If children obtain half of their genes from one parent and half from the other parent, why aren’t siblings identical?
• This is, because the particular combination of chromosomes (and hence gene alleles) one inherits from his/her parents is random due to random segregation of homologs in meiosis I.
Two black Labrador retrievers were mated and over several years produced 15 black and 6 brown offspring. Explain these results by giving the genotypes of the parents and progeny. We also discuss several other topics like What are the three aspects of the industrial revolution?
• The parents were both heterozygous for the fur color gene (B/b) with black (B) being dominant over brown (b). The black progeny were either B/B (about 5 of them) or B/b (about 10 of them), while the brown progeny were all b/b.
In jimsonweed, purple flower (P) is dominant to white (p), and spiny pods (S) are dominant to smooth (s). In a cross between a jimsonweed homozygous for white flowers and spiny pods (plant A) and one homozygous for purple flowers and smooth pods (plant B), determine the phenotype of plants resulting from two backcrosses of the F1 to the two
• The F1 will have purple flowers and spiny pods but be heterozygous for both traits (Pp Ss). Backcrossing these plants to the parent A (pp SS) will result in 50% purple flowers, spiny pods and 50% white flowers, spiny pods. For the backcross to plant B (PP ss), the progeny will be 50% purple flowers, spiny pods and 50% purple flowers, smooth pods.
Don't forget about the age old question of What exactly is inequality?
You perform a cross between a heterozygous tall pea plant and a homozygous short pea plant and obtain 30 tall plants and 20 short plants in the F1 generation. Assuming standard Mendelian inheritance of this character, how many tall and short plants would you have expected among 50 offspring? Is the observed deviation from this expectation large enough to reject the null hypothesis of single gene, Mendelian (dominant/recessive) inheritance? Provide Chi-square value and degrees of freedom.
• The expectation is 25 tall and 25 short plants (1:1 ratio). To calculate the chi square value, determine the square of the difference between expected and observed (25 for both phenotypes), divide this number by the respective expected values(25/25), and add the resulting numbers (1+1) to obtain Chi-Square = 2. With one degree of freedom, the probability of obtaining such a chi-square value by chance is > 0.05. Thus, there is not enough evidence to suggest the null hypothesis. We also discuss several other topics like What is additive model?
What is meant by “Independent Assortment”? Describe two mechanisms that ensure the Independent Assortment of Alleles described by Gregor Mendel? • Independent assortment means that alleles from different genes are segregated
independently of one another between generations. Independent assortment occurs because the two chromosomes of a pair of homologs are randomly segregated into daughter cells during meiosis (I). Thus, alleles from different genes can appear in new combinations. Crossing over further shuffles the allele combinations contained on individual chromosomes, thus leading to even more more independent assortment.
Field mice can have either black or grey fur color. It is known that this is a single-gene
trait where black is dominant over grey. You catch a black mouse and want to determine whether it is homozygous or heterozygous. What do you do? Explain your reasoning and the anticipated outcome.
• I cross the black mouse to a grey mouse. Since grey is recessive, the grey mouse will be homozygous for the recessive allele. If the black mouse was homozygous, all the offspring would be black. If the black mouse was heterozygous, then half of the offspring would be black while the other half would be grey.
From the data provided, can you explain which allele is dominant? Based on this conclusion, define gene symbols and give the possible genotypes of the parents of each cross. We also discuss several other topics like What causes democracy and democratization?
• From cross (d), we can assume that yellow (Wf) is dominant to white (wf): (a) Wf wf × wf wf.
(b) Wf Wf × Wf Wf or Wf Wf × Wf wf
(c) wf wf × wf wf
(d) Wf wf × Wf wf
(e) Wf Wf × wf wf
Chromatids exist only for a limited time during the cell cycle. Describe how when and how they are formed, how you can recognize them, how long they exist, and what happens to them when they are no longer found in a cell.
• Chromatids are formed during S-phase when DNA is replicated. They can be identified by being attached to their sister chromatid at the centromere. They exist until anaphase/telophase when they turn into chromosomes. We also discuss several other topics like How many people are there in the us base on the 2010 census?
The phenotype of vestigial (short) wings (vg) in Drosophila melanogaster is caused by a recessive mutant gene that independently assorts with a recessive gene for hairy (h) body. Assume that a cross is made between a fly with normal wings and a hairy body and a fly with vestigial wings and normal body hair. The wild-type F1 flies were crossed among each other to produce 1024 offspring. Which phenotypes would you expect among the 1024 offspring, and how many of each phenotype would you expect?
• Phenotypes: wild-type, vestigial, hairy, vestigial hairy; Numbers expected: WT (576), vestigial (192), hairy (192), vestigial hairy (64)
Fur colors in horses are controlled by a single gene that can lead to multiple
chestnut = brown
cremello = almost white
palomino = golden coat with lighter mane and tail
A number of matings between these varieties were carried out with the following results: Cross Parents Offspring
1 chestnut x chestnut chestnut
2 cremello x cremello cremello
3 chestnut x cremello palomino Don't forget about the age old question of What are the two theories that explain where regulation is and is not imposed?
