Draw a Punnet square indicating the F2 genotypes from the mating of an F1 male (QQBbHh) with an F1 female (QQBBHH). Note: Make sure to write all the heterozygous genotypes as Gg, not gG.
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- Assume that the trihybrid cross AABBrr x aabbRR is made in a plant species. Assume that A and B are dominant alleles, but there is no dominance effect of alleles at the R locus. a) How many different gametes are possible in the F1generation? What are the genotypes of these gametes? b) What is the probability of the parental aabbRR genotype in the F2 progeny? c) What proportion of the F2 progeny would be expected to be homozygous for all three genes?Consider two blood polymorphisms that humans have in addition to the ABO system. Two alleles LM and LN determine the M, N, and MN blood groups. The dominant allele R of a different gene causes a person to have the Rh+ (rhesus positive) phenotype, whereas the homozygote for r is Rh− (rhesus negative). Two men took a paternity dispute to court, each claiming three children to be his own. The blood groups of the men, the children, and their mother were as follows:From this evidence, can the paternity of the children be established?Five human matings numbered 1–5 are shown in the following table. Included are both maternal and paternal phenotypes for ABO and MN blood-group antigen status. Each mating resulted in one of the five offspring shown to the right (a–e). Match each offspring with one correct set of parents, using each parental set only once. Is there more than one set of correct answers?
- Identify the four possible gametes produced by the following individuals (Y = yellow, y = blue, S = smooth, s = wrinkled): YY Ss: Yy Ss: Create a Punnett square using these gametes as P1 and determine the genotypes of the F1. What are the phenotypes? What is the ratio of those phenotypes?For the following problem: Identify the gametes for each parent, build a Punnett Square of the probability of offspring. Show the resulting genotype ratios and the resulting phenotype ratios. The submission file on Canvas will typically have a multiple-choice component or may be a fill-in-the-blank question related to genotype and/or phenotype outcomes. Be prepared to address all of these details in the questions. a )In Pansy flowers, a gene for follower color is “incompletely dominant”. All flowers resulting from the cross of a homozygous red (R) flowering plant with a homozygous white (r) flowering plant were pink. If a pink-flowering plant is bred with another pink-flowering plant, determine the probable genotype and phenotype ratios of the offspring. b) The ratio of genotypes is ________________ .c) The ratio of phenotypes (red:pink:white) is ______________.Five human matings numbered 1–5 are shown in the followingtable. Included are both maternal and paternal phenotypes forABO and MN blood-group antigen status. Parental Phenotypes Offspring(1) A, M * A, N (a) A, N(2) B, M * B, M (b) O, N(3) O, N * B, N (c) O, MN(4) AB, M * O, N (d) B, M(5) AB, MN * AB, MN (e) B, MN Each mating resulted in one of the five offspring shown to theright (a–e). Match each offspring with one correct set of parents,using each parental set only once. Is there more than one set ofcorrect answers?
- The genetic identity of the female parent is RrGg and the genetic identity of the male parent is Rrgg. They produce 320 offspring together from a single mating: 57 red-eyed females with grey bodies, 61 red-eyed females with yellow bodies, 22 brown-eyed females with grey bodies, 20 brown-eyed females with yellow bodies.59 red-eyed males with grey bodies, 63 red-eyed males with yellow bodies, 20 brown-eyedmales with grey bodies, 18 brown-eyed males with yellow bodies. Show the simultaneous transmission of the two genes involved to give rise to the progeny given with the use of genetic diagrams and summaries as required. (Hint: you are only requiredto show the simultaneous transmission from the P to F1 generations)Consider the following dihybrid testcross: B/b • E/e × b/b • e/e For the progeny from this testcross, determine the relative proportions (from 0% to 100%) of each genotype if the two genes: a) are linked (dominant alleles in cis conformation) with no crossing over: Be/be: be/be: BE/be: bE/be: b) assort independently. B/b; E/e: B/b; e/e: b/b; E/e: b/b; e/e: c) are linked (dominant alleles in cis conformation) and 20 map units apart. Be/be: be/be: BE/be: bE/be:The following two genotypes are crossed: AaBbCcX+X x AaBBCCX+Y, where a, b, and c represent autosomal genes and X and X' represent X-linked alleles in an organism with XY sex determination. What is the probability of obtaining genotype AaBBCCX+X+? Additionally, what must we assume about these genes to calculate probability? 1/16, assume all genes are located on the same chromosome 1/64, assume all genes are located on the same chromosome O 1/32, assume all genes are located on the same chromosome 1/32, assume genes assort independently 1/16, assume genes assort independently 1/64, assume genes assort independently
- 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. 0.233In onion, male sterility is produced when the nuclear genotype is aa and the mitochondrial gene S (sterile) are present. Any other combination of nuclear genotype and mitochondrial gene (including gene F for fertile) will result in a male fertile plant. Give the genotypic ratio and the phenotypic ratio or the percentage of male sterile and male fertile offspring that will be produced in the following crosses. 1. Aa + S male x aa + F female 2. Reciprocal cross of number 1. (Note that when we do reciprocal cross, we interchange/swap the genotypes of the parents (if there is a nuclear gene involved, you interchange the nuclear genotype as well). 3. Aa + S female x Aa + F male 4. Reciprocal cross of number 3.From the BbHh x BbHh dihybrid cross where BB or Bb represented full melanin and bb represented no melanin, and where HH or Hh represented black fur and hh represented brown fur, interpret the 9 : 3 : 3 : 1 phenotypic result from the F generation. Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a 9 had no melanin and black fur, 3 had no melanin and brown fur, 3 had no melanin and black fur, 1 had no melanin and brown fur b 9 had no melanin and brown fur, 3 had full melanin and brown fur, 3 had no melanin and black fur, 1 had full melanin and black fur c 9 had no melanin and black fur, 3 had full melanin and brown fur, 3 had no melanin and black fur, 1 had no melanin and brown fur d 9 had full melanin and black fur, 3 had full melanin and brown fur, 3 had no melanin and black fur, 1 had no melanin and brown fur