Pomblems in Genetics Plz read additional info for the question?

Question

Free earlobes are dominant over attached earlobes. If Ron has free earlobes and five of his children also do, while two do not, what must be Ron's genotype? What is Ron's wife probable genotyupe?

NOTE: This is not homework, im studying for a quiz and have stumble across this question

Answer 1

First we have to know what type of inheritence we're dealing with. I'm pretty sure this is a trait showing complete dominance, meaning that the heterozygous condition (Ff) will result in free earlobes. It's important to know the kind of inheritance because incomplete or codominance will give different phenotypic ratios.

Let's look at the number of kids with the traits. Five free, two attached...5+2=7...almost 8! Punnett square has 4 blocks, 2 x 4 = 8. So you're looking for a phenotypic ratio of 3:1.

Two of the kids have attached earlobes. This is the recessive form of the trait (f). They have to get one allele (gene for the trait) from EACH parent. Therefore, each parent has at least one f. If Ron's phenotype is free earlobes, he has to have one dominant allele (F). All that's left is to figure out which is the other allele of his wife. Try doing Punnett squares here with mom being each of the two possibilities (Ff or ff) and see which one gives you the 3:1 phenotypic ratio shown by the kids.

(This is how to think about and study for problems like these. This will get you the most likely result. Doesn't work 100% in real life, just goes with the probability.)

Answer 2

Ron is necessarily a heterozygote...otherwise there is no way his children with attached earlobes could have the second recessive allele necessary for that recessive phenotype. His wife is either a heterozygote or homozygote recessive (the attached earlobes need one gene from her). She is probably homozygous as well, because the ratio suggests this. If she were a homozygous recessive, then the ratio of free:attached would be closer to 1:1.

Hope that helps!

Answer 3

extraordinary.. while it tells you that via crossing Ponderosa and pink Peach style you get all mushy culmination then you definately can infer that the sleek allele is the dominant one. subsequently: ss = pink Peach (furry) SS/Ss = Ponderosa (mushy) attempt the two crosses to examine which one provides all mushy ....S...S s Ss Ss s Ss Ss SS x ss provides a hundred% probability of mushy culmination (considering in basic terms one greater case S implies mushy) ....S...s s Ss ss s Ss ss Ss x ss provides 50% probability of mushy culmination and 50% of furry ones. So now you fairly understand that Ponderosa is SS. then it says that via crossing the F1 you get 174 furry and 520 mushy. it relatively is approximately 25%-seventy 5%. So make the pass between the F1 (first sq.) .....S...s S SS Ss s Ss ss Ss x Ss provides a 25% threat of furry culmination and seventy 5% of mushy. it relatively is through the fact through fact the sleek allele is dominant, that's not had to have a homozygous genotype to recent the function interior the phenotype. meaning.. that's mandatory to have the two decrease case -s- for the fruit to have furry texture. The genotype (genes) must be homozygous (the two the comparable) so as that the phenotype (actual function) supplies the trait. That explains why the F1 is all mushy yet remains waiting to offer furry culmination. that's mushy through fact it has one dominant allele, even though it additionally has the recessive one. desire that helped.. sorry that's see you later..