4 palomino x palomino 1/4 chestnut, 1/4 cremello, 1/2 palomino First, assign gene symbols for the genetic control of coat color on the basis of these data. Then, give the genotypes of parents and offspring for the last two matings.
• Chestnut is the dominant allele, cremello is the recessive allele, and palomino is heterozygous and is the incomplete dominance allele. Chestnut = CC, cremello = cc, and palomino = Cc.
In cross 3, the parents' cross is chestnut x cremello, so the genotype cross of the parents is CC x cc. The genotype for all offspring in cross 3 is Cc, so all offpsring will be palomino.
In cross 4, the genotypes of the parents are Cc x Cc. 1/4 of the offspring of cross 4 will have the genotype CC, 1/2 will have the genotype Cc, and 1/4 will have the genotype cc.
In cats, the black and yellow pigments of their fur are controlled by an X-linked pair of alleles. Females heterozygous for these alleles have areas of black and areas of yellow in their coat (called tortoise-shell, or calico if there are also patches of white hair). A calico cat has a litter of eight kittens: one yellow male, two black males, two yellow females, and three calico females. Assuming there is a single father for the litter, what is his probable color? Explain your reasoning!
• The father must be yellow since the two yellow female kittens must have received two copies of the yellow allele from their parents – one from the father and one from the mother.
A couple arrives at your genetic counselling service and asks about the risk of inherited diseases. In this case, the woman has a male cousin on her mother’s side with Duchenne muscular dystrophy, which is a rare sex-linked recessive disorder. Neither the woman’s aunt (the cousin’s mother) nor her own mother are affected by the disease. What is the probability that the woman’s first child will have Duchenne muscular dystrophy? Explain your reasoning by providing the possible genotypes of the woman, her mother, her aunt, and her grandmother.
• The genotypes are D/d for the grandmother and the aunt. The mother and the woman are D/-, meaning we don’t know the second allele. The woman has a 50 % chance of inheriting the recessive allele, if her mother is heterozygous. Her mother has a 50% chance of inheriting the recessive allele from the grandmother. This means that the woman has a 1/4 (1/2 * 1/2 ) chance of being a carrier. Only her sons will be affected (1/2 of her children), and of those only 1/2. This means that the chance of her first
child having DMD is 1/16.
The cross EE FF x ee ff is made, and the F1 is then backcrossed with the recessive parent. The progeny genotypes are inferred from the phenotypes. The progeny genotypes are in the following proportions:
Ee Ff 2/6 Ee ff 1/6 ee Ff 1/6 ee ff 2/6
These ratios do not conform to typical Mendelian segregation ratios. Give a detailed explanation why some genotypes are more frequent than others.
• The parental allele combinations, E F and e f are more frequent than expected from a 1:1:1:1 ratio because they reside on the same chromosome, i.e. they are linked. The recombinants, E f and e F, add up to one-third of the progeny, giving a 33.3% recombination frequency. Thus, the two genes are 33.3 map units (m.u.) apart.
In barley, three genes, L, M, and R, lie near each other on chromosome 5 according to the following map:
L (10 m.u.) M (20 m.u.) R
You create a trihybrid plant (L M R/l m r) by crossing true-breeding L M R/L M R and l m r/l m r parents. In testcross of the trihybrid, how many L m R/l m r plants would be expected in a progeny of 1000 if there were no interference? If the coefficient of coincidence were 0.3, how would this change your answer?
• This requires a double cross-over in the trihybrid parent. Without interference, we can simply multiply the cross-over frequencies (=map units) of the flanking genes: 0.1 * 0.2 = 0.02, i.e. 20 of the 1000 progeny will be double recombinant, but only half of them will be L m R (the others are l M r), i.e. we can expect 10 L m R plants. If the coefficient of coincidence is 0.3, observed crossovers/expected crossovers = 0.3 = obs/0.02, therefore obs = 0.02*0.3 = 0.006, or 6 individuals, but only half of them are L m R, so the answer is 3.
A recessive X-linked character appears in 40 percent of males and 16 percent of females in a randomly interbreeding population. Assume only two alleles are present. What are the allele frequencies? How many females are heterozygotes? How many males are heterozygotes?
• In males, the allele frequencies for the males are 40% for the recessive allele and 60% for the dominant allele. The allele frequencies for the females will be 40% for the recessive allele and 60% for the dominant allele. 48% of the females will be heterozygous and there will be no heterozygous males because they can only receive one X chromosome for the X-linked allele
You backcross an allotetraploid plant to one of its progenitor species. The allotetraploid species has 2n = 30 chromosomes, whereas the progenitor species has 2n = 16 chromosomes. The offspring are all sterile. How many chromosomes does the sterile progeny possess, and what are the chromosome numbers of the two progenitor species? Explain your answer.
