Answer:
- 120 individuals yySs
- 60 individuals yyss
Explanation:
<u>Available data</u>:
- Two autosomal diallelic genes Y and S
- Y is the dominant allele that codes for yellow
- y is the recessive allele that expresses black
- S is the dominant allele that expresses stars
- s is the recessive allele that codes for starless
- Lethal genotypes: YYss and yySS
1st cross: A true-breeding yellow star bellied sneetch with a true-breeding black starless sneetch
Parentals) YYSS x yyss
F1) 100% YySs
N = 50
2nd Cross:
Parentals) YySs x YySs
Gametes) YS, Ys, yS, ys
YS, Ys, yS, ys
Punnett square) YS Ys yS ys
YS YYSS YYSs YySS YySs
Ys YYSs YYss YySs Yyss
yS YySS YySs yySS yySs
ys YySs Yyss yySs yyss
F1) N = 840
- 14/16 survivers
- 2/16 death YYss and yySS
- 9/16 yellow with stars
- 2/16 yellow starless Yyss
- 2/16 black with stars yySs
- 1/16 black and starless, yyss
14 ------------------ 100% of the progeny --------------840 individuals
2 yySs -----------X = 14.28% ------------------------------X = 120 individuals yySs
1 yyss ------------X = 7.14% ------------------------------X = 59.97 individuals yyss
Answer:
Juan father has the sickle cell trait. Gina's maternal grandmother shows the trait and Gina's mother would have inherited it, since she is heterozygous. Gina's paternal grand father shows the trait and Gina's father inherited it as he is also heterozygous.
The chances of Gina to inherit the trait is 25%
(1/2 from both parents)
1/2 + 1/2 = 1/4 equivalent to 25%
Since Gina do not show the trait. A 50% probability for Gina to have it cannot be ruled out since she's not homozygous and there is a 25% probability of being free from the trait.
Take, both Gina and juan to be carriers
HbA HbS x HbAHbS
(Shown in image 1)
We could see a 25% of probability of the trait, 50% chances and 25% of lacking the trait.
Assume Gina is free from the trait and Juan is a the carrier,
HbA HbS x HbAHbA
(Shown in image 2)
The resulting offspring would be at a 50% normal and 50% carriers and none are affected.
My advise for Juan and Gina is to be tested to affirm the likelihood of the trait being present or not as this will allow them plan for giving birth offspring rightly.
<span>Evolution and natural selection is continually applied to all animals. We are constantly evolving its just we don't see it now because change takes millions of year, evolutionary change that is. People who study our generation will be the ones who note the evolutionary changes, its hard to see them now because the change is so minuscule compared to other things... hope that helped</span>
A peplomer is a glycoprotein spike on a viral capsid or viral envelope. These protrusions will only bind to certain receptors on the host cell; they are essential for both host specificity and viral infectivity.
Answer:
7.4
Explanation:
7.4 is the normal main ph level of intracellular fluid.