Overview
Breeding cheap animals quickly is good, but how do we get the most profit from breeding any species? In my guide, I want to tell you how genetics works at Planet zoo and how to get animals with 3 or more than 100% indicators.
How does game genetics work?
In the Planet Zoo game, 4 genes are available to us:
A – the size of the animal
B – longevity
C – fertility
D – immunity
Each animal has 12 pairs of genetic codons (6 are responsible for fertility and 6 for immunity) – its genotype. Each pair is a set of parental codons (1 from the father and 1 from the mother), the cub gets a random combination of each pair. There can be 10 options for a pair:
AA AB AC AD
BB BC BD
CC CD
DD
The genetics of the game works in such a way that each pair that occurs in the genotype (AA, BB, CC, DD) reduces the percentage of fertility and / or immunity indicators, depending on which lineage this repetition was found in.
Repetitions of genes A and B do not affect the decrease in the corresponding indicators (size, lifespan). They are inherited from the percentage of parents. Also, according to the developer, these indicators are influenced by the quality of the animal’s keeping.
It is very important to monitor indicators of size and longevity, even in the absence of double codons (C and D – 100%), we can get a decrease in genes A and B
Analysis
So, we figured out the genetics, but how do we get animals with 100% indicators?
We will do everything through the “genealogy” section, it can be opened in the zoo-animals menu (below), after choosing an animal, it will become active, or through an animal card –
in the “pedigree book” tab.
First, let’s analyze the animals we have. We look at the indicators of size and longevity. We need 100% metrics. Let’s select such animals, call them conventional numbers / letters and leave them in the zoo, we always sell the rest. It is possible to obtain offspring with 4 100% indicators from such animals only if one of the parents has 100% genes for size and / or longevity, a combination is possible:
parent 1 – size 100, longevity 92
parent 2 – size 92, longevity 100
Also, both parents can have 100% genes for size, for example, and only one of them has 92% longevity.
Fertility and immunity indicators can be any, but the higher the percentage, the better, and fertility should not be 0%.
Now we need to analyze the genotypes of the selected animals. We open the genealogy of each and write down for ourselves either the entire genotype, or only the repetitions of genes and where they occur. It is better to write them down because it is impossible to remember them.
Matching a pair
Unfortunately, the developers do not allow us to look at the genealogy of animals on the market, we can only choose the animals that are presumably suitable for us by genotype.
Life hack: You can breed animals of the same species in several aviaries and / or zoos, so you can always view the genealogy of a particular animal and more accurately select a pair. It is also economically beneficial since you do not have to buy animals and guess whether it will fit, you will know for sure.
The goal is to select a pair of animals whose offspring will not receive double codons in the genotype. To do this, we first of all need that the pairs of letters of one parent do not coincide with pairs of letters of the other in the corresponding places.
Be sure to put a filter in the “attractiveness” tab for genes of size and longevity by 100%, if we don’t select it, we will reduce it to 92%. We look at the “compare partners” tab, it is not very informative, but it will give us the opportunity to weed out some animals. After the purchase, we can see the genealogy of a potential parent only by sending him to the aviary. We’ll have to buy a few individuals before we can find a more or less suitable one. We send the rest back for sale.
We also compare all the codons of the parents and assume where we may have pairs of letters. If several letters match, then the probability of a pair will be:
1 letter – 50% (AB and CB for example)
2 letters (different) – 50% (AB and BA for example)
3 letters – 75% (AA and AB for example)
4 letters – 100% (AA and AA in the same codons)
It is extremely difficult to find the ideal one, so we buy presumably suitable animals and analyze. If you have a maximum of 5 codons, this is a good option, but the less the better. You always take risks, but with such a scheme, you will breed animals with 3 or more 100% indicators much more often.
Example
The lioness was bred in my zoo, the lion was bought from a third-party.
Let’s analyze the lioness: she has 100% indicators of size, longevity and immunity, fertility-83% (repeat BB in codon 6). Therefore, we need to find a lion that does not have the letters B at all in the 6 fertility codon.
By the way, the letter B is responsible for the longevity gene and we see its repetition, but at the same time the indicator itself is 100%. This once again proves and shows that the genes for size (A) and longevity (B) are inherited differently, regardless of the genes for fertility and immunity.
Let’s analyze the lion: it has 100% indicators of size, longevity and fertility, immunity-50% (repeat in 2 (CC), 4 (BB) and 6 (AA) codons).
Let’s analyze the genotypes and assume the likely outcomes.
Fertility.
DC DA DC CB AC AD – LION
DB CA DC CB CB BB – THE LIONESS
In each, except for the last codon, the probability of a pair occurring is 50%, this is a big risk, but we will definitely remove the last repeat and we do not have a single intersection with a probability of 75% (DD and DB in the example).
Counter of probable pairs: 5
Immunity.
BA CC DB BB CB AA – LION
CD AB CB DC AD BD – THE LIONESS
Here, there is a 50% probability of a pair occurring only in codon 3, we will remove all repetitions of the lion with a 100% probability, which is the most important thing for us now.
Counter of probable pairs: 6
This is a good option, we have a high chance of getting lion cubs with 100% indicators
Here’s what we got:
1 of 3 lion cubs (the first litter) turned out to have 4 100% indicators, the rest had pairs of codons in the places where we expected. To get better results, you need to minimize the number of possible repetitions. My goal was to show that the result can be achieved and not under ideal conditions.
Mistakes and tips
1. Do not allow crossing of related individuals, you can get a gene mutation and get zero immunity and / or fertility
2. Remove other animals from the enclosure so as not to accidentally cross from an unsuitable pair
3. For one male, you can pick up several females, so you can increase the number of offspring.
4. Do not sell all the offspring, plan in advance who you will leave for breeding and match them to pairs
5. Do not be afraid to breed animals with low fertility (but not 0%) and immunity
6. When crossing animals in which all 4 indicators are 100% there is no guarantee to get the same offspring. Moreover, more often you will receive a decrease in performance
7. Even with an ideal pairing, there is always a chance of gene mutation, often this can happen when crossing albino animals
8. Even if the first offspring from your pair did not meet your expectations, try again, the genotype of the offspring will be different each time
9. Not all animals can be bought immediately with good performance. For such species, it is necessary to search for the optimal pair, which, to begin with, will give a minimum of paired codons
Conclusion
Thanks for reading my guide, I hope it helps you)
This is not my native language and I used programs for translation, I hope it did not disappoint me. If you find an error – write to me