• The sterile progeny will have 23 chromosomes (15 from the allotetraploid parent, and 8 from the progenitor species). The two progenitor species had n1=8 and n2=7 chromosomes (i.e. 2n1=16 and 2n2=14, respectively).
An Hfr strain of genotype a+ b+ c+ strS is mated with an F- strain of genotype a- b- c- strR. At various times, a small sample of the culture is removed and shaken vigorously to separate the mating pairs. The cells in these samples are then plated on agar of the following three types:
Medium Str A B C
1 + + + -
2 + - + +
3 + + - +
(Str = streptomycin, an antibiotic; A, B, C = nutrients; a plus indicates the presence of the compound in the agar; a minus indicates its absence)
Nutrient A allows growth of a- cells, Nutrient B allows growth of b- cells, and nutrient C allows growth of c- cells.
Colonies appear on type 1 agar only if the mating had been allowed to proceed for at least 7 minutes. For type 2 agar, the mating had to continue for at least 17 minutes. In the case of type 3 agar, colony growth required a minimum mating duration of 12 minutes. a) What donor genes are being selected on each type of agar?
b) In what order are the genes a+, b+, and c+ arranged behind the F plasmid on the chromosome?
• Type 1 agar selects for c+ cells. Type 2 agar selects for a+ cells. Type 3 agar selects for b+ cells.
• b) The order of the genes is c+ b+ a+ since the first colonies appear on agar type 1 which selects for c+ cells, and the last colonies appear on the type 2 agar which selects for a+ cells.
In your microbiology lab, you are working with auxotroph mutants of E. coli. Your disorganized lab partner gives you two unlabeled tubes with E. coli cultures and says that one that contains an arginine auxotroph and the other that contains a histidine auxotroph. Unfortunately, he doesn’t remember which is which. Develop an experiment that allows you to determine which tube contains which kind of auxotroph. Be sure to explain the anticipated outcome of the experiment.
• Grow each sample on 1) minimal media (negative control), 2) complete media (positive control), 3) minimal + arginine, and 4) minimal + histidine and observe growth on the various media. The arginine auxotroph will only grow on complete and minimal + arginine medium. The histidine auxotroph will only grow on complete and minimal + histidine medium.
Your friend obtained several pairs of golden brown rabbits in order to establish his own breeding colony. To his amazement, the F1 generation contained not only 56 golden brown rabbits, but also 37 brown, 35 very light brown, 4 almost white and 3 dark brown individuals. Explain to your friend what mode of inheritance could most easily explain these results. Make sure to include an explanation of the observed phenotype ratio.
• This is an example of polygenic inheritance: The different phenotypes can be considered as different “degrees of brownness” and could be explained by a system of two genes with two alleles each that work in an equal and additive manner. The parents would be A/a; B/b. A cross between two such individuals would produce a ratio of 1:4:6:4:1 in terms of “doses” if we consider the dominant alleles are “functioning” and the recessive ones as “non-functioning” in terms of producing pigment.
In Arabidopsis thaliana, the genes for jagged leaves and short stem are both on chromosome 5 approximately 20 map units apart. Assume that a jagged leaf plant was crossed with a plant with short stems and that the resulting F1 (round leaves, tall stem) were crossed to jagged-leafed, short plants. What would be the expected phenotypes, and in what numbers would they be expected among 1000 offspring?
• jagged-tall = 400; round-short = 400; wild type (round-tall) = 100; jagged, short = 100.
You cross two true-breeding mutant plants with white flowers and obtain progeny that all have the wild-type purple flower color. Explain the mechanism of inheritance and predict which phenotypic ratio you expect if the F1 plants were allowed to self. • This is complementation of two genes in the same pathway. The next generation should have 9/16 wild-type (purple) and 7/16 white plants.
Inbreeding is shunned in all human societies since genetic defects, particularly those caused by rare recessive alleles, are more frequent in matings among relatives than in the general population. Explain why this is the case.
• The likelihood that two individuals inherit the recessive allele from a common ancestor, who was either a carrier or homozygous recessive, is much higher than the probability among two random individuals in the population being carriers.
Autosomal dominant osteoporosis type II is caused by a rare allele of the CLCN7 gene on chromosome 16. In a mating involving a man and a woman that are affected by the disease, what would be the probability that their three children have normal bone development? Assume that both parents are heterozygous since the condition is rare. Explain your answer.
• Since the parents are heterozygotes, the probability of each child being healthy (i.e. homozygous recessive) is 1/4. For all three children to have normal bones, these probabilities have to be multiplied 1/4 * 1/4 * 1/4 = 1/64.