wwwwwwww wwwwwwwww If the progeny of the cross aaBB x AAbb is testcrossed, and the following genotypes are observed among the progeny of the testcross, what is the frequency of recombination between these loci? AaBb 135 wwwwwwww Aabb 430 www aaBb 390 wwwwwwww aabb 120 wwwwww
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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. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?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?FAlpQLSfiOhfAvlhxzCSiUll_6rt-nU5b0WI73UmWOxkOw8OCwk01ng/formResponse B 1 2 Bb x Bb b 4 The fur in both parents in this cross is * 1 B B Bb x Bb b 3 4 brown black O homozygous dominant homozygous recessive 3. 近
- d/1n5NtidRwTwUzcDkDPi5Z9P SHPZ9IA-XH-pfftLbhNc/edit 1) @ Is Add-ons Help Last edit was 15 minutes ago | Calibri в I UA 12 + 3I | II 6 1 I 7. Construct a Punnett square for a cross between two heterozygous pea plants with violet flower color. a. What genotypes would you expect in the offspring? b. What percentage or ratio of each genotype would you expect in the offspring? !!!(a) DdGGww x DDGgWw ow being (b) ddggWw x DdGgww (c) DdGgWw x DdggWw (d) DdGgWW x DdGgWW de 10. What phenotypes are expected and in what proportion from each of the following crosses? (Assume D, G and Ware dominant over d, g and w, respectively.) mo to esc Ils ovit 101 SUO hamnow brid Jesus 1860 88 HOW bildicion baild am lemon a Bal 101 mem nem lemon5 & :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
- GsnKivd010j2gIRWLIZOMZZ-VibKYvBbo61ylATAQ/viewform RECOMBINATION". For numbers 7-35, reler to the given data below. Glven the following testcross data for com In whlch the genes for fine stripe (f), bronze gleurone (bz) and knotted leaf (Kn) are involved: + = wild type f fine stripe +=wild type bz = bronze gleurone +=wild type Kn knotted leaf %3D Genotype Ko f Number 451 Ko 134 97 436 bz bz bz Ko 18 119 f 24 Kn f bz 86 Total: Your answer 7-8. What would be the recombination frequency or the frequency of the recombinant type between +/Kn and +/f genes? Oa. 16% O b. 16 map units + + + + +Classes SBI3C1-2 rr x rr Meet - rz pQLSeUir31BTTSeUl8EYpVNYpajrmzBg_g0n6oMivineMfM4k0w/viewform rr x Rr Classwork O Rrx Rr ORR X Rr Genet X SBI3C1-2 Genetics Two parents were known to be right-handed. Assuming that right-handed (R) is dominant to left-handed (r), what would be the genotypes of the parents if their son is left-handed? Google M Post Atte Sp * 1 poir/d/1n5NtidRwTwUzcDkDPi5Z9P_SHPZ91A-XH-pfftLbhNc/edit (1) O pols Add-ons Help Last edit was seconds ago BIUA ミ: 12 + ext Calibri I|1 6 I 2 Section 5: Trihybrid cross and Laws of probability For a trihybrid cross, in which inheritance of alleles for three genes is tracked, drawing a Punnett square that combines all three genes may not be practical. Instead the laws of probability may be used. The product law of probabilities says that when alleles for separate genes segregate independently, we can figure out the probability of a particular combined genotype by multiplying the probability of the alleles for each gene. 13. We cross a homozygous tall pea plant with yellow, round seeds to a homozygous dwarf pea plant with green, wrinkled seeds. All the F1 offspring are all tall plants with yellow, round seeds. a. What are the expected F2 ratios (use fractions) of tall and dwarf plants? b. What are the expected F2 ratios (use fractions) of yellow and green seeds? C. What are the expected F2…
- 97 Assume that the trihybrid cross MM SS tt X mm ss TT is made in a plant species in which Mand S are dom- inant but there is no dominance between T and t. Consider the F2 progeny from this cross, and assume independent assortment. (a) How many phenotypic classes are expected? (b) What is the probability of genotype MM SS t? obib (c) What proportion would be expected to be 15 2nod bolleo homozygous for all genes? ablla Insninmobo.(omCH 1-4 X с Maya S x Credib x app.wizer.me/learn/00E0AB wizer.me a Maya X A To-do 25% The table gives the genoty Assign x M Dashb x 50% Figurat X C In cows, brown (B) is codominant with white (W). The heterozygous phenotype is brown and white speckles. A farmer decided to cross a brown cow and a white cow in hopes of making all brown and white speckles. What percentage of the offspring will be brown with white speckles? Figurat X 75% Dashboard Incomp X d Interac X Enter class code Go 100%A strain of Neurospora with the genotype H ⋅ I is crossedwith a strain with the genotype h ⋅ i. Half the progeny areH ⋅ I, and the other half are h ⋅ i. Explain how this outcome is possible.
