A) Note that if recombination is not an option then the gametes will be each chromosome per se. These are the resulting gametes:
ABFG/KNQR
abfqr/KNQR
ABFG/kng
abfqr/kng
B) We have to make a punnet square for this, then eliminate any individual without the two G/g copies and calculate frequencies for such resulting individuals.
ABFG/KNQR | abfqr/KNQR | ABFG/kng | abfqr/kng | |
abfg/knqr | AaBbFfGg/KkNnQqRr | aabbffgqr/KkNnQqRr | AaBbFfGg/kknngqr | aabbffgqr/kknngqr |
Note that the highlighted offspring lacks the two copies for gene G, that means such individual will never develop. So the observed frequencies will be:
1/3 = AaBbFfGg/KkNnQqRr
1/3 = AaBbFfGg/kknngqr
1/3 = aabbffgqr/kknngqr
We can state this because we know that alleles segregate in reproductive events, just like the second law of Mendel states (Law of segregation). We can actually get to this conclusions because we know that some of these genes are linked because they are in the same chromosome. That actually goes against the third law that states all alleles segregate independently (Law of independent assortment), but science actually refused such law around a century ago
C) We can see there is a trisomy for the first pair in question. Trisomies occur when there are problems during segregation in either Anaphase I or Anaphase II. If the issue in question is about segregating the homologous chromosomes (which actually is the case in this exercise) then it occured during Anaphase I, if the issue is about segregating sister chromatides then it occured during Anaphase II.
In this case a spermatocyte had a problem segregating homologous chromosomes during Anaphase I, leading to a final sperm cell with both homologous chromosomes for that pair. Such sperms managed to fecundate an egg and produced this individual with such trisomy.
2. You have a fly strain heterozygous for a reciprocal translocation between an acrocentric chromosome (solid)...
Assortment of genes on same chromosome In the fruit fly Drosophila, there is a dominant gene for normal wings and its recessive allele for vestigial wings. At another gene locus. there is a dominant gene for red eyes and its recessive allele for purple eyes. A female that was heterozygous at both gene loci was mated with a male that is homozygous for both recessive alleles. Knowing this, complete the sentences with the correct terms. 94% crossing over independent assortment...
Nameindulia Los Drosophila Genetics Predictions-L113 (20 pts.) Part I. Meiosis and Punnett Squares Remember, whenever you use Punnett Squares to solve genetics problems, be sure you are completing each of the following steps: 1) Identify the genotypes of the parents. 2) For the specific traits of interest, figure out what kinds of haploid gametes each parent can make. In each gamete, there should be one allele for each trait of interest. If there is more than one trait, make sure...
1. The following diagram shows the genetic map for a chromosome. A B C D 10cM 5cM 40cM If a diploid female heterozygous for A and D (A D//a d) is testcrossed to a homozygous recessive male (a d/a d), what percent of the progeny are predicted to have the genotype A d/a d? A. 100 B. 75 C. 50 D. 25 E. Can’t be determined 2. Yellow body (y), vermillion eye color (v) and miniature body (m) are recessive...
**Please answer ALL 10 QUESTION** 1. A species of flowering plant includes plants with blooms that range from bright golden yellow to very pale yellow; no flowers carrying the yellow pigment allele appear white. The allele for yellow flower color: a. must be a suppressor of another allele. b. demonstrates incomplete penetrance. c. demonstrates variable expressivity. d. must be a modifier of another allele. e. demonstrates variable penetrance. 2. Coat color in mice is determined by two alleles acting at...
QUESTION 2 Which type of chromosome abnormality would be most likely to result in the least severe phenotype? a. A duplication of 1/4 of a chromosome b. A monosomy G. A trisomy d. A deletion of 1/4 of a chromosome e. A balanced Translocation QUESTION 3 A woman with albinism (an X-linked recessive gene) has a child with a man who has normal pigment. Their child is heterozygous for albinism with Klinefelter's syndrome (XXY). Where did the non-disjunction event occur?...