Can someone explain why those values were chosen to be divided by the total? I thought it was the four smallest amounts, so I would have done 3+64+67++7/1600. Can someone explain why that would have been wrong 16.) In a diploid plant species, an F₁ with thegenotype Gg Li Tt is test-crossed to a pure-breeding recessive plant with the genotype gg II tt.The offspring genotypes are listed in the table.Genotype NumberGg LI Tt621Gg Ll tt3Gg ll Tt64Gg il tt109gg LI Tt103gg Ll tt678g ll Tt7gg ll tt6261600B. Calculate the recombination frequencybetween Gand Tpair of gene.C. Calculate the recombination frequencybetween Tand L pair of gene.D. Calculate the recombination frequencybetween Gand L pair of gene as if youwished to map it on a chromosome.E. What is the interference value for thisdata set? Step 2The recombination frequency can be calculated as follows.Number of recombinant gametesTotal number of gametesRecombinant frequencyRecombination frequency between G and T222160013+109+103+71600

Recombination frequency refers to the probability that two linked genes will be separated during…

Can someone explain why those values were chosen to be divided by the total? I thought it was the four smallest amounts, so I would have done 3+64+67++7/1600. Can someone explain why that would have been wrong

Can someone explain why those values were chosen to be divided by the total? I thought it was the four smallest amounts, so I would have done 3+64+67++7/1600. Can someone explain why that would have been wrong

Biology: The Unity and Diversity of Life (MindTap Course List)

14th Edition

ISBN:9781305073951

Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Chapter14: Chromosomes And Human Inheritance

Section: Chapter Questions

Problem 3GP: Human females have two X chromosomes (XX); males have one X and one Y chromosome (XY). a. With...

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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?
In a diploid plant species, an F1 with the genotype Gg Ll Tt is test-crossed to a pure-breeding recessive plant with the genotype gg ll tt. The offspring genotypes are listed in the table. Genotype Number Gg Ll Tt 621 Gg Ll tt 3 Gg ll Tt 64 Gg ll tt 109 gg Ll Tt 103 gg Ll tt 67 gg ll Tt 7 gg ll tt 626 1600 Calculate the recombination frequency between G and T pair of genes. A. 0.227 B. 0.139 C. 0.454 D. 0.233
+ ec +/Y + + w/Y y ec +/Y + ec +/y ec w ++ w/y ec w у ес +у ес и Determine the order in which the three loci y, ec, and w Occur on the chromosome and prepare a linkage map. 7.22 A cross involving X-linked genes was made between yellow, bar, vermilion female fies and wild males, and the F1 females were crossed with y B v males. The following phenotypes were obtained when 1000 progeny were exam- ined: Dra ord ma the 7.2 546 244 160 50 + + + + Bv y Bv y+ + y+v y B+ and an and and and +B + re + + v ge Determine the order in which the three loci occur on the chromosome and prepare a linkage map. 7.23 Female Drosophila heterozygous for ebony (e"le), scarlet (st*/st), and spineless (ss*/ss) were testcrossed, and the following progeny were obtained: PROGENCY PHENOTYPES NUMBER ir Wild type Ebony Ebony, scarlet Ebony, spineless Ebony, scarlet, spineless Scarlet 67 8. 68 347 78 368 Scarlet, spineless Spineless (a) Are these genes linked? Justify your answer. (b) Write the genes given on a…
34 a,b Note that the starting heterozygous plant does not have its parents genotypes listed. However, we can determine that the starting heterozygous plant has S,U,tu alleles on one chromosome and s,u,Tu alleles on the other because they are most common combinations in the offspring.
n corn, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of the corn plants (i.e the tassel) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male fertile corn plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male sterile lines sing the cardboard chips, simulate the crosses indicated below. Give the genotypes and phenotypes of the offsprings in each cross, and properly label the nucleus and the cytoplasm of each individual in the cross Legend male sterile cytoplasm Male fertile cytoplasm FF nucleus Ff nucleus ff nucleus A. Male sterile female x FF male Explain the phenotype of the offspring B. Male sterile female x Ff male Explain the phenotype of the offspring
In tomato, the following genes are located on chromosome 2: + tall plant d dwarf plant + uniformly green leaves m mottled green leaves + smooth fruit p pubescent (hairy) fruit Results of the cross +++ / dmp and dmp / dmp were: + + + = 470 + m p = 1 + + p = 14 d + p = 25 d + + = 0 d m p = 441 + m + = 19 d m + = 30 Identify the single and double crossovers among the progeny.
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
In a diploid plant species, an F1 with the genotype Mm Rr Ss is test crossed to a pure breeding recessive plant with the genotype mm rr ss. The offspring genotypes are as follows: Genotype Number Mm Rr Ss 687 Mm Rr ss 5 Mm rr Ss 68 Mm rr ss 196 mm Rr Ss 185 mm Rr ss 72 mm rr Ss 8 mm rr ss 679 Total 1900 1. What is the gene order of these linked genes?
1. Please consider the following pedigree. I 1 2 II 1 a) Assume that colour is controlled by a single sex-influenced gene where orange is expressed preferentially in females. Individuals I-1 and I-2 are homozygous for orange and blue respectively. Which individual/s in generation II will be blue? b) If colour is a controlled by cytoplasmic DNA, which individual/s in generation II will be orange?
An individual is heterozygous for a reciprocal translocation, with the following chromosomes: A • B C D E F A • B C V W X R ST • U D E F R ST • U V W X Q. Explain why the fertility of this individual is likely to be less than the fertility of an individual without a translocation.
Arabidopsis thaliana is a diploid plant model organism with 2n 10. Please select the number options to match the following number of copies of each gene an Arabidopsis thaliana 1 leaf cell has number of chromosomes an Arabidopsis thaliana leaf cell contains 1. two 2. five number of chromosomes an Arabidopsis thaliana gamete cell contains 3. ten pairs of homologous chromosomes an Arabidopsis thaliana 2 leaf cell contains