Answer:
(Score for Question 2: ____ of 2 points)
2. Which offspring are the recombinant offspring in this cross?
Answer:
The recombinant offspring in this cross are Yybb and yyBb
(Score for Question 3: ___of 2 points)
3. How far apart are Y and B?
Answer:
Y and B are 16 map units apart
Explanation:
Hope this kina help :)
It regulates the body's growth, metabolism (the physical and chemical processes of the body) and sexual development and function.
Answer:
<em>Embryo</em><em> </em><em>retention</em><em>,</em><em> </em><em>a</em><em> </em><em>cutic</em><em>le</em><em>,</em><em> </em><em>stomat</em><em>a</em><em>,</em><em> </em><em>and</em><em> </em><em>vascu</em><em>lar</em><em> </em><em>tissu</em><em>es</em><em>.</em>
Alcoholic fermentation and indeed most forms of fermentation are low-energy producing. The waste products are often medium-sized hydrocarbons/alcohols that, when fully oxidized to carbon dioxide provide more energy.
Cells ferment sugars to lactic acid B. During training. This is why your muscles are sore and sore during exercise. Essentially, building this lactic acid causes your muscles to scream for oxygen and use a simplified fermentation process to meet their energy needs, which is not efficient at all.
Yes, animals will ferment for a short time if they are not supplied with enough oxygen. Humans perform lactic acid fermentation when the body urgently needs a lot of energy. If you sprint as fast as you can, your cells only have a few seconds of ATP stored. When stored ATP is depleted, muscles begin to produce ATP through lactic acid fermentation.
Learn more about Lactic acid here:-brainly.com/question/490148
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Answer:
0.549 is the frequency of the F allele.
0.495 is the frequency of the Ff genotype.
Explanation:
<em>FF </em>or <em>Ff</em> genotypes determine freckles, <em>ff</em> determines lack of freckels.
In this class of 123 students, 98 have freckles (and 123-98= 25 do not).
If the class is in Hardy-Weinberg equilibrium for this trait, then the genotypic frequency of the ff genotype is:
q²= 25/123
q²=0.203
q= 
q= 0.451
<em>q </em>is the frequency of the recessive f allele.
Given <em>p</em> the frequency of the dominant F allele, we know that:
p+q=1, therefore p=1-q
p=0.549 is the frequency of the F allele.
The frequency of the Ff genotype is 2pq. Therefore:
2pq=2×0.549×0.451
2pq=0.495 is the frequency of the Ff genotype.
