3, Bob and Becky both have blood type AB. They have 4 children born in the follo
ID: 149705 • Letter: 3
Question
3, Bob and Becky both have blood type AB. They have 4 children born in the following order: Freddie, Frani, Frankie, and Joe. They recently examined their children's medical records and noted the blood types of their offspring: Fred: AB Frani: A Frankie: CO Joe: B Assuming they are the biological parents of all four children, what is the probability the couple would have four children with those 4 different phenotypes? 4. In corn kernels, purple (P) is dominant over the color red (p). The Y allele produces yellow corn and is epistatic to P& p. The recessive y allele has no effect on color. If a heterozygous purple plant is crossed with a yellow plant that is heterozygous for both genes, what are the phenotypic and genotypic probabilities of the otfspring? Calculate the probability that a cross between these 2 plants will produce 1 purple offspring and I red offspring. 5. Five different coat color phenotypes have been identified in hamsters: black (B), brown (R), gray (G), tan (T) and white (W). There are four alleles for coat color. The hamsters were crossed to understand inheritance of the genes associated with coat color. The results are shown below: tan x black tan x brown brown x black black x white tan x white brown x white 100% tan 100% tan 100% brown 100% 100% tan 100% brown F2 offspring phenotypes 3/4 tan: 1/4 black 3/4 tan: 1/4 brown 3/4 brown: 1/4 black 1/4 black: 1/4 white: 1/2 gray 3/4 tan: 1/4 white 3/4 brown: 1/4 white gray Based on the results of the crosses, determine the dominance (allelic) series for these alleles that control coat color.Explanation / Answer
3. Since Bob and Beckie both have AB blood type, their genotypes are also AB and AB. (AB blood type is present when there are both A and B genes.)
Let us make a Punnett square.
Only AA, AB and BB genotypes can arise from a cross between two AB blood types. Hence, Biologically, it is not possible for Bob and Beckie to be the parents of Frankie.
Frankie: 0 probability.
Fred: 2/4 probability.
Frani: 1/4 probability
Joe: 1/4 probability.
Total probability = 1/4*1/4*1/2*0 = 0.
5. From the first cross (tan x black), we see that tan is dominant over black. Let us assume that tan is denoted by T and black is denoted by t.
TT (tan) x tt (black)
F1 generation punnett square:
All offspring are tan, so we know tan is dominant over black.
The same square can be used for Tan x Brown also and we will know that tan is dominant over brown.
The same square can be used for Tan x White also and we will know that tan is dominant over white.
when we cross black and brown, the F1 progeny are all brown. Hence, we know brown is dominant over black.
when we cross white and brown, the F1 progeny are all brown. Hence, we know brown is dominant over white.
When we cross black and white, the F1 progeny are all gray. This is a phenomenon of incomplete dominance when the trait expressed is a combination of both the traits.
Let us make the Punnett squares. Let Black be B and white be b.
F1 generation Punnett square:
All are gray (Bb).
F2 generation will be a cross between two Bb genotypes.
hence, 1/4 black, 1/4 white and 1/2 gray.
So the dominance series will be Tan>Brown>black=white.
Grey is not a real genotype so it has not been included in the series.
A B A AA (A type) AB (AB type) B AB BB (B type)Related Questions
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