters represent the genotypes and phenotypes of the rabbits: BB = black hair EE = long ears Ee = long ears ee = floppy ears Bb = black hair bb = brown 1. A male rabbit with the genotype BBee is crossed with a female rabbit with the genotype bbEe The square is set up below. Fill it out and determine the phenotypes and proportions in the offspring. How many out of 16 have black hair Be Be Be Be and long ears? How many out of 16 have black hair bE and floppy ears? be How many out of 16 have brown hair and long ears? bE How many out of 16 have brown hair be and floppy ears? 2. Show the cross: GgBb x ggBb How many out of 16 have black hair and long ears? How many out of 16 have black hair and floppy ears? How many out of 16 have brown hair and long ears? How many out of 16 have brown hair and floppy ears?
5 & :56M ******* 24 DIHYBRID CROSSES DRV 0 Stv T alı A @ zladenA 9160p2-id2 bns obidalbaneoviene da II\ MOD YR 21 $59A ... Create a dihybrid cross and determine the expected phenotypic percentages of the offspring of two corn plants both of which are heterozygous for colour and texture (RrTt X RrTt). Don't forget to include clear let statements, and follow the all six steps taught on solving genetics problems. insig moni nellog: bna. zoom
Answer the following. Letters only. 1. In mice, black color (B) is dominant over brown (b), and a solid pattern (S) is dominant over white spotted (s). Color and spotting are controlled by genes that assort independently. A homozygous black, spotted mouse is crossed with a homozygous brown, solid mouse. All the F1 mice are black and solid. What are the genotypes of the parents? * a. BBSs and bbss b. BBSs and bbSS c. BbSs and bbSs d. BBss and bbss e. BBss and bbSS f. None of the choices 2. In mice, black color (B) is dominant over brown (b), and a solid pattern (S) is dominant over white spotted (s). Color and spotting are controlled by genes that assort independently. A homozygous black, spotted mouse is crossed with a homozygous brown, solid mouse. All the F1 mice are black and solid. What will be the genotypic probability of the first filial (F1) generation offspring? * a. 100% BBSS b. 100% BbSs c. 50% BBSS and 50% BbSs d. 25% BBSS; 50% BbSs and 25% bbss e. 100% black, solid mice…
4:12 5. Set up the punnett square for each of the crosses listed below. The characteristic being studied is seed texture round seeds (dominant) and wrinkled seeds (recessive). Rr x rr Rr x Rr RR x Rr Tall: % Complete the punnett squares below using the given information. 6. A TT (tall) plant is crossed with a tt (short plant). Which trait is dominant? Give the expected probabilities for each genotype and phenotype. TT: Tt: Short: Previous % % tt: What percentage of the offspring would you expect be round? Wrinkled? ZOOM + What percentage of the offspring would you expect be round?_____ Wrinkled? What percentage of the offspring would you expect be round? Wrinkled? % 7. A hybrid tall (Tt) plant is crossed with a hybrid (Tt) plant. Give the expected probabilities for each genotype. phenotype. wvm.instructure.com
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
Show Comments Track Changet O Markup OptionS Track Changes wious Next Genetics Lab Exercise (by Dr. Lapik) Question 2. 田 BB = black Analyze the following dihybrid cross: Bb = black Parents: bbLI bb = white LL = short hair U = short hair Gametes: U= long hair Punnett Square: How many of the offspring are: Black, Short Black, Long White, Short White, Long Autupay Credentials Username ENNIN Citrix
(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.
Topic: Modifications of Mendelian Genetics LEARNING ACTIVITY and ASSESSMENT 1. In addition to the ABO blood group, many others have been identified in humans. One such group is the MN group, controlled by two codominant alleles, M and N, at one locus. What will be the probability of the genotypes and phenotypes that would be produced in crosses involving the following phenotypes: а. Туре М and type N b. Туре М and type MN c. Type N and type N d. Type MN and type MN 2. Could a child of type N result from the mating of M and MN? Justify your answer.