Answer: Only parental 1 A b : 1 aB
Explanation:
Check my work Genes A and B are so close together on the same chromosome that...
15. You are studying com plants. You were able to determine that the gene for kernel color (C-colored and no color) and kernel waxiness (W = waxy and not waxy) are on the same chromosome and are 30 CM apart. You are ultimately interested in determining the potential phenotypic ratio if crossing over occurs. You start by crossing a male who is homozygous dominant for both genes with a female who is homozygous recessive for both genes. The F1 heterozygotes...
1) Consider a single germ cell with the genotype A/a;B/b undergoing meiosis. Which of the following are possible genotypes for the set of four gametes resulting from meiosis of this single germ cell? (assume that there is NOT any nondisjunction) A, a, B, b Aa, Aa, Bb, Bb AB, Ab, aB, ab Ab, Ab, aB, aB aa, AA, BB, bb 2) From true-breeding stocks, a female fly with no eyes (mutation in eyeless (ey) gene) and no wings (mutation in...
1. 2. 13. Which statement explains why the recombination frequency between two genes is always less than 50%? Recombination cannot be more than 50% because chromosomes are only 50 map units in length. The genotype of the F, gametes will always be 50% parental gametes and 50% recombinant gametes. Genes with a recombination frequency near 50% are unlinked and have an equal likelihood of being inherited together or separately. F1 gametes always have 50% of their alleles from each parental...
Genes A and B are located on the same chromosome 10 centimorgans apart. What is the likelihood of AB/ab progeny in a cross between Ab/aB and ab/ab individuals? a. 1/20 b. 1/10 c. 1/4 d. 9/20 e. 9/10
Genes ‘‘a’’ and ‘’b’’ are linked with a recombination frequency of 10%. “A” is completely dominant over ‘’a’’ and ‘’B’’ is completely dominant of ‘’b’’. Assuming you start with the following parental genotypes AAbb and aaBB in a dihybrid cross Indicate the percentage of F2 progeny following an F1 self-cross that have: only parental types, both parental and recombinant types and only recombinant types. a. 90%, 10% 0% b. 45%, 45%, 10% c. 85%, 10%, 5% d. 81%, 18%, 1%...
Genes A and B are linked. Individual plants homozygous for different combinations of these genes (AABB, aabb, AAbb, aaBB) were crossed at random to produce F1 plants. The F1 plants were self pollinated to produce the F2. The F2 progeny from six of the F1 self pollinations are listed below. Number of F2 plants in each phenotypic class F1 Plant-------------------AB--------------- Ab --------------- aB----------------- ab 1 --------------------------0 ----------------98 ---------------- 0------------------ 32 2 ------------------------- 64 --------------10 -----------------8...
Two genes, A and B, are 64 cM apart on a chromosome. How can we measure the genetic distance between A and B? In a test cross between a double heterozygote and a homozygous recessive individual, we should observe 64% recombinant progeny If gene C lies between A and B, we can measure the recombination rate between A-C and C-B. Both of the above will work There is no way to measure genetic distance greater than 50 cM. I know...
Which of these statements is incorrect? Syntenic genes are located on the same chromosome. Independent assortment results in recombinant chromosomes. You can reliably predict the relative genetic distance from genes’ physical distance on a chromosome. Linked genes are always syntenic. What is the relative genetic distance between two linked genes if the recombination frequency is 0.49? 0.49 cM 4.9 cM 49 cM 490 cM What statement best explains the distortion in Mendelian ratios observed by Bateson & Punnett in 1905?...
You have three genes on the same chromosome - A, B and C. Each gene has two alleles in a dominant/recessive relationship. For these genes the homozygous recessive has the mutant phenotype for that trait, the dominant phenotype = wild type for that trait. allele A is dominant to a; phenotype a = mutant for trait a; phenotype A = wild type for trait A allele B is dominant to b; phenotype b = mutant for trait b; phenotype B...
answeres are provided. please explain how to do them. :) A small number of type O individuals have the Bombay blood phenotype. Individuals with the Bombay blood phenotype always appear to have type O blood because they are homozygous recessive for the gene, H, which is epistatic to gene I. As long as at least one dominant H alele is present, the ABO blood type associated with the person's ABO genotype will be expressed. Individuals who are homozygous hh always...